I am in the midst of confusion in regards of this question:
I have to find an expression for the differential dV, and hence ${\frac{dV}{dt}}$.
The ${\frac{dV}{dt}}$ part I know how to solve (shown at the bottom). But I don't know how to find the expression for dV.
The question is as follows:
The base radius r (mm) of a right circular cone increases at 40mm/s and its height h (mm) increases at 50mm/s. Given that the volume of such a cone is $$V = 1/3 \pi r^2 h$$ Find an expression for the differential dV, and hence ${\frac{dV}{dt}}$.
This is what I have gotten so far: $${\frac{dV}{dt}} = \frac{\partial V}{\partial r}{\frac{dr}{dt} + \frac{\partial V}{\partial h}{\frac{dh}{dt}}}$$
How do I find the expression for dV?
Help is greatly appreciated. Many thanks!
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$\begingroup$It sounds like you are looking for the differential of $V$. I believe the expression $dV$ you are looking for is: $dV = \frac {\partial V} {\partial r} dr + \frac {\partial V} {\partial h} dh = (2/3 \pi rh)dr + (1/3 \pi r^2)dh$.
The next step would be evaluating $\frac {dV} {dt} = (2/3 \pi rh) \frac {dr} {dt} + (1/3 \pi r^2) \frac {dh} {dt}$ by setting $\frac {dr} {dt} = 40$ and $\frac {dh} {dt} = 50$. Now you can plug in any r and h to find out the instantaneous rate of change of the volume at that particular r and h.
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