Thomas' Calculus 13th Edition

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

Chapter 3: Derivatives - Section 3.1 - Tangents and the Derivative at a Point - Exercises 3.1 - Page 109: 30


v(10)= 60 ft/sec

Work Step by Step

The height of the rocket t seconds after liftoff is $3t^{2}$ ft. Using kinematics, we know that the velocity of the rocket is $\frac{d}{dt}$($3t^{2}$ ), since the derivative of a position function with respect to time gives the velocity function. Differentiating the position functions gives us v(t)=2*($3t^{2-1}$)= 6t. v(t)=6t. So at t=10 sec, the rocket is climbing at 6(10)=60 ft/sec.
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