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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

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

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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,

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

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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: $V(t)=\pi R^2(t)\cdot H \implies V'(t)=6\pi R'(r)R(t)$
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$$\frac {dv}{dt}=\frac{dv}{ds}\times \frac {ds}{dt}$$

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.

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