Fishfingers52
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A potter is making an open topped vessel shaped as a right circular cylinder of radius r
and height 2r.
(i) Find the rate at which the volume is increasing when the radius is 2 cm and
increasing at a rate of 0.25 cm/s. [5]
(ii) Given that the volume is increasing at a rate of 5 cm
3
/s when the radius is 5 cm,
find the rate at which the surface area is increasing at this point. [6]

Can someone please talk me through this?
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Fishfingers52
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(Original post by arkanm)
The volume of a cylinder is

V=\pi r^2 h

where r is the radius and h the height.

We know that h=2r so

V=\pi r^2 \times 2r=2\pi r^3.

We want the rate at which the volume is increasing, i.e. dV/dt where t is the time.

It is given to us that when r=2, dr/dt=0.25cm/s, or more conveniently, 1/4 cm/s.

Using the chain rule on the formula for V above yields

\frac{dV}{dt}=\frac{dV}{dr} \times \frac{dr}{dt}.

Finding dV/dr is fairly easy, and we already have dr/dt - can you finish it off from here?
I get that. What about the second part?
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Fishfingers52
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(Original post by arkanm)
Can you tell me what you got for the first part? Just checking that you actually understand the hints. If you still haven't wrote it down I recommend doing that first before moving onto part II; I'm sure someone else will be willing to help on the off chance that I don't check back on TSR after this (going to eat now )
6Pi
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brianeverit
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(Original post by Fishfingers52)
I get that. What about the second part?
Write down a formula for surface area, A, in terms of r (don't forget the area of the base)
Then it is almost the same as the first part. You are given dV/dt and want dA/dt
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Fishfingers52
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(Original post by brianeverit)
Write down a formula for surface area, A, in terms of r (don't forget the area of the base)
Then it is almost the same as the first part. You are given dV/dt and want dA/dt
Can you talk me through it please? I need to see what's happening
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brianeverit
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(Original post by Fishfingers52)
Can you talk me through it please? I need to see what's happening
Surface area of cylinder is A=2\pi rh+\pi r^2=5\pi r^2 \mathrm{\ since\ }h=2r
Now find  \frac{dA}{dr}
You are given \frac{dV}{dt}=5 \mathrm{\ when\ }r=5 so evaluate \frac{dA}{dt} and then use \frac{dA}{dt}=\frac{dA}{dr} \times \frac{dr}{dt} \mathrm{\ and\ }\frac{dr}{dt}=\frac{dr}{dV} \times\frac{dV}{dt}

Can you complete it from here?
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Fishfingers52
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(Original post by brianeverit)
Surface area of cylinder is A=2\pi rh+\pi r^2=5\i r^2 \mathrm{\ since\ }h=2r
Now find  \frac{dA}{dr}
You are given \frac{dV}{dt}=5 \mathrm{\ when\ }r=5 so evaluate \frac{dA}{dt} and then use \frac{dA}{dt}=\frac{dA}{dr} \times \frac{dr}{dt} \mathrm{\ and\ }\frac{dr}{dt}=\frac{dr}{dV} \times\frac{dV}{dt}

Can you complete it from here?
From what you said I got 40, which I'm guessing is wrong :L
I just done this again and got 1/5Pi. No idea what I'm meant to be doing
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brianeverit
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(Original post by Fishfingers52)
From what you said I got 40, which I'm guessing is wrong :L
I just done this again and got 1/5Pi. No idea what I'm meant to be doing
A=5\pi r^2 \Rightarrow \frac{dA}{dr}=10\pi r
V=\pi r^2h=2\pi r^3\Rightarrow\frac{dV}{dr}=6\pi r^2\Rightarrow\frac{dr}{dV}= \frac{1}{6\pi r^2}
r=5 \Rightarrow\frac{dr}{dV}=\frac{1  }{150\pi}, \frac{dA}{dr}=50\pi
so \frac{dA}{dt}=\frac{dA}{dr} \times \frac{dr}{dV} \times \frac{dV}{dt}=50\pi \times \frac{1}{150\pi}\times 5=\frac{1}{3}
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Fishfingers52
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(Original post by brianeverit)
A=5\pi r^2 \Rightarrow \frac{dA}{dr}=10\pi r
V=\pi r^2h=2\pi r^3\Rightarrow\frac{dV}{dr}=6\pi r^2\Rightarrow\frac{dr}{dV}= \frac{1}{6\pi r^2}
r=5 \Rightarrow\frac{dr}{dV}=\frac{1  }{150\pi}, \frac{dA}{dr}=50\pi
so \frac{dA}{dt}=\frac{dA}{dr} \times \frac{dr}{dV} \times \frac{dV}{dt}=50\pi \times \frac{1}{150\pi}\times 5=\frac{1}{3}
I think I follow that one a lot better. Thanks
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Mathsislyf
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Doesn't dv /dt = 5pi so the last calculation would be 50pi X 1/150pi X 5pi which equals 5/3pi? I'm not sure if I'm correct but I have this question for homework so it would be much appreciated if someone replied asap
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