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| Math Central | Quandaries & Queries |
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Question from Ishaak, a student: A hemispherical bowl is filled with water at a uniform rate. When the height of water is h cm the volume is π(rh^2-1/3 h^3 )cm^3, where ri s the radius. Find the rate at which the water level is rising when it is half way to the top, given that r = 6 and the bowl fills in 1 minute. |
Hi Ishaak,
You know that the water is flowing in at a constant rate, but you need to know that rate. You can use the expression given to you,
\[V = \pi \left(r h^2 - \frac13 h^3 \right) \mbox{ cm}^3\]
to find the rate. You know the radius is 6 cm and when the bowl is full $h = \frac{r}{2} = 3$ cm. What is the volume? It takes 1 minute to fill the bowl so at what rate (in $\mbox{ cm}^3$ per minute) is the water flowing in?
In the expression
\[V = \pi \left(r h^2 - \frac13 h^3 \right) \mbox{ cm}^3\]
$r$ is 6, so substitute $r = 6$ into the expression. $V$ and $h$ are both functions of time $t$ and hence you can differentiate both sides with respect to $t.$ Now you have an equation with $\frac{dV}{dt}$ on the left and an expression containing $h$ and $\frac{dh}{dt}$ on the right. When the bowl is half full, $h = \frac{3}{2}\mbox{ cm},$ you know $\frac{dV}{dt}$ and hence you can solve for $\frac{dh}{dt}.$
Penny
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