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2 Exercise. Let C be a curve drawn by: Find the length of the curve drawn by p as t runs from 0 to z: P(t) = (2 cos(t), -2sin(t)) Can you cxplai u fuis answer? length = 2pi

2 Exercise. Let C be a curve drawn by: Find the length of the curve drawn by p as t runs from 0 to z: P(t) = (2 cos(t), -2sin(t)) Can you cxplai u fuis answer? length = 2pi

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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1. Exercise. Suppose that the curve C in the zy-plane is traced out by the vector-valued function T(t) – (2/2, 4t + 1),0sts4. In order to determine if this continuously differentiable curve is parameterized by arclength, we could check either if • ST(7)| dr = t • T'(t)| - 1 for all t. If either of these holds, then the curve uses arclength as a parameter. Furthermore, in order to establish the first result, we would have to compute T(t)| anyways, so let's take this approach. We calculate that T'(t) - ), and hence T'(t)| - Since |T'(t)| + 1 for all t, the curve does not v use arclength as a parameter. 2. Can you cxpai n fuis answer? Exercise. Let C be a curve drawn by: p(t) = (2 cos(t), -2 sin(t)) Find the length of the curve drawn by p as t runs from 0 to : length = 2pi Exercise. Is p parameterized by arc length? yes v Correct no Exercise. Give a new description of C, call it q(s), such that s is an arc length parameter. q(s) – 2cos(s/2) -2sin(s/2) 4. for 0 VS8S 2pi 3. Exercise. The starting position of a particle is given by P(0) = (0,0,0) and its acceleration is described by the vector-valued function a(t) = (-3,0, -6t) for t > 0. Suppose the initial velocity is given by v(0) = (60, 10, 75). Find: • The velocity function: v(t) = • The speed function: s(t) = • The position function: p(t) = • The maximum z-height of the particle when t > 0: maximum height:
Transcribed Image Text:1. Exercise. Suppose that the curve C in the zy-plane is traced out by the vector-valued function T(t) – (2/2, 4t + 1),0sts4. In order to determine if this continuously differentiable curve is parameterized by arclength, we could check either if • ST(7)| dr = t • T'(t)| - 1 for all t. If either of these holds, then the curve uses arclength as a parameter. Furthermore, in order to establish the first result, we would have to compute T(t)| anyways, so let's take this approach. We calculate that T'(t) - ), and hence T'(t)| - Since |T'(t)| + 1 for all t, the curve does not v use arclength as a parameter. 2. Can you cxpai n fuis answer? Exercise. Let C be a curve drawn by: p(t) = (2 cos(t), -2 sin(t)) Find the length of the curve drawn by p as t runs from 0 to : length = 2pi Exercise. Is p parameterized by arc length? yes v Correct no Exercise. Give a new description of C, call it q(s), such that s is an arc length parameter. q(s) – 2cos(s/2) -2sin(s/2) 4. for 0 VS8S 2pi 3. Exercise. The starting position of a particle is given by P(0) = (0,0,0) and its acceleration is described by the vector-valued function a(t) = (-3,0, -6t) for t > 0. Suppose the initial velocity is given by v(0) = (60, 10, 75). Find: • The velocity function: v(t) = • The speed function: s(t) = • The position function: p(t) = • The maximum z-height of the particle when t > 0: maximum height:
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