Explanation Given: Two ellipses in standard position: E 1 : ( x a 1 ) 2 + ( y b 1 ) 2 = 1 ... ( 1 ) E 2 : ( x a 2 ) 2 + ( y b 2 ) 2 = 1 ... ( 2 ) Calculation: Assume e 1 and e 2 are the eccentricity of the two ellipses and we get: b 1 2 = a 1 2 ( 1 − e 1 2 ) b 2 2 = a 2 2 ( 1 − e 2 2 ) Now, b 2 2 b 1 2 = a 2 2 ( 1 − e 2 2 ) a 1 2 ( 1 − e 1 2 ) If the eccentricity of the two ellipses is equal that means e 1 = e 2 , then the above relation will be: b 2 2 b 1 2 = a 2 2 a 1 2 = r 2 Now, b 2 = r b 1 and a 2 = r a 1 This shows that one ellipse is enlargement of the other ellipse by the factor r...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 12.5, Problem 74E

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That $E_{1}$ and $E_{2}$ are similar under scaling if and only if they have the same eccentricity. We need to show that any two circles are similar under scaling.

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Chapter 12.5, Problem 73E

Chapter 12.5, Problem 75E

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(12 points) Let E={(x, y, z)|x²+ y² + z² ≤ 4, x, y, z > 0}. (a) (4 points) Describe the region E using spherical coordinates, that is, find p, 0, and such that (x, y, z) (psin cos 0, psin sin 0, p cos) € E. (b) (8 points) Calculate the integral E xyz dV using spherical coordinates.

(10 points) Let f(x, y, z) = ze²²+y². Let E = {(x, y, z) | x² + y² ≤ 4,2 ≤ z < 3}. Calculate the integral y, f(x, y, z) dV.

(14 points) Let f: R3 R and T: R3. →R³ be defined by f(x, y, z) = ln(x²+ y²+2²), T(p, 0,4)=(psin cos 0, psin sin, pcos). (a) (4 points) Write out the composition g(p, 0, 4) = (foT)(p,, ) explicitly. Then calculate the gradient Vg directly, i.e. without using the chain rule. (b) (4 points) Calculate the gradient Vf(x, y, z) where (x, y, z) = T(p, 0,4). (c) (6 points) Calculate the derivative matrix DT(p, 0, p). Then use the Chain Rule to calculate Vg(r,0,4).

Chapter 12 Solutions

Calculus - Standalone book

Ch. 12.1 - Prob. 1PQ Ch. 12.1 - Prob. 2PQ Ch. 12.1 - Prob. 3PQ Ch. 12.1 - Prob. 4PQ Ch. 12.1 - Prob. 5PQ Ch. 12.1 - Prob. 6PQ Ch. 12.1 - Prob. 7PQ Ch. 12.1 - Prob. 1E Ch. 12.1 - Prob. 2E Ch. 12.1 - Prob. 3E

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That E 1 and E 2 are similar under scaling if and only if they have the same eccentricity. We need to show that any two circles are similar under scaling.

Solution Summary: The author explains that E 1 and E 2 are similar under scaling if they have the same eccentricity.

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That $E_{1}$ and $E_{2}$ are similar under scaling if and only if they have the same eccentricity. We need to show that any two circles are similar under scaling.

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Chapter 12 Solutions

That E 1 and E 2 are similar under scaling if and only if they have the same eccentricity. We need to show that any two circles are similar under scaling.

Solution Summary: The author explains that E 1 and E 2 are similar under scaling if they have the same eccentricity.

Concept explainers

Videos

That E1 and E2 are similar under scaling if and only if they have the same eccentricity. We need to show that any two circles are similar under scaling.

Want to see the full answer?

Chapter 12 Solutions

That $E_{1}$ and $E_{2}$ are similar under scaling if and only if they have the same eccentricity. We need to show that any two circles are similar under scaling.