Related Rates Question (1 Viewer)

Web Addict · Jan 15, 2013

Hi.

The radius of the base of a cone is increasing at the rate of 2% per second. At what percentage is the height diminishing if the volume is kept constant?

Can someone help me do this question please?

Thank you.

asianese · Jan 15, 2013

This might not be right but oh shwell

$\frac{dr}{dt}=1.02 \,u/s\\V=\frac{1}{3}\pi r^2 h \Rightarrow h=\frac{3V}{\pi}\left(\frac{1}{r^2}\right) \text{ since the volume is a constant, the height becomes a function of the radius}\\ \frac{dh}{dr}=\frac{3V}{\pi}\left(\frac{-2}{r^3}\right)\\ \text{Now using the chain rule: } \frac{dh}{dt}=\frac{dh}{dr}\cdot \frac{dr}{dt}\\=-\frac{6V}{\pi r^3}\times 1.02\\=-6.12\frac{V}{\pi r^3} \, u/s$

Web Addict · Jan 15, 2013

Thanks for the help. I got the same answer as you, but the answers says its 4% somehow.

braintic · Jan 15, 2013

I'm afraid that answer is not correct. Your initial statement should be
$\frac {dr} {dt} = 0.02r$
This statement says correctly that the rate of change of the radius is 2% of the radius itself.

When you sub this into your (correct) derivative you get
$\frac {dh} {dr}=-\frac{0.12V}{\pi r^2}$

However our answer is also supposed to be a percentage, so we need to express it as a multiple of h. So we split the answer so that our expression for h appears:
$\frac {dh}{dt}=\frac{3V}{\pi r^2} \times -0.04$

$= -0.04h$

Which says that the height is decreasing at 4% per second.

Related Rates Question (1 Viewer)

Web Addict

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asianese

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braintic

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