My answer isn't correct, but I'm not sure why. Your help is much appreciated!
Coffee is poured at a uniform rate of $2 \text{ cc}/ \text{sec} $ into a cup whose bowl is shaped like a truncated cone. If the upper and lower radii of the cup are $4 \text{ cm}$ and $2 \text{ cm}$ (respectively) and the height of the cup is $6 \text{ cm}$, how fast will the coffee level be rising when the coffee is halfway up?
The formula for the volume is $$V=\frac {h\pi}3(R^2+Rr+r^2)$$ where $r<R$. The lower radius remains constant (it is $2 \text{ cm}$) and the upper radius and height varies as more coffee is poured in. I used similar figures to establish the relationship $R=\frac 23h$.
$$V=\frac {h\pi}3(\frac {4h^2}9 + \frac {4h}3 + 4)$$
Substituting in the expression $\dfrac 23h$ for $R$ and the value $2 \text{ cm}$ for $r$.
Multiply $\dfrac {h\pi}3$ into the parenthesis to avoid a messy product rule, then take the derivative of both sides with respect to time.
$$\frac {dV}{dt}=\frac {dh}{dt}(\frac {4\pi h^2}9 +\frac {8\pi h}9+\frac {4\pi}3)$$
Since we want to know how fast the coffee level is rising when the coffee is halfway up, I'll replace the value $3 \text{cm}$ for $h$. $\dfrac {dV}{dt}$ is $2$. Then, I solve for $\dfrac {dh}{dt}$ and get $\dfrac 1{4\pi}$.
Where did I go wrong? The correct answer is $\dfrac 2{9\pi}$.
Best Answer
I think the only error is your expression $R = \frac{2}{3}h$. This suggests that when $h = 0$ (that is, when the cup is empty) the upper radius is zero, when it should be $2$. Instead, observe that as $h$ varies from $0$ to $6$, $R$ rises from $2$ to $4$; that's a slope of only $\frac{1}{3}$. So $R = \frac{1}{3}h + 2$ should give you the correct answer.