



 
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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