We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
cookie policy and privacy policy.

Rate of change of Volume of Square Pyramid at a time \(t_0\)?

- side of base increasing by a constant 6

- height decreasing by constant -1

- at time 0 the base is 3 and the height is 9

\(V_{\textrm{sq pyramid}} =\frac{1}{3}s^2h\) Volume formula

\(\frac{dV_{\textrm{sq pyramid}}}{dt} = \frac{1}{3}2s\frac{ds}{dt}\frac{dh}{dt}\) Differentiate with respect to time

\(\frac{dV}{dt}=\frac{1}{3}2(3)(6)(-1)\) Plugging

\(\frac{dV}{dt}=-12\)

My answer choices are

-105

-111

111

105

The units for space and time do not change between question and answer nor one number given to the next. Also this is not that type of problem where you are given one rate like surface area to start and have to work backward to volume so with the information given I really do not know what to do. I keep getting -12

:(

Thanks!

Guest Mar 10, 2019