# Rate of change of volume of cylinder with respect to surface area

The surface area of a cylinder is increasing at a rate of 9π m^2/hr. The height of the cylinder is fixed at 3 meters. At a certain instant, the surface area is 36π m^2. What is the rate of change of the volume of the cylinder at the instant (in cubic meters per hour)

My daughter got stuck and asked me for help. I spent an hour trying to figure it out and I'm stuck too! Thanks!

Here is the answer, thanks to Rishi. Looks like there are a few other ways to solve. Thanks to everyone!!

Formulas:

S = 2πrh + 2πr^2

V = πr^2h

Steps:

dS/dt = 9π <-- given

d/dt (2πrh + 2πr^2) = 9π

dr/dt (2πh + 4πr) = 9π

(2r + 3) dr/dt = 9/2 <-- divide by 2π

Equation:

dV/dt = 2πrh dr/dt = 6πr dr/dt

Solve for r at the instant:

6πr + 2πr^2 = 36π

r = 3

Use r to find dr/dt in above equation:

(2x3 + 3) dr/dt = 9/2

dr/dt = 1/2

Put values into equation:

dV/dt = 6π(3) (1/2)

dV/dt = 9π

• Is it change in curved surface area or total surface area – Rishi Sep 27 '19 at 12:30
• We are given $S_{tot}'(t)$ and $S(t)$ then we need to find $V'(t)$. – user Sep 27 '19 at 12:54

Surface area of a cylinder S is $$2\pi rh+2\pi r^2$$ Now $${ds \over dt}=9\pi$$ or $$\frac{d}{dt}(2\pi rh+2\pi r^2)=9\pi$$ or $${dr \over dt}.(2\pi h+4\pi r)=9\pi$$ or $$(2r+3){dr \over dt}={9 \over 2}$$ Since $$v=\pi r^2h$$ $${dv \over dt}=2\pi rh {dr \over dt}=6\pi r{dr \over dt}$$

Now solve for r since $$6\pi r+ 2\pi r^2=36 \pi$$ and find $${dr \over dt}$$ from above equation put values in last equation and finish it.

From the formulas,

$$V = πr^2h$$

$$S = 2πrh + 2πr^2$$

Eliminate $$r$$ to get the relationship between the surface $$S$$ and the volume $$V$$,

$$S = 2\sqrt{\pi h V}+\frac 2h V$$

Then, take the derivatives,

$$\frac{dS}{dt} = \sqrt{\frac{\pi h}{V}} \frac{dV}{dt}+\frac 2h \frac{dV}{dt}= \left(\frac 1r + \frac 2h \right)\frac{dV}{dt}$$

Thus, the rate of change for the volume is simply,

$$\frac{dV}{dt} = \frac{ 1 }{\frac 1r + \frac 2h}\frac{dS}{dt}$$

where the right-hand-side are all known, i.e. $$h=3m$$, $$\frac{dS}{dt}=9\pi m^2/h$$ and $$r$$ is solved from the given surface area at that instant $$36\pi = 6\pi r +2\pi r^2$$, or

$$r^2+3r-18=0$$

which yields $$r=3m$$. As a result, you should get the volume change rate,

$$\frac{dV}{dt} = \frac{ 9\pi }{\frac 13 + \frac 23}=9\pi \>(m^3/hr)$$

HINT

We have that

• area of tha base: $$A(t)=\pi R^2(t)$$
• lateral surface area: $$S(t)=2\pi R(t)H=6\pi R(t)$$
• total surface area: $$S_{tot}(t)=2\pi R^2(t)+6\pi R(t)\implies S'_{tot}(t)=4\pi R'(t)R(t)+6\pi R'(t)$$

then consider

• volume at time t: $$V(t)=\pi R^2(t)\cdot H \implies V'(t)=6\pi R'(r)R(t)$$

$$\frac {dv}{dt}=\frac{dv}{ds}\times \frac {ds}{dt}$$

You have $$\frac {ds}{dt}$$

You need to find a formula for $$v$$ in terms of $$s$$ and find $$\frac {dv}{ds}$$ to finish the problem.