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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 $V=frac 13pi r^2 H$ 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 $dfrac{2pi a H}{3}$.

Help would be very much appreciated as my test is coming up soon.

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_{hto0}frac{(r+h)^2-r^2}{h}right]\
&=frac{pi H}{3}left[lim_{hto0}frac{r^2+2rh+h^2-r^2}{h}right]\
&=frac{pi H}{3}left[lim_{hto0}frac{2rh+h^2}{h}right]\
&=frac{pi H}{3}left[lim_{hto0}2r+hright]\
&=frac{2pi r H}{3}\
end{align}
So, when $r=a$, the instantaneous rate of change is
$$frac{2pi a H}{3}$$

The difference quotient, between, say, $r= r_1$ and $r= r_2$ is $frac{frac{1}{3}r_1^2H- frac{1}{3}r^2H}{r_1- r_2}= frac{1}{3}Hfrac{r_1^2- r_2^2}{r_1- r^2}$.

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 $frac{1}{3}Hfrac{-h(1+ h)}{-h}$.

The "instantaneous rate of change" is the imit of those as $r_2$ goes to $r_1$ or as h goes to 0.

$$V=frac 13pi r^2 H$$

when $H$ is constant it means you are partially differentiating w.r.t. $r,$ leaving $H$ alone, i.e., as a constant.

$$frac{dV}{dr}=frac 23pi r H$$

At $r=a$ the d.c. evaluates to

$$frac{dV}{dr}=frac 23pi a H $$

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,

$$frac{delta V}{delta r}approxfrac23pi aH=frac{d V}{d r},$$

where the notation with $d$ indicates that we only want the term that does not vary with $delta r$.