calculus
My teacher has not taught us derivatives yet, so I need to solve this without their use.
The problem states: "Find the instantaneous rate of change of the volume $V=\frac 13\pi r^2 H$ of a cone with respect to the radius $r$ at $r=a$ if the height $H$ does not change."
I've already rearranged the problem into difference quotient form. According to the book and a calculator I found online the answer is $\dfrac{2\pi a H}{3}$.
Help would be very much appreciated as my test is coming up soon.
hint: $\dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt}$, and you are given: $\dfrac{dV}{dt} = 4, r = 2$, can you find $\dfrac{dr}{dt}$ ?
Below is a (not-to-scale and I-totally-didn't-misread-the-question-at-first-as-6-being-the-diameter-instead-of-the-radius) picture of the cone and cylinder in question (behold my abysmal drawing skills in Microsoft Paint thanks to my expired license on Mathematica!)
In any case, we can let the radius of the inscribed cylinder be $x$. That must therefore mean that the height of the cylinder is $4 - \frac{4}{6}x$. This makes the volume of the cylinder
$$V = \pi x^2 \left(4 - \frac{2}{3} x\right) = \pi \left(4x^2 - \frac{2}{3} x^3\right)$$
To find the maximum volume of the cylinder we take the derivative of $V$ with respect to $x$ and set it to $0$:
$$\frac{dV}{dx} = \pi \left(8x - 2x^2\right) = 0 \Rightarrow 8x - 2x^2 = 2x(4-x) = 0$$
$x = 0$ leads to a trivial case, which leaves us with $x = 4$. This means that the height is $4 - \frac{2}{3} \cdot 4 = \frac{4}{3}$, and the volume is therefore $$V = \pi \cdot 4^2 \cdot \frac{4}{3} = \frac{64}{3}\pi$$
Best Answer
The instantaneous rate of change with respect to $r$ is equal to the following: \begin{align} \lim_{h \to 0} \frac{\frac{1}{3}\pi (r+h)^2H-\frac{1}{3}\pi r^2H}{h}&=\frac{\pi H}{3}\left[\lim_{h\to0}\frac{(r+h)^2-r^2}{h}\right]\\ &=\frac{\pi H}{3}\left[\lim_{h\to0}\frac{r^2+2rh+h^2-r^2}{h}\right]\\ &=\frac{\pi H}{3}\left[\lim_{h\to0}\frac{2rh+h^2}{h}\right]\\ &=\frac{\pi H}{3}\left[\lim_{h\to0}2r+h\right]\\ &=\frac{2\pi r H}{3}\\\ \end{align} So, when $r=a$, the instantaneous rate of change is $$\frac{2\pi a H}{3}$$