Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 87

Answer

1) To prove the formula, - First, write the formula of $a(t)$ as the rate of change of $v(t)$. - Then write the formula of $v(t)$ as the rate of change of $s(t)$. - Finally, use the Chain Rule with $ds$ to modify the first formula, then use the second formula to change it into the formula that needs to be proved. 2) $\frac{dv}{dt}$ means the rate of change of the velocity of the particle according to time. $\frac{dv}{ds}$ means the rate of change of the velocity of the particle according to displacement, or position.

Work Step by Step

1) Show that $$a(t)=v(t)\frac{dv}{ds}$$ We know that the acceleration $a(t)$ of a particle is the rate of change of its velocity $v(t)$. Therefore, $$a(t)=\frac{dv}{dt}\hspace{0.5cm}(1)$$ However, we also know that the velocity $v(t)$ of a particle is the rate of change of its displacement $s(t)$. Therefore, $$v(t)=\frac{ds}{dt}\hspace{0.5cm}(2)$$ So, we would use the Chain Rule to write $(1)$ into $$a(t)=\frac{dv}{ds}\frac{ds}{dt}$$ $$a(t)=\frac{dv}{ds}v(t)$$ The formula is proven. 2) $\frac{dv}{dt}$ means the rate of change of the velocity of the particle according to time. That means, at time $t$, velocity changes at a $a_t$ value, which is the acceleration of the particle at that time $t$ $a(t)$. $\frac{dv}{ds}$ means the rate of change of the velocity of the particle according to displacement, or position. That means, at position $s$, velocity was changing at an $a_s$ value, which is the acceleration of the particle at that position $s$ $a(s)$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.