Vtyjkyuk

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 $dfrac2pi a H3$.

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:
beginalign
lim_h to 0 fracfrac13pi (r+h)^2H-frac13pi r^2Hh&=fracpi H3left[lim_hto0frac(r+h)^2-r^2hright]\
&=fracpi H3left[lim_hto0fracr^2+2rh+h^2-r^2hright]\
&=fracpi H3left[lim_hto0frac2rh+h^2hright]\
&=fracpi H3left[lim_hto02r+hright]\
&=frac2pi r H3\
endalign
So, when $r=a$, the instantaneous rate of change is
$$frac2pi a H3$$

The difference quotient, between, say, $r= r_1$ and $r= r_2$ is $fracfrac13r_1^2H- frac13r^2Hr_1- r_2= frac13Hfracr_1^2- r_2^2r_1- r^2$.

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 $frac13Hfrac-h(1+ h)-h$.

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.

$$fracdVdr=frac 23pi r H$$

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

$$fracdVdr=frac 23pi a H $$