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Water is flowing out at the rate of $6m^3$/min from a reservoir shaped like a hemispherical bowl of radius $R=13m.$The volume of the water in the hemispherical bowl is given by $V=\frac{\pi}{3}y^2(3R-y)$ when the water is $y$ m deep.Find
$(a)$At what rate is the water level changing when the water is $8m$ deep.
$(b)$At what rate is the radius of the water surface changing when the water is $8m$ deep.

$(a)V=\frac{\pi}{3}y^2(3R-y)$

$\frac{dV}{dt}=\frac{\pi}{3}[y^2\frac{d}{dt}(3R-y)+(3R-y)\frac{d}{dt}y^2]$

$\frac{dV}{dt}=\frac{\pi}{3}[y^2(-1)\frac{dy}{dt}+(3R-y)2y\frac{dy}{dt}]$

$6=\frac{\pi}{3}[8^2(-1)\frac{dy}{dt}+(3\times 13-8)(16)\frac{dy}{dt}]$

$\frac{dy}{dt}=\frac{1}{24\pi}$
This answer is correct but i am not able to solve the second part.

Brahmagupta
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1 Answers1

3

Since your hemisphere is the lower half of a circle, you can express the radius as the x-coordinate (see image added in edit below),

$$ R^2 = x^2 + y^2 \implies x = \sqrt{R^2 - y^2} $$

now differentiate to find $\frac{dx}{dt}$,

$$ \frac{dx}{dt} = \frac{1}{2} (R^2-y^2)^{-\frac{1}{2}} (-2y) \frac{dy}{dt} = \frac{-y}{\sqrt{R^2-y^2}} \frac{dy}{dt} $$

and you know $y=8$, $R=13$, and $\frac{dy}{dt}=\frac{1}{24\pi}$, so you should be able to plug these values in and get $\frac{dx}{dt} = \frac{-1}{3\pi\sqrt{105}} \approx -0.0103546$

Edit:

graph of hemisphere

Dando18
  • 5,368
  • Doesn't depth refer to the height of water from the bottom ? Here y should be 5 since you took the height from the top of the container. CMIIW – Qui Gonn Jinn Aug 31 '21 at 14:36