Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 13 - Section 13.3 - Tangent Lines and Derivatives - 13.3 Exercises - Page 922: 39

Answer

$v(a)=12a^{2}+6;$ $t=1\Rightarrow v\left( 1\right) =12\times 1^{2}+6=18\dfrac {m}{s} $ $v\left( 2\right) =12\times 2^{2}+6=54\dfrac {m}{s}$ $ v\left( 3\right) =12\times 3^{2}+6=114\dfrac {m}{s}$

Work Step by Step

$v=\dfrac {\partial s}{\partial t}=\dfrac {\partial l}{\partial t}\left( 4t^{3}+6t+2\right) =12t^{2}+6\Rightarrow t=a\Rightarrow v\left( a\right) =12a^{2}+6;$$t=1\Rightarrow v\left( 1\right) =12\times 1^{2}+6=18\dfrac {m}{s} $ $v\left( 2\right) =12\times 2^{2}+6=54\dfrac {m}{s}$ $ v\left( 3\right) =12\times 3^{2}+6=114\dfrac {m}{s}$
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