Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.9 - Related Rates - 3.9 Exercises - Page 250: 27


The water level is rising at a rate of $\frac{10}{3}~cm/min$

Work Step by Step

Let $b$ be the distance across the bottom of the trough. Then $b = 0.3~m$ We can use similar triangles to write an expression for the distance $B$ across the top of the water in terms of the water level $h$: $B = 0.3+h$ We can differentiate both sides of the equation for the water volume with respect to $t$: $V = \frac{h}{2} (B+b)~L$ $V = \frac{h}{2} (0.3+h+0.3)~(10)$ $V = 5h~ (h+0.6)$ $V = 5h^2+3h$ $\frac{dV}{dt} = (10h + 3)~\frac{dh}{dt}$ $\frac{dh}{dt} = (\frac{1}{10h+3})(\frac{dV}{dt})$ $\frac{dh}{dt} = [\frac{1}{(10)(0.3)+3}](0.2)$ $\frac{dh}{dt} = \frac{0.2}{6}~m/min$ $\frac{dh}{dt} = \frac{10}{3}~cm/min$ The water level is rising at a rate of $\frac{10}{3}~cm/min$
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.