Related Rates - Volume of cylinder

JimmysJohnson

New member
Joined
Apr 18, 2020
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4
G'day again.
I was wondering if i could get some help on my understanding of this question.
"A cylinders radius is decreasing at the rate of 2cm/sec while it height is increasing at a rate of 3cm/sec. How fast is the cylinders volume changing when its radius is 10cm and its height 18cm?"

Since we know V = (pi)r2h
I thought that we could just use implicit differentiation to find the rate of change in respect to time. But in doing so i imagined that we'd simply use the product rule for the three terms (im not sure if this is where the flaw happens, but after a google search it seems as though it is possible).
After which we'll get
dV/dt = (r2h)+((pi)(2r)(dr/dt)(h))+((pi)(r2)(dh/dt))
However when i sub in the respective points to solve for the rate of change of volume, i get very far off from the actual equation.
I understand that in order to solve the equation properly we take pi out of the equation
pi(r2h) and to go from there. I'm just not sure why my method doesn't work since its probably the methodology i'd instantly go for. Is it because pi is a constant and as such we can't differentiate it?
Thank you for your time and any/all help.
 

MarkFL

Super Moderator
Staff member
Joined
Nov 24, 2012
Messages
2,954
I would begin with:

\(\displaystyle V=\pi r^2h\)

And now differentiate with respect to time \(t\):

\(\displaystyle \frac{dV}{dt}=\pi\left(2rh\frac{dr}{dt}+r^2\frac{dh}{dt}\right)\)

And now plug in the given data. If you try to include the constant \(\pi\) as a function to be differentiated, you must use:

\(\displaystyle \frac{d}{dt}(\pi)=0\)
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
6,293
You have "dV/dt = (r2h)+((pi)(2r)(dr/dt)(h))+((pi)(r2)(dh/dt))"
(It would be better to write "r^2" rather than "r2" to make the exponent clearer.)
Where did the "r^2h" come from? Was that the derivative of \(\displaystyle \pi\)? If so why isn't there a "\(\displaystyle (d\pi/dt)\)"?

You ask "Is it because pi is a constant and as such we can't differentiate it?"
Of course we can differentiate a constant! One of the first things you learned in Calculus I is that the derivative of any constant is 0!
 
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