text:
# The Geometrical Meaning of the Tangent of an Eclipse

Original Question:
Let $(x_{0},y_{0})$, where $y_{0} \neq 0$, be a point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Show that the tangent line to the ellipse passing through $(x_{0},y_{0})$ is  $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^2}=1$.

Suggested Solution:
This is not a difficult question to tackle considering the fact that $(x_{0},y_{0})$ is on the circle. Some basic set-ups:  $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$  and $b^2x^2+a^2y^2=a^2b^2$ . Performing implicit differentiation and some further algebraic manipulations, we obtain the gradient of the tangent at $(x_{0},y_{0})$:  $\frac{dy}{dx}=\frac{-b^2x}{a^2y}$ . Substituting the gradient of the tangent into the tangent equation $y-y_{0}=m\times(x-x_{1})$ (where m refers to the gradient) to obtain the equation, simplify and we can get $\frac{x_{0}x}{a^2}+\frac{y_{0}y}{b^2}=\frac{x_{0}^2}{a^2}+\frac{y_{0}^2}{b^2}$. After that, we can easily get the answer $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^2}=1$ by realising $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.

A rather routine question so far, but what if the point $(x_{0},y_{0})$ is not on the circle, and it is placed outside, or inside the circle instead? What is the geometrical meaning of the equation  $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^2}=1$

Let us consider the outside scenario first. (In the diagrams below, I use circles to represent eclipses. Sorry for the inaccuracy.)

According to the diagram on the right, draw two tangents (L1 and L2) from the point $(x_{0},y_{0})$ to the circle.  At the point $B (x_{1},y_{1})$ on the circle, the tangent of the circle L2 is $\frac{x_{1}x}{a^{2}}+\frac{y_{1}y}{b^2}=1$. Then, since $(x_{0},y_{0})$ is on L2, we have $\frac{x_{1}x_{0}}{a^{2}}+\frac{y_{1}y_{0}}{b^2}=1$. Based on the equation, we can conclude that $B (x_{1},y_{1})$ is on the line $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^2}=1$.

Similarly, at the point $C (x_{2},y_{2})$ on the circle, the tangent of the circle L1 is $\frac{x_{2}x}{a^{2}}+\frac{y_{2}y}{b^2}=1$.. Then, since $(x_{0},y_{0})$ is on L1, we have $\frac{x_{2}x_{0}}{a^{2}}+\frac{y_{2}y_{0}}{b^2}=1$. Based on the equation, we can conclude that $B (x_{2},y_{2})$ is on the line $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^2}=1$

With $B (x_{1},y_{1})$ and $C (x_{2},y_{2})$ on L, we can conclude that the geometrical meaning of the equation $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^2}=1$ is the line passing through the two points on the circle such that the line between $(x_{0},y_{0})$ and the point is tangential to the circle.

Finally, the inside scenario, the most tricky case

Locate two points $B (x_{1},y_{1})$ and $C (x_{2},y_{2})$ on the circle such that BC (L3) passes through A$(x_{0},y_{0})$, and A is the mid-point of BC, i.e. $x_{0}=\frac{1}{2}(x_{1}+x_{2}); y_{0}=\frac{1}{2}(y_{1}+y_{2})$ . Let the tangents (L1 and L2) passing through B and C intersect outside the circle at  D(m,n).

By the previous discussion when $(x_{0},y_{0})$ is outside the circle, the equation of L3 can be written as $\frac{mx}{a^{2}}+\frac{ny}{b^2}=1$. Since A is on L3, then we have $\frac{mx_{0}}{a^{2}}+\frac{ny_{0}}{b^2}=1$.  Hence D is on the line L: $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^2}=1$. Re-arranging the equation, we get $y=-\frac{b^2x_{0}}{a^2y_{0}}x+\frac{b^2}{y_{0}}$

Hence, the gradient of L is $-\frac{b^2x_{0}}{a^2y_{0}}$

Consider the line L3, since it passes through $B (x_{1},y_{1})$ and $C (x_{2},y_{2})$, we have $\frac{x_{1}}{a^{2}}+\frac{y_{1}}{b^2}=1$ (*) and $\frac{x_{2}}{a^{2}}+\frac{y_{2}}{b^2}=1$ (**). (*)-(**): $\frac{x_{1}^2-x_{2}^2}{a^2}+\frac{y_{1}^2-y_{2}^2}{b^2}=0$, which is equivalent to $\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{a^2}+\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{b^2}=0$

From the above equation, we can simply the gradient of L3: $\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{-b^2(x_{1}+x_{2})}{a^2(y_{1}+y_{2})}=\frac{-b^2(2x_{0})}{a^2(2y_{0})}=-\frac{b^2x_{0}}{a^2y_{0}}$ = gradient of L

That is to say, the geometrical meaning of $\frac{x_{0}x}{a^{2}}+\frac{y_{0}y}{b^2}=1$ is a line parallel to BC and passes through the intersection of the tangents passing through B and C.

Note that my solution for the final case may not be perfect. What if I cannot construct ABC according to the manner I expect?

More specially, there could be something more to show: (1)Draw any secant BC of the ellipse passing through A (A is not necessarily the midpoint of BC), and draw the tangent lines to the ellipse at B and C.  Then these two lines intersect at a point on the line L.  (2)Conversely, pick any point D on L, and draw the tangent lines to the ellipses.  Let B and C be the tangent points on the ellipse.  Then D lies on BC. Based on the two conditions we can reach the same  conclusion. I believe that this method should be more credible, but considerably more tedious as well.

Feel free to leave your comments if you have critics or alternatives.

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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Appendix A: Answers to Selected Problems (3e) | CK-12 Foundation
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# 26.1: Appendix A: Answers to Selected Problems (3e)

Created by: CK-12

## Ch 1: Units and Problem Solving

1. A person of height $5 \;\mathrm{ft}$. $11 \;\mathrm{in}$. is $1.80 \;\mathrm{m}$ tall
2. The same person is $180 \;\mathrm{cm}$
1. $3 \;\mathrm{seconds} = 1/1200 \;\mathrm{hours}$
2. $3x10^3 \;\mathrm{ms}$
1. $87.5 \;\mathrm{mi/hr}$
2. c. if the person weighs $150 \;\mathrm{lb}$. this is equivalent to $668 \;\mathrm{N}$
3. Pascals (Pa), which equals $\;\mathrm{N/m}^2$
4. $168 \;\mathrm{lb}., 76.2 \;\mathrm{kg}$
5. $5 \;\mathrm{mi/hr/s}$
6. $15.13 \;\mathrm{m}$
7. $11.85 \;\mathrm{m}$
8. $89,300 \;\mathrm{mm}$
9. f. $2025 \;\mathrm{mm}^2$
10. b. $196 \;\mathrm{cm}^2$
11. c. $250 \;\mathrm{cm}^3$
12. $8:1,$ each side goes up by $2 \;\mathrm{cm}$, so it will change by $2^3$
13. $3.5 \times 10^{51}:1$
14. $72,000 \;\mathrm{km/h}$
15. $0.75 \;\mathrm{kg/s}$
16. $8 \times 2^N \;\mathrm{cm}^3/\;\mathrm{sec}$; $N$ is for each second starting with $0$ seconds for $8 \;\mathrm{cm}^3$
17. About $12$ million
18. About $1 \frac{1}{2}$ trillion $(1.5 \times 10^{12})$
19. $[\mathrm{a}] = \;\mathrm{N/kg} = \;\mathrm{m/s}^2$

## Ch 2: Energy Conservation

1. d
2. (discuss in class)
1. $5.0 \times 10^5 \;\mathrm{J}$
2. $3.7 \times 10^5 \;\mathrm{J}$
3. Chemical bonds in the food.
4. $99 \;\mathrm{m/s}$
1. $5.0 \times 10^5 \;\mathrm{J}$
2. $108 \;\mathrm{m/s}$
1. $450,000 \;\mathrm{J}$
2. $22,500 \;\mathrm{J}$
3. $5,625 \;\mathrm{J}$
4. $21.2 \;\mathrm{m/s}$
5. $9.18 \;\mathrm{m}$
3. .
4. b. $KE = 504,600 \;\mathrm{J}; U_g = 1,058,400 \;\mathrm{J}; E_{total} = 1,563,000 \;\mathrm{J}$
1. $34 \;\mathrm{m/s \ at \ B}; 28 \;\mathrm{m/s \ at \ D}, 40 \;\mathrm{m/s \ at \ E}, 49 \;\mathrm{m/s \ at \ C \ and \ F}; 0 \;\mathrm{m/s \ at \ H}$
2. $96 \;\mathrm{m}$
1. $1.7 \;\mathrm{J}$
2. $1.3 \;\mathrm{m/s}$
3. $0.4 \;\mathrm{J}, 0.63 \;\mathrm{m/s}$
1. $1.2 \;\mathrm{m/s}^2$
2. $130 \;\mathrm{J}$
1. $6750 \;\mathrm{J}$
2. $2.25 \times 10^5 \;\mathrm{J}$
3. $1.5 \times 10^5 \;\mathrm{J/gallon \ of \ gas}$
5. $0.76 \;\mathrm{m}$

## Ch 3: One-Dimensional Motion

1. .
2. .
3. .
4. .
1. Zyan
2. Ashaan is accelerating because the distance he travels every 0.1 seconds is increasing, so the speed must be increasing
3. Ashaan
4. Zyan
5. Ashaan
5. .
6. .
7. 6 minutes
8. d. $20 \;\mathrm{meters}$ e. $40 \;\mathrm{meters}$ f. $2.67 \;\mathrm{m/s}$ g. $6 \;\mathrm{m/s}$ h. Between $t = 15 \;\mathrm{s}$ and $t = 20$ sec because your position goes from $x = 30 \;\mathrm{m}$ to $x = 20\mathrm{m}$. i. You made some sort of turn
1. $7.7 \;\mathrm{m/s}^2$
2. $47 \;\mathrm{m}, 150 \;\mathrm{feet}$
3. $34 \;\mathrm{m/s}$
1. $1.22 \;\mathrm{m}$
2. $4.9 \;\mathrm{m/s}$
3. $2.46 \;\mathrm{m/s}$
4. $-4.9 \;\mathrm{m/s}$
9. b. 1 second c. at 2 seconds d. $4\mathrm{m}$
1. $250 \;\mathrm{m}$
2. $13 \;\mathrm{m/s}, -13 \;\mathrm{m/s}$
3. $14 \;\mathrm{s}$ for round trip
10. Let’s say we can jump $20 \;\mathrm{feet} \ (6.1 \;\mathrm{m})$ in the air. ☺ Then, on the moon, we can jump $36.5 \;\mathrm{m}$ straight up.
11. $-31\mathrm{m/s}^2$
1. $23 \;\mathrm{m/s}$
2. 3.6 seconds
3. $28 \;\mathrm{m}$
4. $45\mathrm{m}$
1. $25 \;\mathrm{m/s}$
2. $30 \;\mathrm{m}$
3. $2.5 \;\mathrm{m/s}^2$
12. $2 \;\mathrm{m/s}^2$
1. $v_0 = 0$
2. $10 \;\mathrm{m/s}^2$
3. $- 10 \;\mathrm{m/s}^2$
4. $60 \;\mathrm{m}$
1. $0.3 \;\mathrm{m/s}^2$
2. $0.5 \;\mathrm{m/s}$

## Ch 4: Two-Dimensional and Projectile Motion

1. .
2. .
3. .
4. .
5. .
6. .
1. $13 \;\mathrm{m}$
2. $41$ degrees
3. $v_y = 26 \;\mathrm{m/s}; v_x = 45 \;\mathrm{m/s}$
4. $56$ degrees, $14 \;\mathrm{m/s}$
7. .
8. $32 \;\mathrm{m}$
1. $0.5 \;\mathrm{s}$
2. $0.8 \;\mathrm{m/s}$
9. $104 \;\mathrm{m}$
10. $t = 0.60 \;\mathrm{s}, 1.8 \;\mathrm{m}$ below target
11. $28 \;\mathrm{m}$.
1. $3.5 \;\mathrm{s}$.
2. $35 \;\mathrm{m}; 15 \;\mathrm{m}$
12. $40 \;\mathrm{m}; 8.5 \;\mathrm{m}$
13. $1.3$ seconds, $7.1$ meters
14. $50 \;\mathrm{m}; v_{0y} = 30 \;\mathrm{m/s}; 50^0$; on the way up
15. $4.4 \;\mathrm{s}$
16. $19^\circ$
17. $0.5 \;\mathrm{s}$
18. $2.3 \;\mathrm{m/s}$
19. $6 \;\mathrm{m}$
20. $1.4$ seconds
1. yes
2. $14 \;\mathrm{m/s}$ @ $23$ degrees from horizontal
21. $22 \;\mathrm{m/s}$ @ $62$ degrees

## Ch 5: Newton’s Laws

1. .
2. .
3. .
4. Zero; weight of the hammer minus the air resistance.
5. $2$ forces
6. $1$ force
7. No
8. The towel’s inertia resists the acceleration
1. Same distance
2. You go farther
3. Same amount of force
9. .
10. a. $98 \;\mathrm{N}$ b. $98\;\mathrm{N}$
11. .
12. $32\;\mathrm{N}$
13. $5.7 \;\mathrm{m/s}^2$
14. .
15. .
16. $F_x = 14 \;\mathrm{N}, F_y = 20\;\mathrm{N}$
17. Left picture: $F = 23\mathrm{N} \ 98^\circ$, right picture:$F = 54 \;\mathrm{N} \ 5^\circ$
18. $3 \;\mathrm{m/s}^2 \;\mathrm{east}$
19. $4 \;\mathrm{m/s}^2; 22.5^\circ \;\mathrm{NE}$
20. $0.51$
21. $0.2$
22. The rope will not break because his weight of $784\;\mathrm{N}$ is distributed between the two ropes.
23. Yes, because his weight of $784\;\mathrm{N}$ is greater than what the rope can hold.
24. Mass is $51 \;\mathrm{kg}$ and weight is $82\;\mathrm{N}$
1. While accelerating down
2. $686\;\mathrm{N}$
3. $826\;\mathrm{N}$
1. $390\;\mathrm{N}$
2. $490\;\mathrm{N}$
25. $0.33$
26. $3.6 \;\mathrm{kg}$
27. $\mathrm{g} \sin \theta$
28. b.$20\;\mathrm{N}$ c. $4.9\;\mathrm{N}$ d. $1.63 \;\mathrm{kg}$ e. Eraser would slip down the wall
1. $1450\;\mathrm{N}$
2. $5600\;\mathrm{N}$
3. $5700\;\mathrm{N}$
4. Friction between the tires and the ground
5. Fuel, engine, or equal and opposite reaction
29. b. $210\;\mathrm{N}$ c. no, the box is flat so the normal force doesn’t change d. $2.8 \;\mathrm{m/s}^2$ e.$28 \;\mathrm{m/s}$ f. no g. $69\;\mathrm{N}$ h. $57\;\mathrm{N}$ i. $40\;\mathrm{N}$ j. $0.33$ k. $0.09$
30. .
1. zero
2. $-kx0$
31. b. $f_1= \mu_km_1\mathrm{g} \cos\theta; f_2 = \mu_km_2\mathrm{g} \cos\theta$ c. Ma d. $T_A= (m_1 + m_2) (a + \mu \cos\theta)$ and $T_B = m_2a + \mu m_2 \cos\theta$ e. Solve by using $d = 1/2at^2$ and substituting $h$ for $d$
1. Yes, because it is static and you know the angle and $m_1$
2. Yes, $T_A$ and the angle gives you $m_1$ and the angle and $T_C$ gives you $m_2, m_1 = T_A \cos 25/\mathrm{g}$ and $m_2 = T_C \cos 30/\mathrm{g}$
32. a. $3$ seconds d. $90 \;\mathrm{m}$
33. .
34. .
35. .
1. $1.5 \;\mathrm{N}; 2.1 \;\mathrm{N}; 0.71$

## Ch 6: Centripetal Forces

1. .
2. .
3. .
4. .
1. $100 \;\mathrm{N}$
2. $10 \;\mathrm{m/s}^2$
1. $25 \;\mathrm{N}$ towards her
2. $25 \;\mathrm{N}$ towards you
1. $14.2 \;\mathrm{m/s}^2$
2. $7.1 \times 10^3 \;\mathrm{N}$
3. friction between the tires and the road
5. $.0034\mathrm{g}$
1. $6.2 \times 10^5\;\mathrm{m/s}^2$
2. The same as a.
6. $3.56 \times 10^{22}\mathrm{N}$
7. $4.2 \times 10^{-7} \;\mathrm{N}$; very small force
8. $g = 9.8 \;\mathrm{m/s}^2$; you’ll get close to this number but not exactly due to some other small effects
1. $4 \times 10^{26} \;\mathrm{N}$
2. gravity
3. $2 \times 10^{41} \;\mathrm{kg}$
9. $.006 \;\mathrm{m/s}^2$
1. $.765$
2. $4880 \;\mathrm{N}$
1. $\sim 10^{-8} \;\mathrm{N}$ very small force
2. Your pencil does not accelerate toward you because the frictional force on your pencil is much greater than this force.
10. a. $4.23 \times 10^7\mathrm{m}$ b. $6.6 \ R_e$ d. The same, the radius is independent of mass
11. $1.9 \times 10^7\mathrm{m}$
12. You get two answers for $r$, one is outside of the two stars one is between them, that’s the one you want, $1.32 \times 10^{10}\mathrm{m}$ from the larger star.
13. .
14. .
1. $v = 28\;\mathrm{m/s}$
2. $v-$down, $a-$right
3. $f-$right
4. Yes, $640\mathrm{N}$

## Ch 7: Momentum Conservation

1. .
2. .
3. .
4. .
5. .
6. .
7. .
8. $37.5 \;\mathrm{m/s}$
9. $v_1 = 2v_2$
1. $24 \frac{kg-m}{5}$
2. $0.364 \;\mathrm{m/s}$
3. $22 \frac{kg-m}{5}$
4. $109 \;\mathrm{N}$
5. $109 \;\mathrm{N}$ due to Newton’s third law
10. $2.0 \;\mathrm{kg}, 125 \;\mathrm{m/s}$
11. $21 \;\mathrm{m/s}$ to the left
12. $3250 \;\mathrm{N}$
1. $90 \;\mathrm{sec}$
2. $1.7 \times 10^5 \;\mathrm{sec}$
1. $60 \;\mathrm{m/s}$
2. $.700 \;\mathrm{sec}$
3. yes, $8.16 \;\mathrm{m}$
13. $0.13 \;\mathrm{m/s}$ to the left
1. $11000 \;\mathrm{N}$ to the left
2. tree experienced same average force of $11000 \;\mathrm{N}$ but to the right
3. $2500 \;\mathrm{lb}$.
4. about $2.5$ “g”s of acceleration
1. no change
2. the last two cars
1. $0.00912 \;\mathrm{s}$
1. $0.0058 \;\mathrm{m/s}^2$
2. $3.5 \;\mathrm{m/s}^2$
1. $15 \;\mathrm{m/s}$
2. $49^\circ \;\mathrm{S}$ of $\;\mathrm{E}$
14. b. $4.6 \;\mathrm{m/s} \ 68^\circ$

## Ch 8: Energy & Force

1. .
2. .
3. .
4. .
5. .
1. $7.18 \times 10^9 \;\mathrm{J}$
2. $204 \;\mathrm{m/s}$
1. $34 \;\mathrm{m/s}$ @ $B; 28 \;\mathrm{m/s}$ @ $D; 40 \;\mathrm{m/s}$ @ $E; 49 \;\mathrm{m/s}$ @ $C$ and F; $0 \;\mathrm{m/s}$ @ $H$
2. $30 \;\mathrm{m}$
3. Yes, it makes the loop
6. a. $2.3 \;\mathrm{m/s}$ c. No, the baby will not clear the hill.
1. $29,500 \;\mathrm{J}$
2. $13 \;\mathrm{m}$
7. .
1. $86\;\mathrm{m}$
2. $220\;\mathrm{m}$
1. $48.5 \;\mathrm{m/s}$
2. $128 \;\mathrm{N}$
8. $0.32 \;\mathrm{m/s}$ each
1. $10 \;\mathrm{m/s}$
2. $52\;\mathrm{m}$
1. $1.1 \times 10^4 \;\mathrm{N/m}$
2. $2\;\mathrm{m}$ above the spring
9. $96$%
10. .
1. $.008\;\mathrm{m}$
2. 5.$12^\circ$
11. $8 \;\mathrm{m/s}$ same direction as the cue ball and $0 \;\mathrm{m/s}$
12. $\mathrm{v}_{golf} =-24.5 \;\mathrm{m/s}; \mathrm{vpool} = 17.6 \;\mathrm{m/s}$
13. $2.8\;\mathrm{m}$
1. $0.57 \;\mathrm{m/s}$
2. Leonora’s
3. $617 \;\mathrm{J}$
1. $19.8 \;\mathrm{m/s}$
2. $8.8 \;\mathrm{m/s}$
3. $39.5\;\mathrm{m}$
1. $89 \;\mathrm{kW}$
2. $0.4$
3. $15.1 \;\mathrm{m/s}$
14. $43.8 \;\mathrm{m/s}$
15. .
16. .
17. .
1. $3.15 \times 10^5 \;\mathrm{J}$
2. $18.0 \;\mathrm{m/s}$
3. $2.41\;\mathrm{m}$
4. $7900 \;\mathrm{J}$
1. $v_0 /14$
2. $mv_{0}{^{2}}/8$
3. $7mv_{0}{^{2}}/392$
4. $71$%

## Ch 9: Rotational Motion

1. .
1. $9.74 \times 10^{37} \;\mathrm{kg \ m}2$
2. $1.33 \times 10^{47} \;\mathrm{kg \ m}^2$
3. $0.5 \;\mathrm{kg \ m}^2$
4. $0.28 \;\mathrm{kg \ m}^2$
5. $0.07 \;\mathrm{kg \ m}^2$
2. a. True, all rotate $2\pi$ for $86,400 ;\mathrm{sec}$ which is 24 hours, b. True, $\omega = 2\pi/\mathrm{t}$ and $t=86,400 \;\mathrm{s}$ f. True, $L$ is the same g. $L = I\omega$ and $I = 2/5 \;\mathrm{mr}^2$ h. True, $K = \frac{1}{2} I\omega^2$ & $I = 2/5 \;\mathrm{mr}^2 \;\mathrm{sub-in} \ K = 1/5 \;\mathrm{mr}^2\omega^2$ i. True, $K = \frac{1}{2} I\omega^2$ & $I = \;\mathrm{mr}^2 \;\mathrm{sub-in} \ K = \frac{1}{2} mr^2\omega^2$
1. $250 \;\mathrm{rad}$
2. $40 \;\mathrm{rad}$
3. $25 \;\mathrm{rad/s}$
4. Force applied perpendicular to radius allows $\alpha$
5. $0.27 \;\mathrm{kg \ m}^2$,
6. $K^5 = 84 \;\mathrm{J}$ and $K^10 = 340 \;\mathrm{J}$
3. .
4. Moment of inertia at the end $1/3 \;\mathrm{ML}^2$ at the center $1/12 \;\mathrm{ML}^2$, angular momentum, $L = I\omega$ and torque, $\tau = I\alpha$ change the in the same way
5. .
6. Lower
7. Iron ball
1. $200 \;\mathrm{N}$ team
2. $40 \;\mathrm{N}$
3. $0.02 \;\mathrm{rad/s}^2$
4. $25 \;\mathrm{s}$
1. Coin with the hole
2. Coin with the hole
1. weight
2. $19.6 \;\mathrm{N}$
3. plank’s length $(0.8\mathrm{m})$ left of the pivot
4. $15.7 \;\mathrm{N \ m}$,
5. Ba. weight, Bb. $14.7 \;\mathrm{N}$, Bc. plank’s length $(0.3\mathrm{m})$ left of the pivot, Bd. $4.4 \;\mathrm{N \ m}$, Ca. weight, Cb. $13.6\;\mathrm{N}$, Cc. plank’s length $(1.00 \;\mathrm{m})$ right of the pivot, Cd. $13.6 \;\mathrm{N \ m}$, f) $6.5 \;\mathrm{N \ m \ CC}$, g) no, net torque doesn’t equal zero
1. $7.27 \times 10^{-6} \;\mathrm{Hz}$
2. $7.27 \;\mathrm{Hz}$
1. $100 \;\mathrm{Hz}$
2. $1.25 \times 10^5 \;\mathrm{J}$
3. $2500 \;\mathrm{J-s}$
4. $12,500 \;\mathrm{m-N}$
8. $28 \;\mathrm{rev/sec}$
9. $2300\;\mathrm{N}$
10. b. $771\;\mathrm{N}, 1030\;\mathrm{N}$ c. $554 \;\mathrm{kgm}^2$ d. $4.81\mathrm{rad/sec}^2$
1. $300\;\mathrm{N}$
2. $240N, -22\;\mathrm{N}$
3. $.092$
1. $2280\;\mathrm{N}$
2. $856 \;\mathrm{n}$ toward beam, $106\;\mathrm{N}$ down
3. $425 \;\mathrm{kgm}^2$
4. $3.39 \;\mathrm{rad/sec}^2$
1. $-1.28 \;\mathrm{Nm}$
2. $\;\mathrm{CCW}$
11. a. $1411 \;\mathrm{kg}$ c. $17410\;\mathrm{N}$ d. angular acc goes down as arm moves to vertical

## Ch 10: Simple Harmonic Motion

1. Buoyant force and gravity
2. $T = 6 \;\mathrm{s}, f = 1/6 \;\mathrm{Hz}$
1. $9.8 \times 10^5 \;\mathrm{N/m}$
2. $0.5 \;\mathrm{mm}$
3. $22 \;\mathrm{Hz}$, no.
1. $3.2 \times 10^3 \;\mathrm{N/m}$
2. a. $110 \;\mathrm{N/m}$ d. $v(t)=(25) \cos(83\mathrm{t})$
3. .
4. .
1. $0.0038 \;\mathrm{s}$
2. $0.0038 \;\mathrm{s}$
5. .
6. .
7. $4$ times
8. $0.04 \;\mathrm{m}$
1. $16 \;\mathrm{Hz}$
2. $16$ complete cycles but $32$ times up and down, $315$ complete cycles but $630$ times up and down
3. $0.063 \;\mathrm{s}$
1. $24.8 \;\mathrm{J},165 \;\mathrm{N}, 413 \;\mathrm{m/s}^2$
2. $11.1\mathrm{m/s}, 0, 0$
3. $6.2 \;\mathrm{J}, 18.6 \;\mathrm{J}, 9.49 \;\mathrm{m/s}, 82.5 \;\mathrm{N}, 206 \;\mathrm{m/s}^2$
4. $.169 \;\mathrm{sec}, 5.9 \;\mathrm{Hz}$
9. b. $.245 \;\mathrm{J}$ c. $1.40\mathrm{m/s}$ d. $1.00 \;\mathrm{m/s}$ f. $2.82 \;\mathrm{N}$ g. $3.10 \;\mathrm{N}$

## Ch 11: Wave Motion and Sound

1. $390 \;\mathrm{Hz}$
1. $4 \;\mathrm{Hz}$
2. It was being driven near its resonant frequency.
3. $8 \;\mathrm{Hz}, 12 \;\mathrm{Hz}$
4. (Note that earthquakes rarely shake at more than $6 \;\mathrm{Hz}$).
2. .
3. .
1. $7$ nodes including the $2$ at the ends
2. $3.6 \;\mathrm{Hz}$
4. $1.7 \;\mathrm{km}$
1. $1.7 \;\mathrm{cm}$
2. $17 \;\mathrm{m}$
1. $4.3 \times 10^{14} \;\mathrm{Hz}$
2. $2.3 \times 10^{-15} \;\mathrm{s} -$ man that electron is moving fast
1. $2.828 \;\mathrm{m}$
2. $3.352 \;\mathrm{m}$
3. $L = 1/4 \ \lambda$ so it would be difficult to receive the longer wavelengths.
5. Very low frequency
6. b. Same as closed at both ends
7. .
8. $1.9 \;\mathrm{Hz}$ or $2.1 \;\mathrm{Hz}$.
9. $0.53 \;\mathrm{m}$
10. $2.2 \;\mathrm{m}, 36 \;\mathrm{Hz}; 1.1 \;\mathrm{m}, 73 \;\mathrm{Hz}; 0.733 \;\mathrm{m}, 110 \;\mathrm{Hz}; 0.55 \;\mathrm{m}, 146 \;\mathrm{Hz}$
11. $430\;\mathrm{Hz}; 1.3 \times 10^3 \;\mathrm{Hz}; 2.1 \times 10^3 \;\mathrm{Hz}; 3.0 \times 10^3 \;\mathrm{Hz};$
1. The tube closed at one end will have a longer fundamental wavelength and a lower frequency.
2. If the temperature increases the wavelength will not change, but the frequency will increase accordingly.
12. struck by bullet first.
13. $80 \;\mathrm{Hz}; 0.6 \;\mathrm{m}$
1. $0.457 \;\mathrm{m}$
2. $0.914 \;\mathrm{m}$
3. $1.37 \;\mathrm{m}$
14. $2230 \;\mathrm{Hz}; 2780 \;\mathrm{Hz}; 2970 \;\mathrm{Hz}$
15. $498 \;\mathrm{Hz}$
16. $150 \;\mathrm{m/s}$

## Ch 12: Electricity

1. .
2. .
3. .
4. .
5. .
6. .
7. .
8. .
9. .
10. .
11. b. $1350 \;\mathrm{N}$ c. $1350 \;\mathrm{N}$
1. $1.1 \times 10^9 \;\mathrm{N/C}$
2. $9000 \;\mathrm{N}$
12. $F_g = 1.0 \times 10^{-47} \;\mathrm{N}$ and $F_e = 2.3 \times 10^{-8} \;\mathrm{N}$. The electric force is $39$ orders of magnitudes bigger.
13. $1.0 \times 10^{-4} C$
14. .
15. a. down b. Up $16c, 5.5 \times 10^{11} \;\mathrm{m/s}^2$ e. $2.9 \times 10^8 \;\mathrm{m/s}^2$
1. Toward the object
2. $3.6 \times 10^4 \;\mathrm{N/C}$ to the left with a force of $2.8 \times 10^{-7} \;\mathrm{N}$
16. Twice as close to the smaller charge, so $2 \;\mathrm{m}$ from $12\mu \mathrm{C}$ charge and $1 \;\mathrm{m}$ from $3\mu \mathrm{C}$ charge.
17. $0.293 \;\mathrm{N}$ and at $42.5^\circ$
18. $624 \;\mathrm{N/C}$ and at an angle of $-22.4^\circ$ from the $+x-$axis.
1. $7500\mathrm{V}$
2. $1.5 \;\mathrm{m/s}$
1. $6.4 \times 10^{-17}\;\mathrm{N}$
2. $1300\mathrm{V}$
3. $2.1 \times 10 ^{-16} \;\mathrm{J}$
4. $2.2 \times 10^7 \;\mathrm{m/s}$
19. b. $0.25\mathrm{m}$ c. $F_T = 0.022\;\mathrm{N}$ d. $0.37 \mu \mathrm{C}$

## Ch 13: Electric Circuits – Batteries and Resistors

1. $4.5\mathrm{C}$
2. $2.8 \times 10^{19}$ electrons
1. $0.11 \;\mathrm{A}$
2. $1.0 \;\mathrm{W}$
3. $2.5 \times 10^{21}$ electrons
4. $3636 \;\mathrm{W}$
1. $192 \ \Omega$
2. $0.42 \;\mathrm{W}$
1. $5.4 \;\mathrm{mV}$
2. $1.4 \times 10^{-8} \;\mathrm{A}$
3. $7.3 \times 10^{-11} \;\mathrm{W}$, not a lot
4. $2.6 \times 10^{-7} \;\mathrm{J}$
1. left = brighter, right = longer
1. $224 \;\mathrm{V}$
2. $448 \;\mathrm{W}$
3. $400 \;\mathrm{W}$ by $100 \ \Omega$ and $48 \;\mathrm{W}$ by $12 \ \Omega$
2. b. $8.3 \;\mathrm{W}$
3. $0.5\mathrm{A}$
4. .
5. $0.8\mathrm{A}$ and the $50 \ \Omega$ on the left
1. $0.94 \;\mathrm{A}$
2. $112 \;\mathrm{W}$
3. $0.35 \;\mathrm{A}$
4. $0.94 \;\mathrm{A}$
5. $50, 45, 75 \ \Omega$
6. both $50 \ \Omega$ resistors are brightest, then $45 \ \Omega$, then $75 \ \Omega$
1. $0.76 \;\mathrm{A}$
2. $7.0 \;\mathrm{W}$
6. b. $1000 \;\mathrm{W}$
7. .
1. $9.1 \ \Omega$
2. $29.1 \ \Omega$
3. $10.8 \ \Omega$
4. $26.8 \ \Omega$
5. $1.8\mathrm{A}$
6. $21.5\mathrm{V}$
7. $19.4\mathrm{V}$
8. $6.1\mathrm{V}$
9. $0.24\mathrm{A}$
10. $16 \;\mathrm{kW}$
1. $3.66 \ \Omega$
2. $0.36\mathrm{A}$
3. $1.32 \;\mathrm{V}$
8. .
9. .
10. .
11. .
12. .
13. .
1. $10\mathrm{V}$

## Ch 14: Magnetism

1. No: if $v = 0$ then $F = 0$; yes: $F = qE$
2. .
3. .
1. Into the page
2. Down the page
3. Right
4. Both pointing away from north
5. .
6. .
7. $7.6 \;\mathrm{T}$, south
8. Down the page; $60 \;\mathrm{N}$
1. To the right, $1.88 \times 10^4 \;\mathrm{N}$
2. $91.7 \;\mathrm{m/s}$
3. It should be doubled
9. East $1.5 \times 10^4 \;\mathrm{A}$
10. $0.00016 \;\mathrm{T}$; if CCW motion, B is pointed into the ground.
11. $1.2 \times 105 \;\mathrm{V}$, counterclockwise
1. $15 \;\mathrm{V}$
2. Counter-clockwise
1. $2 \times 10^{-5} \;\mathrm{T}$
2. Into the page
3. $2.8 \;\mathrm{N/m}$
4. CW
1. $2.42 \times 10^8 \;\mathrm{m/s}$
2. $9.69 \times 10^{-12} \;\mathrm{N}$
3. $.0055 \;\mathrm{m}$
12. E/B
1. $8 \times 10^{-7} \;\mathrm{T}$
2. $1.3 \times 10^{-6} \;\mathrm{C}$
1. $0 .8 \;\mathrm{V}$
2. CCW
3. $.064 \;\mathrm{N}$
4. $.16 \;\mathrm{N/C}$
5. $.13 \;\mathrm{w}$
13. a. $1.11 \times 10^8 \;\mathrm{m/s}$ b. $9.1 \times 10^{-30} \;\mathrm{N} < < 6.4 \times 10^{-14} \;\mathrm{N}$ d. $.00364 \;\mathrm{T}$ e. $.173 \;\mathrm{m}$ f. $7.03 \times 1016 \;\mathrm{m/s}^2$ g. $3.27^\circ$
14. $19.2 \;\mathrm{V}$
1. $8.39 \times 10^7 \;\mathrm{m/s}$
2. $2.68 \times 10^{-13} \;\mathrm{N}, -y$
3. $2.95 \times 10^17 \;\mathrm{m/s}^2$
4. $.00838 \;\mathrm{m}$
5. $1.68 \times 10^6 \;\mathrm{N/C}$
6. $16,800 \;\mathrm{V}$
1. $1.2 \times 10^{-6} \;\mathrm{T}, +z$
2. $1.5 \times 10^{-17} \;\mathrm{N}, -y$
3. $96 \;\mathrm{N/C}, -y$

## Ch 15: Electric Circuits—Capacitors

1. .
1. $4 \times 10^7 \;\mathrm{V}$
2. $4 \times 10^9 \;\mathrm{J}$
2. .
1. $100 \;\mathrm{V}$
2. A greater voltage created a stronger electronic field, or because as charges build up they repel each other from the plate.
3. $21 \;\mathrm{V}, \;\mathrm{V}$ is squared so it doesn’t act like problem $4$
1. $3.3 \;\mathrm{F}$
2. $54 \ \Omega$
1. $200 \;\mathrm{V}$
2. $5 \times 10^{-9} \;\mathrm{F}$
3. $2.5 \times 10^{-9} \;\mathrm{F}$
1. $6\mathrm{V}$
2. $0.3\mathrm{A}$
3. $18\mathrm{V}$
4. $3.6 \times 10^{-4}\mathrm{C}$
5. $3.2 \times 10^{-3}\mathrm{J}$
1. $80 \mu \mathrm{F}$
2. $40\mu \mathrm{F}$
3. $120 \mu \mathrm{F}$
1. $26.7\mu \mathrm{F}$
2. $166.7\mu \mathrm{F}$
1. $19.0 \times 10^3 \;\mathrm{N/C}$
2. $1.4 \times 10^{-15} \;\mathrm{N}$
3. $1.6 \times 10^{15} \;\mathrm{m/s}^2$
4. $3.3 \times 10^{-11} \;\mathrm{s}$
5. $8.9 \times 10^{-7} \;\mathrm{m}$
6. $5.1 \times 10^{-30}$

1. $4.9 \times 10^{-5} \;\mathrm{H}$
2. $-9.8 \times 10^{-5} \;\mathrm{V}$
1. Zero
1. Yes
2. No
3. Because they turn current flow on and off.
1. $0.5 \;\mathrm{V}$
2. $0.05 \;\mathrm{A}$
3. $0.05 \;\mathrm{A}$
4. $5.5 \;\mathrm{V}$
5. $8.25\mathrm{V}$
6. $11 \times$
1. $On$
2. $On$
3. $On, on, off, on, off, off, on, on$
2. 6b. $10.9\mu \;\mathrm{F}$ c. $195 \ \Omega$ d. $169 \ \Omega$ e. $1.39 \;\mathrm{A}$ f. $-42^\circ$ g. $115\mathrm{Hz}$

## Ch 17: Light

1. .
2. .
3. $2200$ blue wavelengths
4. $65000 \ x-$rays
5. $6 \times 10^{14} \;\mathrm{Hz}$ $6. 3.3 \;\mathrm{m}$
6. .
7. .
8. b. vacuum & air c. $1.96 \times 10^8 \;\mathrm{m/s}$
9. $6.99 \times 10^{-7}\;\mathrm{m}; 5.26 \times 10^-7\;\mathrm{m}$
10. .
11. .
12. Absorbs red and green.
13. $25^\circ$
14. .
15. $33.3^\circ$
1. $49.7^\circ$
2. No such angle
3. $48.8^\circ$
16. b. $11.4\;\mathrm{m}$ c. $11.5\;\mathrm{m}$
17. $85 \;\mathrm{cm}$
18. c. $+4$ units e. $-1$
1. $6$ units
2. bigger; $M = 3$
19. c. $1.5$ units d. $2/3$
20. c. $3$ units e. $- 2/3$
21. c. $5.3$ units
22. .
23. b. $22.5 \;\mathrm{mm}$
24. .
25. $32\;\mathrm{cm}$
1. $10.2^\circ$
2. $27\;\mathrm{cm}$
3. $20\;\mathrm{cm}$
1. $0.72\;\mathrm{m}$
26. .
27. $54\;\mathrm{cm}, 44\;\mathrm{cm}, 21\;\mathrm{cm}, 8.8\;\mathrm{cm}$
28. .
29. $13.5^\circ$
30. $549 \;\mathrm{nm}$

## Ch 18: Fluids

1. $0.84$
2. $1.4 \times 10^5 \;\mathrm{kg}$
1. $90$% of the berg is underwater
2. $57$%
3. b. $5.06 \times 10^{-4} \;\mathrm{N}$ c. $7.05 \;\mathrm{m/s}^2$
4. $4.14 \;\mathrm{m/s}$
5. $40$ coins
6. b. upward c. $4.5 \;\mathrm{m/s}^2$ d. Cooler air outside, so more initial buoyant force e. Thin air at high altitudes weighs almost nothing, so little weight displaced.
7. a. At a depth of $10 \;\mathrm{cm}$, the buoyant force is $2.9 \;\mathrm{N}$ d. The bottom of the cup is $3 \;\mathrm{cm}$ in radius
1. $83,000 \;\mathrm{Pa}$
2. $104 \;\mathrm{N}$
3. $110 \;\mathrm{N}$
1. $248 \;\mathrm{kPa}$
2. $591 \;\mathrm{kPa}$
3. $1081 \;\mathrm{kPa}$
8. .
9. $.0081$
1. $12500 \;\mathrm{J/m}^3$
2. $184 \;\mathrm{kPa}$
3. $1.16 \;\mathrm{kW}$
4. $2.56 \;\mathrm{kW}$
5. $11.8 \;\mathrm{A}$
6. \$$12.60$
1. $611 \;\mathrm{kPa}$
2. $6 \;\mathrm{atm}$
10. b. $500,000 \;\mathrm{N}$
1. $27 \;\mathrm{m/s}^2, (2.7 \;\mathrm{g})$ upward
2. $1600 \;\mathrm{N}$
3. $2200 \;\mathrm{N}$
1. $10 \;\mathrm{N}$
2. $10.5 \;\mathrm{N}$
3. $11 \;\mathrm{N}$
4. $11 \;\mathrm{N}$
11. a. “The Thunder Road” b. $2.0 \;\mathrm{m}$ (note: here and below, you may choose differently) c. $33.5 \;\mathrm{m}^3$ e. $3.5 \;\mathrm{million} \;\mathrm{N}$ f. $111 \;\mathrm{MPa}$

## Ch 19: Thermodynamics and Heat Engines

1. .
2. .
3. .
4. .
5. .
6. .
7. .
8. .
9. .
10. .
11. .
12. .
13. .
14. .
15. .
16. .
17. .
18. $517 \;\mathrm{m/s}$
19. $1.15 \times 10^{12}\;\mathrm{K}$
20. .
21. $40 \;\mathrm{N}$
22. $\approx \ 10^{28}$ molecules
1. $21,000 \;\mathrm{Pa}$
2. Decreases to $61,000 \;\mathrm{Pa}$
3. $5.8 \;\mathrm{km}$
1. No
2. allowed by highly improbable state. More likely states are more disordered.
23. a. $8.34 \times 10^{23}$ b. $6.64 \times 10^{-27}\;\mathrm{kg}$ c. $1600 \;\mathrm{m/s}$ d. $744 \;\mathrm{kPa}$ e. $4.2 \times 10^{20}$ or $0.0007 \;\mathrm{moles}$ g. $0.00785 \;\mathrm{m}^3$
1. $1.9 \;\mathrm{MW}$
2. $0.56 \;\mathrm{MW}$
3. $1.3 \;\mathrm{Mw}$
1. $54$%
2. $240 \;\mathrm{kW}$
3. $890 \;\mathrm{kW}$
4. $590 \;\mathrm{kW}$
5. $630 \;\mathrm{kg}$
1. $98$%
2. $4.0$%
3. $12$%
24. $14800 \;\mathrm{J}$
25. $12,000 \;\mathrm{J}$
26. b. $720 \;\mathrm{K}, 300 \;\mathrm{K}, 600 \;\mathrm{K}$ c. isochoric; isobaric d. $\;\mathrm{C}$ to $\;\mathrm{A}; \;\mathrm{B-C}$ e. $0.018 \;\mathrm{J}$
27. b. $300 \;\mathrm{K}, 1200 \;\mathrm{K}$
1. $1753 \;\mathrm{J}$
2. $-120 \;\mathrm{J}$
3. $80 \;\mathrm{J}$
4. $35 \;\mathrm{J}$
5. $-100 \;\mathrm{J}, 80 \;\mathrm{J}, 80 \;\mathrm{J}$

## Ch 20: Special and General Relativity

1. longer
2. $\gamma = \infty$, the universe would be a dot
3. $76.4 \;\mathrm{m}, 76.4\;\mathrm{m}$
4. .
5. $\gamma = 1.002$
6. $9.15 \times 10^7\;\mathrm{m/s}$
7. $2.6 \times 10^8\;\mathrm{m/s}$
1. $0.659 \;\mathrm{km}$
2. $22.4$
3. $4.92\times10^{-5}\;\mathrm{m/s}$
4. $14.7 \;\mathrm{km}$
8. $2900\;\mathrm{m}$
9. $1.34 \times 10^{-57}\;\mathrm{m}$
10. $0.303 \;\mathrm{s}$
11. $2.9 \times 10^{-30}\mathrm{kg}$, yes harder to accelerate
1. f
2. c
12. $4.5 \times 10^{16} \;\mathrm{J}; 1.8 \times 10^{13}$ softballs
1. $1.568 \times 10^{-13} \;\mathrm{J}$
2. $3.04 \times 10^6 \;\mathrm{J}$

## Ch 21: Radioactivity and Nuclear Physics

1. .
2. .
3. .
4. .
5. .
1. Substance $A$ decays faster than $B$
2. Substance $B$ because there is more material left to decay.
1. $^{219}{_{88}}\mathrm{Ra} \rightarrow ^{215}{_{86}}\mathrm{Rn} + ^{4}{_{2}}\mathrm{He}$
2. $^{158}{_{63}}\mathrm{Eu} \rightarrow ^{158}{_{64}}\mathrm{Gd} + ^{0}{_{-1}}e^-$
3. $^{53}{_{22}}\mathrm{Ti} \rightarrow ^{53}{_{23}}\mathrm{Va} + ^{0}{_{-1}}e^-$
4. $^{211}{_{83}}\mathrm{Bi} \rightarrow ^{207}{_{81}}\mathrm{Tl} + ^{4}{_{2}}\mathrm{He}$
1. $5 \times 10^{24}$ atoms
2. Decay of a lot of atoms in a short period of time
3. $2.5 \times 10^{24}$ atoms
4. $\frac{1}{2}$
5. $26.6$ minutes
6. The one with the short half life, because half life is the rate of decay.
1. Substance $B = 4.6 \;\mathrm{g}$ and substance $A = 0.035 \;\mathrm{g}$
2. substance $B$
7. $1.2 \;\mathrm{g}$
8. $125 \;\mathrm{g}$
9. $0.46$ minutes
10. $t = 144,700$ years
11. $0.0155 \;\mathrm{g}$
12. $17$ years
13. $49,000$ years

## Ch 22: Standard Model of Particle Physics

1. strange
2. some type of meson
3. Electron, photon, tau$\ldots$
4. Neutron, electron neutrino, $Z^0$
5. Neutron, because it doesn’t have electrical charge
6. No, because it doesn’t have electrical charge
7. Two anti-up quarks and an anti-down quark
8. Lepton number, and energy/mass conservation
9. Yes, $W^+, W^-$, because they both have charge
10. The weak force because it can interact with both quarks and leptons
11. Yes; a,b,c,e; no; d,f
12. The standard model makes verifiable predictions, string theory makes few verifiable predictions.

## Ch 23: Feynman Diagrams

1. Allowed: an electron and anti-electron(positron) annihilate to a photon then become an electron and anti-electron(positron) again.
2. Not allowed: electrons don’t go backward though time, and charge is not conserved
3. Not allowed: lepton number is not conserved
1. Allowed: two electrons exchange a photon
2. Not allowed: neutrinos do not have charge and therefore cannot exchange a photon.
1. Allowed: an electron and an up quark exchange a photon
2. Not allowed: lepton number not conserved
4. Not allowed: quark number not conserved
5. Allowed: electron neutrino annihilates with a positron becomes a $W^+$ then splits to muon and muon neutrino.
6. Allowed: up quark annihilates with anti-up quark becomes $Z^0$, then becomes a strange quark and anti-strange quark
7. Not allowed: charge not conserved
8. Allowed: this is a very rare interaction
9. Not allowed: electrons don’t interact with gluons
10. Not allowed: neutrinos don’t interact with photons
11. Allowed: the electron and the positron are exchanging virtual electron/positron pairs
12. Allowed: this is beta decay, a down quark splits into an up quark an electron and an electron neutrino via a $W^-$ particle.
13. Allowed: a muon splits into an muon neutrino, an electron and an electron neutrino via a $W^-$ particle.

## Ch 24: Quantum Mechanics

1. $6.752 \times 10^{-26} J, 2.253 \times 10^{-34} \;\mathrm{kgm/s}$
2. $5.96 \times 10^{-20} J, 1.99 \times 10^{-28} \;\mathrm{kgm/s}$
3. $4.90 \times 10^{-28} J, 1.63 \times 10^{-36} \;\mathrm{kgm/s}$
1. $1.94 \;\mathrm{eV}, 1.04 \times 10^{-27} \;\mathrm{kgm/s}$
2. $12.7 \;\mathrm{eV}, 6.76 \times 10^{-27} \;\mathrm{kgm/s}$
3. $5.00 \;\mathrm{eV}, 2.67 \times 10^{-21} \;\mathrm{kgm/s}$
1. $.0827 \;\mathrm{nm}$
2. $4.59 \times 10^{-4}\;\mathrm{nm}$
3. $.942 \;\mathrm{nm}$
1. $1.03 \times 10^{-20}\;\mathrm{m}$
1. $36 \;\mathrm{nm}$
2. no
3. $380 \;\mathrm{nm}, 73 \;\mathrm{nm}, 36 \;\mathrm{nm}, 92 \;\mathrm{nm}, 39 \;\mathrm{nm}$
2. $.80 \;\mathrm{V}$
3. $.564 \;\mathrm{nm}$
1. $.124 \;\mathrm{nm}$
2. $.00120 \;\mathrm{nm}$
4. $24,600 \;\mathrm{m/s}$
5. $1.84 \times 10^8\;\mathrm{m/s}$
1. $.491 \;\mathrm{m/s}$
2. $3.14 10^7\;\mathrm{J}$
3. $64 \;\mathrm{Mw}$
4. $1.55 \;\mathrm{pm}$
6. $3.27 \;\mathrm{eV}$
7. .
8. b. $15$ c. $182 \;\mathrm{nm}, 188 \;\mathrm{nm}, 206 \;\mathrm{nm}, 230 \;\mathrm{nm}$
9. $-10.3 \;\mathrm{eV}, -3.82 \;\mathrm{eV}, -2.29 \;\mathrm{eV}, -1.83 \;\mathrm{eV}$
1. $4.19 \times 10^7\;\mathrm{m/s}$
2. $1.70 \times 10^{-11}\;\mathrm{m}$
3. $1.95^\circ$
4. $.068 \;\mathrm{m}$
1. $1.89 \;\mathrm{V}$
2. $1.60 \;\mathrm{A}$
3. $1.25 \ \Omega$
1. $4.40 \times 10^{-24}\;\mathrm{kgm/s}$
2. $1.17 \times 10^{-24}\;\mathrm{kgm/s}$
3. $3.23 \times 10^{-24}\;\mathrm{kgm/s}$
4. $3.76 \times 10^7\;\mathrm{m/s}$
1. $1.1365 \times 10^{-22}\;\mathrm{kgm/s}$
2. $5.860 \;\mathrm{pm}$
3. $^242\;\mathrm{Cu} \rightarrow ^4\mathrm{He} + ^238\mathrm{Pu}$
4. $238.0497 \;\mathrm{amu}$
5. $17.7 \;\mathrm{cm}$
6. $-y$
7. $+y, 34.2 \;\mathrm{N/C}$

Feb 23, 2012

Sep 15, 2014

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OtakuBoards

# Request Me New Like Linkin Park Want Sig and avatar!

## Recommended Posts

Hi Im A HUUUUGE Linkin Park fan.When im home i listen it right away the whole day
and i wanted a sig and avatar both animated i like all songs but these the most:
Somewhere I belong
Faint
Numb
Papercut
And n.12 on Hybrid cd
Breaking The Habit

P.S.:Can U Make An Animated Pic For MSN Too?
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I understand that you are new and all but I really think this goes in the art forum....
I love linkin park also though.lol
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im so sorry i didn't knew that my apologise: :( :( :( :(
SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY SORRY (This is Spam huh?)
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And now, everyone can tell, what a complete idiot you are.
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Hey I Know I Am So Whats Ur Point?(This Is Getting Off-Topic)
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[COLOR=SlateGray][SIZE=1]Shinji...Learn to be nice.

Raizio; I'll try, but I'm not very good with animations. So it might take me awhile. Also could you pick a song and tell me what lyrics/verse you'd like to have on the banner?

(I don't normally listen to LP, nor do I know their song lyrics.)[/SIZE][/COLOR]
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Tnx For Helping Ruby And I've Got All The Time Of The World So Dont Make Any Haste This is The Verse U May Short It Out:
There's something inside me that pulls beneath the surface
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[COLOR=SlateGray][SIZE=1]Out of three banners, this is the one I like move. Let me know if you'd like any changes, or to see one of the other banners.

PS - Sorry this one isn't animated. The file was to large to be uploaded.[/SIZE][/COLOR]
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[size=1]Yes, that was spam.

Raizio, I was going to ban you, but seeing as Ruby went through the trouble of helping out with your request, I'm going to give you a chance to use them.

But first, read the rules. Several times. If you act like this again, you'll be banned without a second thought from anyone.

Do [i]not[/i] post the same word one hundred times.

Do [i]not[/i] post in CAPS.

Do [i]not[/i] Capitalize Every First Letter Of Every Word Of Your Sentence.

If you have questions, PM myself or another staff member. [/size]


text:
# Sentiment Analysis of Blogs by Combining Lexical Knowledge with Text Classification

by Melville et al. (KDD 2009)

### Motivation:

• before the rise of the Web 2.0 companies published product information and reviews on Web sites that were under their direct sphere of influence.
• nowadays, the focus of such discussions has shifted away from the company controlled Web sites into the blogosphere and social media where essentially anyone can comment on products and, therefore, influence purchase decisions.
• to cope with such distributed discussions we need to address the following three challenging data mining tasks:
• detect relevant discussion on products and relevant higher level concepts
• identify the most authoritative and influential contributors
• determine the sentiment of the discussion

### Method

#### General Remarks

• Pang et al. have shown that the use of lexicons is not as effective as learning models from training examples.
• Pooling information is a general approach from the field of Risk Analysis that combines information from multiple "experts", where each expert is usually represented by a probability distribution.
• typical pooling approaches are linear and logarithmic pooling which compute, depending on the used weights, the mean or the geometric mean of the involved distributions.
• Melville et al. adapt the expert's weights with a sigmoid weighting scheme, that considers the expert's past error rate: $\alpha_k = log \frac{1-err_k}{err_k}$

#### Naive Bayes

• The authors present an approach for deriving $$P(w_i|+)$$ and $$P(w_i|-)$$ from the entries of a sentiment lexicon, and
• use Lidstone smoothing with $$\epsilon = 10^{-6}$$ which tends to yield better probabilities than the standard Laplace smoothing: $P(w_i|c_j) = \frac{t_{ij}+\epsilon}{\sum_i t_{ij} + \epsilon|\mathcal{V}|}$ where $$\mathcal{V}$$ is the size of the vocabulary of the domain.
In the evaluation the author show that the combined approach up-weights and down-weights the sentiment values of lexicon entries. For instance, terms such as dark, social, complex, anger, alien and capture are up-weighted for the Movie domain, while talent, reason, promise, save, fair and redeem are down-weighted for this domain.

Tags:

Categories:

Updated:


text:
## anonymous one year ago I need help guys with the elimination method for 4y+2x=18 3x+6y=26

1. jdoe0001

$$\bf \begin{array}{llll} 4y+2x=18\\ 3x+6y=26 \end{array}\implies \begin{array}{llll} 2x+4y=18\\ 3x+6y=26 \end{array}\impliedby \textit{arranging the variables} \\ \quad \\ \begin{array}{llll} 2x+4y=18&{\color{brown}{ \times 3}}\\ 3x+6y=26&{\color{brown}{ \times -2}} \end{array}$$ if you multiply those expressions, by THAT MUCH, what do they become?

2. jdoe0001

that is, you mutiply the top by 3 and the bottom one by -2

3. anonymous

what if I add both equations down like.. 2x+4y=18 +3x+6y=26 equals 5x+10y=44? is this a right step?

4. anonymous

Is this an option?


text:
Go to content
Buffon's needle experiment is an example of a Monte Carlo method, whereby a random process is used to find an approximation to a mathematical value, in this case pi.
A needle is dropped on a floor, with floorboards of width equal to the needle's length. The number of times the needle falls on a crack is counted. This total, when compared to the number of times the needle has been dropped, can give an estimate of pi.
Begin by choosing the number of needle drops then click on Simulate.

an applet from
HREF="http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/bookapplets/index.html">Dartmouth
College, Hanover, NH

## Glossary

### pi

the ratio of the circumference of a circle to its diameter. It is approximately 3.14159265...

Full Glossary List

## This question appears in the following syllabi:

SyllabusModuleSectionTopicExam Year
AQA A-Level (UK - Pre-2017)S1ProbabilityExperimental-
CCEA A-Level (NI)S1ProbabilityExperimental-
Edexcel A-Level (UK - Pre-2017)S1ProbabilityExperimental-
I.B. Higher Level5ProbabilityExperimental-
I.B. (MSSL)3ProbabilityExperimental-
I.B. Standard Level5ProbabilityExperimental-
OCR A-Level (UK - Pre-2017)S1ProbabilityExperimental-
OCR-MEI A-Level (UK - Pre-2017)S1ProbabilityExperimental-
WJEC A-Level (Wales)S1ProbabilityExperimental-
CIE A-Level (UK)S1ProbabilityExperimental-
Methods (UK)M15ProbabilityExperimental-
Pre-U A-Level (UK)ProbProbabilityExperimental-


text:
ISSN 0254-0096　CN 11-2082/K

• •

### 考虑桩土差异的能源桩传热模型及其热响应半径计算

1. 1. 广西大学土木建筑工程学院,南宁 530004;
2. 工程防灾与结构安全教育部重点实验室,南宁 530004;
3. 广西金属尾矿安全防控工程技术研究中心,南宁 530004
• 收稿日期:2021-03-22 出版日期:2022-10-28 发布日期:2023-04-28
• 通讯作者: 江 杰（1979——）,男,博士、研究员,主要从事环境岩土工程与建筑节能方面的研究。jie_jiang001@126.com
• 基金资助:
国家自然科学基金（52068004; 51978179; 51768006）; 广西重点研发项目（AC20238002）

### HEAT TRANSFER MODEL AND THERMAL RESPONSE RADIUS CALCULATION OF ENERGY PILE CONSIDERING PILE-SOIL DIFFERENCE

Ou Xiaoduo, Huang Zhongzheng, Qin Jinxi, Jiang Jie, Lyu Zhengfan

1. 1. School of Civil Engineering and Architecture, Guangxi University, Nanning 530004, China;
2. Key Laboratory of Disaster Prevention and Structural Safety of Ministry of Education, Guangxi University, Nanning 530004, China;
3. Guangxi Engineering Research Center for Metallic Tailings Security Prevention and Control, Nanning 530004, China
• Received:2021-03-22 Online:2022-10-28 Published:2023-04-28

Abstract: In order to solve the problem of error caused by treating pile and soil as the same medium in heat transfer analysis of energy pile, an unsteady heat transfer model of U-shaped buried pipe energy pile is established, which can consider the thermophysical properties difference between pile and soil. The accuracy of the model is verified by comparing with the linear heat source model. On this basis, the approximate simplified expression of thermal response radius of energy pile is obtained by series expansion. Finally, the heat exchange per unit pile length, the thermal diffusion coefficient, pile diameter and the soil type are analyzed, and the "heat storage ratio" is used to evaluate the influence of the above parameters on the heat transfer process of the energy pile. The results show that： compared with the linear heat source model, the model can describe the heat transfer process of energy pile more accurately, which can effectively avoid the problems of underestimating the excess temperature of pile wall in the initial stage of heat transfer and overestimating the pile temperature in the stable stage of heat transfer; in the typical operation cycle of energy pile, the error of the calculation method of thermal response radius proposed in this study is within 0.1 ℃, which meets the engineering requirements; during the heat transfer process of energy pile, the heat storage ratio of soil increases with the decrease of the heat capacity of pile and the relative difference of thermal diffusion coefficient between pile and soil; the excess temperature of pile wall and the heat storage ratio of soil decrease with the increase of pile diameter; with the increase of heat transfer time, the excess temperature difference of pile wall corresponding to different pile diameter increases gradually, and the difference of heat storage ratio of soil decreases gradually; under the same heat transfer power, the change rate of excess temperature of energy pile wall hardly changes with the increase of heat transfer time; after 90 days of heat transfer, the pile diameter has little effect on the distribution of energy transfer.


text:
Lemma 4.33.7. Assume $p : \mathcal{S} \to \mathcal{C}$ is a fibred category. Assume given a choice of pullbacks for $p : \mathcal{S} \to \mathcal{C}$.

1. For any pair of composable morphisms $f : V \to U$, $g : W \to V$ there is a unique isomorphism

$\alpha _{g, f} : (f \circ g)^\ast \longrightarrow g^\ast \circ f^\ast$

as functors $\mathcal{S}_ U \to \mathcal{S}_ W$ such that for every $y\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$ the following diagram commutes

$\xymatrix{ g^\ast f^\ast y \ar[r] & f^\ast y \ar[d] \\ (f \circ g)^\ast y \ar[r] \ar[u]^{(\alpha _{g, f})_ y} & y }$
2. If $f = \text{id}_ U$, then there is a canonical isomorphism $\alpha _ U : \text{id} \to (\text{id}_ U)^*$ as functors $\mathcal{S}_ U \to \mathcal{S}_ U$.

3. The quadruple $(U \mapsto \mathcal{S}_ U, f \mapsto f^*, \alpha _{g, f}, \alpha _ U)$ defines a pseudo functor from $\mathcal{C}^{opp}$ to the $(2, 1)$-category of categories, see Definition 4.29.5.

Proof. In fact, it is clear that the commutative diagram of part (1) uniquely determines the morphism $(\alpha _{g, f})_ y$ in the fibre category $\mathcal{S}_ W$. It is an isomorphism since both the morphism $(f \circ g)^*y \to y$ and the composition $g^*f^*y \to f^*y \to y$ are strongly cartesian morphisms lifting $f \circ g$ (see discussion following Definition 4.33.1 and Lemma 4.33.2). In the same way, since $\text{id}_ x : x \to x$ is clearly strongly cartesian over $\text{id}_ U$ (with $U = p(x)$) we see that there exists an isomorphism $(\alpha _ U)_ x : x \to (\text{id}_ U)^*x$. (Of course we could have assumed beforehand that $f^*x = x$ whenever $f$ is an identity morphism, but it is better for the sake of generality not to assume this.) We omit the verification that $\alpha _{g, f}$ and $\alpha _ U$ so obtained are transformations of functors. We also omit the verification of (3). $\square$

Comment #6306 by Andrew on

In Hypothesis (1), I'm confused as to how the morphisms $f$ and $g$ could fail to be composable.

Comment #6418 by on

OK, I think there usually is no problem with having redundant information.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
12:16 AM
@AkivaWeinberger DogAteMy: I hadn't known about that. Wiki mentions his "deteriorating mental condition." Maybe it's a death wish to study stat mech?

I don't know any more than you do

Ted, I hate to bother you again, but residues are killing me. I am wanting to determine what $\frac{1}{2\pi z}\int_\gamma \frac{z f'(z)}{f(z)-a}\,dz$ counts, as in the 5th question here: its.caltech.edu/~ivrii/practice-exam.pdf

Your textbook author is suggesting you might be in trouble, DogAteMy.

But this means I have to determine residues of solutions to $f(z) = a$ of that function $\frac{zf'(z)}{f(z)-a}$
But I don't know how to start approaching that.

It adds up the roots of $f(z)=a$, @anakhro.

12:19 AM
@TedShifrin Clearly

So that means I have to show that both the residue and the winding numbers are 1?

Well, if $z$ is a simple root of $f(z)=a$, what is the residue (according to what I told you earlier)?
I'm assuming $\gamma$ is known to be a simple curve that winds once around all the roots. (Or else you add up the roots appearing inside $\gamma$.)

If it is simple, then it is $a$.
err
the root

Right, the root.

So then I have to consider higher order roots?
And I suppose there is a formula for this I should find in a book

12:23 AM
Now, if $z_0$ is a double root, say, then what will the Laurent series be there?
Hint: Let $u=z-z_0$.
Ignore all the stupid formulas with derivatives. I hate those.

Where $u$ is a change of variables?

The Laurent expansion is naturally going to be in $u$, right? That's easier than writing $(z-z_0)$ everywhere.

I am afraid I am lost. Maybe I need to read up on Laurent expansions again.

OK.
Or Taylor.

So I let $u=z-z_0$ and write $\frac{zf'(z)}{f(z)-a} = \frac{(u+z_0)f'(u+z_0)}{f(u+z_0)-a}$
I feel like just naively filling it out has gotten me in a rut
Normally there is something to use the power series for.
But because of this f'/(f-a) stuff I am not sure how I would go about it in the normal way.

12:34 AM
But use that $z_0$ is a double-root of $f(z)-a$.
Write $f(z)-a = (z-z_0)^2 g(z) = u^2 g(u+z_0)$.

Should I avoid expanding everything out? I took the liberty of also determining $f'$ with this

Yes, avoid. We're just trying to find the coefficient of the $1/u$ term.

Curious what your thoughts are on this. I found several not-obviously-equivalent solutions:
> Hardest problem on 1st year mid-term logic and sets was: "Find a countably infinite collection of sets such that any two intersect in exactly one element, and any three in none." What's the first solution you can think of (that fits in a tweet)? Students found several. — Andrej Bauer on Twitter
(Not a test I took, to be clear)

Hmmm ... Seems like a reasonable question. I don't have an automatic response.

I am not sure what I am trying to do algebraically. Certainly not just pull a term 1/u out.

12:43 AM
Well, depends on the level of the students and course, I guess. I would never have put that on an exam for our Intro to Higher Math course ... unless we'd done it in class.
The denominator has a $u^2$, @anakhro, so, as usual, you want the $u$ coefficient upstairs.

I have something that looks like $$\frac{(u+z_0)(2ug(u+z_0) + u^2g'(u+z_0))}{u^2g(u+z_0)}.$$

OK, so what is the coefficient of $u$ upstairs?

I can get rid of one u in the top and bottom
So it is $\frac{(u+z_0)(2g(u+z_0) + ug'(u+z_0))}{ug(u+z_0)}$

Now what's the constant term upstairs?
And what's the resulting coefficient of $1/u$ in that thing?

$2z_0g(u+z_0)$

12:46 AM
So we get $2z_0$. Bingo.
Double root.
I leave it to you to generalize to $m$th order root.

How do you conclude $2z_0$ from that? How does the $g$ part disappear?

Look at the denominator.
Hmm @DogAteMy.

OH
I see.

You should practice doing computations like this ... if you're preparing for a complex analysis qualifying or something.

12:50 AM
Yeah, definitely. I 100% agree.
I just feel so pressed.

Practice is the only way to improve speed.

Yeah. I just wish I didn't have to take so many courses so I would have more time to practice.

DogAteMy: Something indexed by primes, maybe?

I appreciate it ever so much.

Anyhow, you should only take a few seconds to do the $m$th order case, @anakhro, now.

12:53 AM
It should follow in a similar manner, or do you mean just guessing the answer?

Both.

Well I think I have the guess in a second, let me try the algorithm.
Yeah I did it
Heh

DogAteMy: What is my set $A_p$ is the set of all prime multiples of $p$?
See, @anakhro: Told you so :P

I thought I was in for some real hard thinking but the moment I started I realized what changes.

So you have $A_p=\{pq:q\text{ prime}\}$ for prime $p$?

12:55 AM
Yes.

That works

I take it that's not the canonical answer.

It's very close, it's like one step removed

Hmm ...

I never said these sets are subsets of $\Bbb N$

12:56 AM
Yes, I realize.
That seemed a natural (pun intended) starting place to me.
Oh. Axes in $\Bbb R^\omega$.
This is a great homework question, but I wouldn't have put it on an exam on an introductory course.

Elaborate?

Oh, no, what I said is dopey.
I have to look at affine subspaces.

You know, since you already gave an answer I might as well tell you it

Well, don't ruin it for everyone else.

OK I won't then
Take "$A_p=\{pq:q\text{ prime}\}$ for prime $p$" and get rid of the number theory

1:06 AM
LOL, how do I get rid of number theory in that?
Oh, $y=n(x+1)$.

I'm thinking of $A_m=\{\{m,n\}:n\in\Bbb N\}$ for $m\in\Bbb N$

@TedShifrin I was wondering if you used LaTex for the "book" on differential geometry

$A_n$ is the set of unordered pairs of integers containing $n$

@Bob, yes, for all my books.
DogAteMy, so the affine lines in the plane do it.
Ah, so you're sticking to integers, but maybe it's the same idea.

@TedShifrin Don't they all contain $(-1,0)$?

1:10 AM
I see, unordered pairs, rather than ordered. Sneaky.
Oh, posh. How 'bout $y=nx+1$, and remove $(0,1)$ from the plane?

@TedShifrin thanks

OR my example removing $(-1,0)$.
@Bob: In truth, had LaTeX not been around, I never would have written any books. Probably would have been a lot better for my research paper production, but oh well.

Do you wish you had done more in research?
Or do you like the balance you struck?

My one regret is that my work with two other guys (where we produced a bunch of papers) had one more giant paper on which we had worked for years. But we were all daunted by the realization that we'd have to write 3 50-60 page papers to incorporate what we'd done, and none of us had the energy to do that.
Each of us had multiple folders full of notes.
I enjoyed collaborative work a lot.
But in honesty I know my books (and YouTube lectures) are impacting orders of magnitudes more people than the research.
4
(Not to mention generating hatred.)

@TedShifrin I don't quite understand. Where do $A_1$ and $A_2$ intersect?
if not at the point you just removed

1:16 AM
Oh crap @DogAteMy. I need to make parabolas instead.
Or make the intersection point move, which was my intention.
Duh. $y=n(x-n)$.
Yes, I was trying to make the intersection point move. I think I succeeded. :P

So where do $A_m$ and $A_n$ intersect
$m(x-m)=n(x-n)$

They're non-parallel lines, so they have to intersect.
And surely that point won't work for a different index :D

They intersect at $(m+n,mn)$, I think
which works

Or I can resort to primes again :P

My idea would be something like this: Put a line with positive slope through the origin in R^2, move a bit to the right, add a line with a steeper slope going through that point, continue. Don't have the time to work it out formally right now, but I think it should work if one makes the increases gradually smaller.

1:20 AM
Yeah, $(m+n,mn)$ is right.

That's basically what he just did I think @Thorgott

That's effectively what I've done, @Thorgott.
Oops.

These are the lines through $(n,0)$ with slope $n$

Oh lol, I didn't try to visualize what you were posting :p

Do I pass, DogAteMy? :D

1:21 AM
Yeah lol

Whew!

The answer I had was, $A_n$ is the set of unordered pairs of integers containing $n$, as $n$ ranges over the integers @Thorgott
(which geometrically looks like a vertical line starting at the x-axis and bouncing off the x=y line)

You know that once you banned "number theory" I had to go "geometric." :D

One of my geometric answers was,
look at the set of circles of unit radius containing the origin
Delete the origin from each circle
Now we need a countable collection of them, taking care not to include two circles that are exactly opposite each other

I think if you decrease the distance between the x-intercepts of the lines and increase their slopes, you get something resembling the tangent bundle of a quarter circle

1:23 AM
which can be done in a variety of ways (e.g. restrict their centers to those with rational positive coordinates)

Oh, that's nice, DogAteMy. Probably morally equivalent to mine if we blow up the origin.

Don't these usually intersect in two points?

@Thorgott Yeah I saw someone else had the set of tangents to a circle (again, countable collection, no two exactly opposite each other, can be solved in the same way eg restrict to quarter circle with countable slope)

But he's making one of them fixed and removing it.

Ah, that's nice

1:25 AM
Oh, I'm behind, I think.

Oh wait, you want the origin to be on the circle, not inside

Yeah

No, @Thorgott, because in the tangent bundle fibers are distinct.

You can also do the graph of $y=\sin(x-t)$ over $x\in[0,\pi)$, for a countable collection of $t$

You mean the envelope of actual tangent lines.

1:26 AM
(Notice that that's a half-period)
You could also do great circles on a sphere of a given inclination (latitude of highest point), provided you identify opposite points

Yeah, that's what I meant

DogAteMy: I admit I'm now troubled by the fact that I first tried to give a discrete ("number theory" based) solution rather than a geometric one. I am getting senile.

Actually, can't we even get an uncountable family using tangents of the curve?

Of course.
My example with the lines is easily made uncountable, too.

I would say my first answer is "set theoretic"

1:32 AM
I claim that DogAteMy biased us in the direction of $\Bbb N$ with the countable restriction.

and thus works with any cardinality

We're firing you.

I was being paid?!

You can be fired from a volunteer non-paid position.

Fair

1:34 AM
As can I.
waits eagerly

Once you have a countably infinite solution, you can take enough copies and get a solution of any cardinality

If we want any three to intersect in one point and any four to have empty intersection, we can let $A_n$ be the set of unordered triples containing $n$
and thus $A_i\cap A_j\cap A_k=\{i,j,k\}_3$
(…I don't actually know of any notation for unordered triple)
(…though it doesn't matter since we don't need to deal with repeated elements, so I can just use sets)

A set is an unordered n-tuple.
Right.

In general you'd want $\{1,1,2\}_3\ne\{1,2,2\}_3$.
The natural framework for "roots of a polynomial" is unordered tuple, because of multiplicities
Basically unordered tuple = set with multiplicity

You can just work with cartesian product minus diagonals (or not minus) ... modulo symmetric group.

1:39 AM
Sure

I'd write $[(i,j,k)]$, as in the equivalence class

I wonder if there's a plane through $(i+j+k,ij+jk+ik,ijk)$
er

I was about to type exactly that.
Lots of planes.

for fixed $i$ and variable $j,k$
What's $A_i:=\{(i+j+k,ij+jk+ik,ijk)\in\Bbb R^3:j,k\in\Bbb R\}$

It's a varying conic.
Put $s$ and $t$ in there instead of $j$ and $k$.

1:44 AM
$(x-i)i^2-yi+z=0$?
which is a plane

Ah, I was playing with quadratics. Is that correct?

So if $A_n$ is the plane $xn^2-yn+z=n^3$ then this solves the triple version of the problem geometrically

Are we sure no three intersect? Generically, three planes will intersect in a point.
Oh, no four.
"Never mind."

Compare this to the 2D version, $y=n(x-n)$, aka $xn-y=n^3$

Yeah, got it.

1:48 AM
I'm kinda curious what these look like

Mathematica can draw those.

You could also generalize the other version to tangent planes of a section of the 2-sphere

@Thorgott: You don't need a sphere, of course.

@Thorgott Can you? You can easily get 4 to intersect in a point
A point gives us an infinite collection of tangent planes through it

1:51 AM
Are you restricting to a circle, DogAteMy?

Restricting what?

The tangent planes of the sphere along a circle?
Generically, Thorgott's right, I think, but if we specialize ...

I don't think any three intersect in a point then

The planes all intersect in a line.
Oh, no, that's wrong.
Agh. Enough for me.

In the original problem, we could do parabolas:
$y=(x-n)^2$
Right?
I think in the 3D case we can do horizontal translates of paraboloids
Wait
No never mind

1:57 AM
Oh right. even three of them need not intersect in a point. Would've been too nice.

2:14 AM
> Scott Aaronson has written about how easy it is to predict people trying to “be random”:

>> In a class I taught at Berkeley, I did an experiment where I wrote a simple little program that would let people type either “f” or “d” and would predict which key they were going to push next. It’s actually very easy to write a program that will make the right prediction about 70% of the time. Most people don’t really know how to type randomly. They’ll have too many alternations and so on. There will be all sorts of patterns, so you just have to build some sort of probabilistic model. Even a very
@Secret This reminds me of you for some reason
> …Being genuinely random is important in pursuing mixed game theoretic strategies. Henrich’s view is that divination solved this problem effectively.

I’m reminded of the Romans using augury to decide when and where to attack. This always struck me as crazy; generals are going to risk the lives of thousands of soldiers because they saw a weird bird earlier that morning? But war is a classic example of when a random strategy can be useful. If you’re deciding whether to attack the enemy’s right vs. left flank, it’s important that the enemy can’t predict your decision and send his best defend
This is so neat I have to repeatedly stop myself from posting more quotes here
Let me just put one:
> Rationalists always wonder: how come people aren’t more rational? How come you can prove a thousand times, using Facts and Logic, that something is stupid, and yet people will still keep doing it?

Henrich hints at an answer: …
So much of this is quotable

1 hour later…
3:43 AM
@TedShifrin heya Ted

4:10 AM
0

I have done dual degree with bachelors in computer science. I am currently a software developer (India). I always have had interest in mathematics. But once was I got into college I did not work towards making myself mathematically mature. In college I just used to participate in coding contests...

4:23 AM
Is there a standard graduate-level textbook on regular languages and automata theory (e.g., one that introduces both algebraically) that is currently in print? Eilenberg is the standard reference, but his books run for $150 on Amazon? um computer science IS a field of applied mathematics? And I won't lie, Impact measured in quantity of people seems like a fallacy 2 hours later… 6:13 AM Morning! 6:28 AM What are you doing awake at dawn 7:10 AM 5. Suppose that$u(x, y)$has a local maximum at$(0,0) .$Show that for any direction$\vec{V}$$\frac{\partial u(0,0)}{\partial \vec{V}}=0$ 0 Suppose that$u(x, y)$has a local maximum at$(0,0) .$Show that for any direction$\frac{\partial u(0,0)}{\partial \vec{V}}=0$. So I want to show that directional derivative is zero. So finally I get${u(ah,bh) - u(0,0)}$/$h$as limit$h$tends to zero ; how to proceed after that. 7:50 AM @Alessandro frantically reading about cyclotomic fields Is there an exam coming up? 8:04 AM Yeah I have exams on the 6th and 10th of Feb, though I think I'm going to delay them until March I see I have one on the 10th and one the 11th I think, not sure about the second And one in March Nice, feeling prepared? For the March one definitely not (that's why I postponed it lol) Lol fair, I'm feeling okay for alg. number theory but modular forms is kicking my ass For the other two I'm definitely feeling more prepared for set theory than global analysis, but there's still time to study 8:11 AM Aye, I just feel like if I postpone until March I'll do really well, rather than cramming for the feb. exams and failing lol Makes sense Do you only have those two exams? Yeah I only took two modules this semester so I could ease myself back into mathematics after a year out lol unfortunately my poor mental health has hamstrung me somewhat, which is the main reason for postponing :P @EdwardEvans sounds like a smart approach Truuuuth plus cyclotomic fields are super interesting and cool so I wanna learn them properly instead of cramming lol Fair For my ANT class two years ago I gave a presentation on how to factor ideals in cyclotomic extensions based on Neukirch's 1.10 That's also how far I've ever gotten in Neukirch 8:26 AM Nice :D hahaha I'm just looking at ramification phenomena in cyclotomic extensions I used to know very little ANT, but now I forgot that too I regret doing algebraic geometry instead of class field theory in my first semester in Bonn though Obviously I think it's amazing, but a lot of the formalities at the start of an ANT course can get kinda boring Í'm hoping CFT will be offered next semester but I've heard on the grapevine that it might not be :( Oh, that sucks On the other hand I like that here they offer different courses every semester instead of only having the usual big ones indeed, I think maybe some courses on lie theory will be offered and I'll be taking algebra 2 for the homological algebra lol yeah that's cool You won't get bored Can you take anything you want as long as you reach 120 credits or are you forced to do stuff from different areas of maths? 8:34 AM But I'm mainly interested in doing lots of number theory so I'll be slightly disappointed if CFT doesn't get offered during my time here lol we have to do 2 applied mathematics modules and then some "interdisciplinary skills" course so I'm gonna take Russian language next semester Applied meaning? Numerical/probability/mathematical physics/discrete? @EdwardEvans I'd suggest Italian, but Russian is also cool :P lol it's slightly blurred, functional analysis 2 counts as applied but yeah I'll probs take some physics @EdwardEvans what's in the syllabus? errr hang on not a clue but the lecturer is a physicist I think That doesn't really tell me much hahaha 8:39 AM yeah idk lol his website doesn't have any info about it I. Differentialkalkül in Banachräumen und Satz von der impliziten Funktion II. Banachmannigfaltigkeiten und Fredhomabbildungen, Satz von Sard-Smale III. Der Abbildungsgrad modulo 2 IV. Der Abbildungsgrad von Brouwer, topologische Anwendungen V. Der Leray-Schauder-Grad VI. Lösungszweige und Verzweigungen von Lösungen VII. Monotone Operatoren I think that's it :P Non-linear functional analysis Uh Banach manifolds Weird stuff Looks like an interesting course though maybe idk what's being offered next semester tho excited to take some alg. geo in 3rd semester tho lol and algebraic topology I was looking forward to AG but then didn't like it. I think you will though yeah maybe we'll see hehe 9:01 AM Hello Anyone worked through baby Rudin here? Hello 9:29 AM @AlessandroCodenotti wanna study an endgame with me? I'm busy with lectures today 10:07 AM 4.5 hours of cyclotomic fields this morning already hahaha 10:35 AM @EdwardEvans you need a break 10:52 AM I'm just watching the proof of quadratic reciprocity tho well that was nice 2 hours later… 12:52 PM World's 3rd Superpower 1 hour later… 2:12 PM @AkivaWeinberger You probably recalled my discussions about indeterminable sequences some time ago. Otherwise, that whole review, despite being cultural anthropology, does have some deep philosophical ideas that interests me for my philosopher hobby For example, the central idea on how we are basically kept alive by randomness, and how that leads to traditions to evolve as a safeguard from killing us and to obey authority, the reason why there is power struggle, and coercion, all boils down to one reason: > In giving humans reason at all, evolution took a huge risk. Surely it must have wished there was some other way, some path that made us big-brained enough to understand tradition, but not big-brained enough to question it. Maybe it searched for a mind design like that and couldn’t find one. So it was left with this ticking time-bomb, this ape that was constantly going to be able to convince itself of hare-brained and probably-fatal ideas. This is a problem more ancient and much much harder than the millenium prize problems: How to get rid of reason completely and as we can see from history, nature still failed to solve it Put it in another way. That this problem is still open and nature is picking a partial solution to is, is the origin of injustice 3:02 PM 0 Definition: A measure space$(X, Σ, μ)$is said to be perfect if, for every$Σ$-measurable function$f : X → \mathbb{R}$and every$A ⊆ \mathbb{R}$with$f^{−1}(A) \in Σ$, there exist Borel subsets$A_1$and$A_2$of$\mathbb{R}$such that${\displaystyle A_{1}\subseteq A\subseteq A_{2}{\mbox{ a...

I am not sure whether I did it correctly.

3:31 PM
@Subhasis What you are saying is that $G$ is a measurable subset of $\mathbb{R}$ and $\mathbb{R}$ is an inner regular measure space, so $G$ is an inner regular set. This is not the same as saying that any subset of $G$ is an inner regular set with respect to the Lebesgue measure on $G$.
What you want to do is look at the subsets of $G$ and see if you can approximate them within $G$. Then use that the Lebesgue measure on $G$ is the restriction of the Lebesgue measure on $\mathbb{R}$ and that this measure is inner regular.

$G$ is an inner regular set w.r.t the Lebesgue measure on $\mathbb{R}$ only (by the arguments I provided), right?

It should also be noted that the Borel-algebra on [a,b] is the same as the trace algebra of [a,b] in the Borel-algebra on R
Yes
The idea of proof is similar, but it's important to be precise on the details in measure theory

@Thorgott Does it directly imply that the subset $A$ of $G$ can be approximated from within?
Since, $A \subset G \subset \mathbb{R}$?

3:54 PM
@Thorgott Approximating a set $G$ in $\mathbb{R}$ means that $\forall \ \epsilon>0$, $\exists$ an open set $U$ such that $\lambda(U \setminus G)< \epsilon$. Is that right?

Hey chat! What textbook would you recommend as a good reading companion to Huffman/Kunze's Linear Algebra?

4:13 PM
Yes, that's the idea. Let $A\subseteq G$ be measurable with respect to the Borel-algebra on $G$. Then $A\subseteq\mathbb{R}$ is measurable with respect to the Borel-algebra on $\mathbb{R}$ (Why?). This measure is inner regular, so $\lambda(A)=\sup\{\lambda(K)\colon K\subseteq A\text{compact}\}$. Now all the $K$ the supremum is taken over are subsets of $A$, hence subsets of $G$.
Now note that this is the same as saying that $A$ is an inner regular set in $G$, because a) a compact subset of $G$ is the same as a compact subset of $\mathbb{R}$ that is contained in $G$ and b) the restriction of

@StupidQuestionsInc, I would recommend Elementary Linear Algebra Applications by Howard Anton

@Thorgott A finite measure $μ$ is inner regular if and only if, for all $ε > 0$, there is some compact subset $K$ of $X$ such that $μ(X \setminus K) < ε$. Let $X=[a,b]$. Consider $K=[a+\epsilon/4 , b-\epsilon/4]$. Evidently, $\mu(X\setminus K)=\epsilon/2 <\epsilon$
can the special case of the inner-regularity of $[a,b]$ be proved this way?

No, you are mixing the concept of inner regular measure and inner regular set up.
This just shows $X$ is an inner regular set (both within $\mathbb{R}$ and within itself).
But to show $X$ (with the inherited sigma-algebra and measure) is a perfect measure space, you need to show that every measurable subset of $X$ is inner regular in $X$.

@Thorgott EVERY
@Thorgott indeed. It is very confusing.
@Thorgott here we are using $C$ is compact in $Y$ iff $C$ is compact in $X$, where $Y \subseteq X$. Right?

Yeah, though that's pretty much just the definition of subspace topology.

4:26 PM
Thank you very much. It is much more clear now.

All of what I said is essentially just carefully unraveling definitions.

Could you post that as an answer?

@CroCo will look into it

Actually, I misspoke at the end. I said this argument shows that restrictions of perfect measures are perfect, but that's not what I meant. What the argument shows is that the restriction of inner regular measures is inner regular. The statement about perfect measures is still true, but it requires a different argument. Also sure, I can.

@Thorgott I actually know nothing about measure theory. But I can sort of understand what you did.

4:44 PM
heya guys, what's the reason(/motivation) behind taking $x,y$ and $x',y'$ distinct in the definition of 2-transitivity (where we define 2-transitivity to be a group action where for any distinct $x,y$ and $x',y'$ we have $\sigma_g(x)=x'$ and $\sigma_{g'}(y)=y'$ for some $g,g'\in G$, where $\sigma\colon G\to S_X$ is our group action)
I would think that if we don't force $x,y$ and $x',y'$ to be distinct, then we would also have transitivity
usually 2-transitvity is called "stronger", but by having $x,y$ and $x',y'$ distinct, it seems like it's not stronger, but just different from transitivity
or am I wrong, and is a 2-transitive action still transitive?


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## Дробильное, шлифовальное и горное оборудование.

### Горячий поиск

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text:
## Chemistry: Molecular Approach (4th Edition)

The theoretical yield of Ammonia is: $29.4$ kg.
** 1 kg = $10^3$ g $5.22 kg \times \frac{10^3g}{1kg} = 5.22 \times 10^3 g$ $31.5 kg \times \frac{10^3g}{1kg} = 31.5 \times 10^3 g$ 1. Calculate the number of moles of $H_2$: 1.008* 2 = 2.016g/mol $5.22 \times 10^3g \times \frac{1 mol}{ 2.016g} = 2.589 \times 10^3mol (H_2)$ According to the balanced equation: The ratio of $H_2$ to $NH_3$ is 3 to 2: $2.589 \times 10^3mol (H_2) \times \frac{ 2 mol(NH_3)}{ 3 mol (H_2)} = 1.726\times 10^3 mol (NH_3)$ --------- 3. Calculate the number of moles of $N_2$: 14.01* 2 = 28.02g/mol $31.5 \times 10^3g \times \frac{1 mol}{ 28.02g} = 1.124\times 10^3mol (N_2)$ According to the balanced equation: The ratio of $N_2$ to $NH_3$ is 1 to 2: $1.124 \times 10^3 mol (N_2) \times \frac{ 2 mol(NH_3)}{ 1 mol (N_2)} = 2.248 \times 10^3 mol (NH_3)$ --------------------------------------------------- Since $H_2$ made the least amount of $NH_3$, it is the limiting reactant, so we should follow its calculations with $1.726 \times 10^3$ moles of Ammonia: 2. Calculate the mass of $NH_3$: 14.01* 1 + 1.008* 3 = 17.03g/mol $1.726 \times 10^3 mol \times \frac{ 17.03 g}{ 1 mol} = 29.4 \times 10^3 g (NH_3) = 29.4 kg (NH_3)$