University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 12 - Section 12.1 - Curves in Space and Their Tangents - Exercises - Page 649: 25


$2 \sqrt 5 i+\sqrt 5 j$

Work Step by Step

We are given that the velocity vector is tangent to the graph of $y^2=2x$ at the point$(2,2)$ with length $5$. We need to find the direction of the velocity vector. $y^2=2x \implies 2y\dfrac{dy}{dx}=2$ Thus, the slope at the point$(2,2)$ is $\dfrac{dy}{dx}=\dfrac{1}{2}$ So, the tangent vector lies in the direction of the vector $i+\dfrac{1}{2} y$ Hence, the velocity $=5(\dfrac{i+\dfrac{1}{2} y}{\sqrt{1+1/4}})=2 \sqrt 5 i+\sqrt 5 j$
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