Explanation Given: Curve = r ( t ) = 〈 4 t 1 2 , ln t , 2 t 〉 Point, a = 1 Formula used: s ( t ) = ∫ a t | | r ' ( u ) | | d u Calculation: r ' ( t ) = 〈 4 × 1 2 × t − 1 2 , 1 t , 2 〉 = 〈 2 t − 1 2 , 1 t , 2 〉 | | r ' ( t ) | | = ( 2 t − 1 2 ) 2 + ( 1 t ) 2 + ( 2 ) 2 = 4 t − 1 + 1 t 2 + 4 = 4 t + 1 t 2 + 4 = 4 t + 1 + 4 t 2 t 2 = ( 1 ) 2 + ( 2 t ) 2 + 2 × 2 t t 2 = ( 1 + 2 t ) 2 t 2 &#

CALCULUS EARLY TRANSCENDENTALS W/ACHIEVE

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ISBN: 9781319057664

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 13.3, Problem 12E

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Compute the arc length of the curve 〈4t12,lnt,2t〉.

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Chapter 13.3, Problem 11E

Chapter 13.3, Problem 13E

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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 13 Solutions

CALCULUS EARLY TRANSCENDENTALS W/ACHIEVE

Ch. 13.1 - Prob. 1PQ Ch. 13.1 - Prob. 2PQ Ch. 13.1 - Prob. 3PQ Ch. 13.1 - Prob. 4PQ Ch. 13.1 - Prob. 5PQ Ch. 13.1 - Prob. 6PQ Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E

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Compute the arc length of the curve 〈 4 t 1 2 , ln t , 2 t 〉 .

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Compute the arc length of the curve 〈4t12,lnt,2t〉.

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Chapter 13 Solutions