Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 13: Vector-Valued Functions and Motion in Space - Practice Exercises - Page 777: 8


$v \cdot j =12; a \cdot j =26$

Work Step by Step

$v=\dfrac{dx}{dt}i+\dfrac{dy}{dt}j$ $\implies v \cdot j=\dfrac{dy}{dt}=\dfrac{1}{3}x^2 \dfrac{dx}{dt}$ Since, $\dfrac{dx}{dt}=4$ $v \cdot j (3,3)=\dfrac{4}{3}x^2=\dfrac{4}{3}\times 3^2=12$ Now, $a(t)=\dfrac{2}{3}x(\dfrac{dx}{dt})^2+\dfrac{1}{3} x^2 \dfrac{d^2 x}{dt^2} $ So, $a \cdot j=\dfrac{2}{3}x(\dfrac{dx}{dt})^2+\dfrac{1}{3} x^2 \dfrac{d^2 x}{dt^2} =\dfrac{2}{3}x(4)^2+\dfrac{1}{3} (3)^2 (-2) =26$ So, $v \cdot j =12; a \cdot j =26$
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