## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 13 - Section 13.2 - Derivatives and Integrals of Vector Functions - 13.2 Exercise - Page 861: 49

#### Answer

$f'(2)=35$

#### Work Step by Step

Given: $f(t)=u(t) \cdot v(t), u(2)=\lt 1,2,-1 \gt,u'(2)=\lt 3,0,4 \gt$ and $v(t)=\lt t, t^2,t^3 \gt$ Our aim is to calculate $f'(2)$. In order to find this, we will use the dot product rule. $f'(t)=u'(t) \cdot v(t)+u(t) \cdot v'(t)$ $v(t)=\lt t, t^2,t^3 \gt \implies v'(t)=\lt 1, 2t,3t^2 \gt$ and $v'(2)=\lt 1,4,12 \gt$ Thus, $f'(2)=\lt 3,0,4 \gt \cdot \lt 2,4,8 \gt+\lt 1,2,-1 \gt \cdot \lt 1,4,12\gt$ $=6+0+32+1+8-12$ $=35$ Hence, $f'(2)=35$

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