University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.1 - Tangents and the Derivative at a Point - Exercises - Page 118: 30

Answer

The rocket climbs at a speed of $60ft/s$ at $10$ sec after liftoff.

Work Step by Step

The function of the height of the rocket after $t$ seconds of liftoff is $f(t)=3t^2$ ft. The velocity of the rocket at $t=10$ seconds, therefore, is the rate of change of the height $f(t)$ with respect to $t=10$, or in other words, the derivative $f'(10)$, which can be calculated by $$f'(10)=\lim_{h\to0}\frac{f(h+10)-f(10)}{h}=\lim_{h\to0}\frac{3(h+10)^2-(3\times10^2)}{h}$$ $$f'(10)=\lim_{h\to0}\frac{3(h^2+20h+100)-300}{h}=\lim_{h\to0}\frac{3h^2+60h+300-300}{h}$$ $$f'(10)=\lim_{h\to0}\frac{3h^2+60h}{h}=\lim_{h\to0}(3h+60)$$ $$f'(10)=3\times0+60=60(ft/s)$$ The rocket climbs at a speed of $60ft/s$ at $10$ sec after liftoff.
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