Instantaneous rate of change of the volume of a cone with respect to the radius, if the height is fixed 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.
 A: 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 
$$\frac{2\pi a H}{3}$$
A: The difference quotient, between, say, $r= r_1$ and $r= r_2$ is $\frac{\frac{1}{3}r_1^2H- \frac{1}{3}r^2H}{r_1- r_2}= \frac{1}{3}H\frac{r_1^2- r_2^2}{r_1- r^2}$.
Some people prefer to use $r_1= 1$ and $r_2= r+ h$.  In that case, $r_1^2- r_2^2= r_1^2- r_1^2- 2r_1h- h^2= -h(r_1+ h)$ and $r_1- r_2= r_1- r_1- h$= -h so the difference quotient is $\frac{1}{3}H\frac{-h(1+ h)}{-h}$.
The "instantaneous rate of change" is the imit of those as $r_2$ goes to $r_1$ or as h goes to 0.
A: $$V=\frac 13\pi r^2 H$$
when $H$ is constant it means you are partially differentiating w.r.t. $r,$ leaving $H$ alone, i.e., as a constant.
$$\frac{dV}{dr}=\frac 23\pi r H$$
At $r=a$ the d.c. evaluates to
$$\frac{dV}{dr}=\frac 23\pi a H $$
A: For simplify of the notation, I will consider a "reduced volume" defined as follows:
$$v:=\frac{3V}{\pi H}$$ so that $$v=r^2.$$ (Any computation about $V$ amounts to one about $v$, to a constant factor.)
Now you give the radius a tiny variation, let $\delta r$, and see how the reduced volume changes.
We have
$$\delta v=(a+\delta r)^2-a^2=(2a+\delta r)\delta r.$$
Because of the factor $\delta r$, the variation of the volume is also tiny and this is not so informative. This is why we want a rate of variation, i.e. the ratio of the volume variation over the radius variation,
$$\frac{\delta v}{\delta r}=2a+\delta r.$$
Now, you understand that when $\delta r$ is made smaller and smaller, the first term stays constant, but the second vanishes. The rate is considered instantaneous when this second term can be completely neglected.
Finally,
$$\frac{\delta V}{\delta r}\approx\frac23\pi aH=\frac{d V}{d r},$$
where the notation with $d$ indicates that we only want the term that does not vary with $\delta r$.
