Question

If Q is a quantity that varies with time. then the derivative $\frac{dQ}{dt}$ gives the rate of change of that quantity with respect to time. A conical paper cup $20$ cm across the top and $15$ cm deep is full of water. The cup springs a leak at the bottom and loses water at $5$ cu cm per minute. $\frac{d^{2}h}{d{t}^2}$ when the water is exactly 7.5 cm deep and $\frac{d^{2}V}{d{t}^2}=-\frac{4}{9}$ cu cm/min${}^2$ is (in cm/min${}^2$)

• A. $-\frac{2}{5}$
• B. $-\frac{2}{125{\pi }^3}$
• C. $-\frac{2}{5{\pi }^3}$
• D. none of these

Answer 1

Answer: (D)

none of these

If V is the volume of water and h, r are height and radius at time t
then $\frac{r}{h}=\frac{2}{3}$, $V=\frac{1}{3}\pi r^{2}h$
So $v=\frac{1}{3}\pi {\left(\frac{2h}{3}\right)}^2$ $h=\frac{4}{27}\pi h^3$
$\frac{dV}{dt}=\frac{4}{9}\pi h^2\frac{dh}{dt}$.
Putting $h=7.5$ and $\frac{dV}{dt}=-5$, we have $\frac{dh}{dt}=-\frac{1}{5\pi }$
Now
$\frac{d^{2}V}{d{t}^2}=\frac{4}{9}\pi [h^2\frac{d^{2}h}{d{t}^2}+2h{\left(\frac{dh}{dt}\right)}^2]$
Putting $\frac{d^{2}V}{d{t}^2}=-\frac{4}{9}$ and $\frac{dh}{dt}=-\frac{1}{5\pi }$, $h=7.5$,
we get $\frac{d^{2}h}{d{t}^2}=-\frac{4(5\pi +3)}{1125{\pi }^2}$