Calculus 1 Related Rates Problem....

[HWilliams44]

Hi! So, today I started on some homework problems and ran across a related rates problem. I'm not sure if I did it correctly or not. Can someone check my work for me, and tell me if I did anything incorrectly?

I've attached a scan of the homework problem. My work is on the lined paper at the bottom of the page, and the problem itself is at the top. Thanks!

 

[wjm11]

Hi! So, today I started on some homework problems and ran across a related rates problem. I'm not sure if I did it correctly or not. Can someone check my work for me, and tell me if I did anything incorrectly?

I've attached a scan of the homework problem. My work is on the lined paper at the bottom of the page, and the problem itself is at the top. Thanks!

V = (1/3)(pi)(r^2)(h)
h = r/2
V = (1/3)(pi)(r^2)(r/2)
V = (1/6)(pi)(r^3)
dV/dt = (1/2)(pi)(r^2)(dr/dt)
(1/2)(pi)(10cm)^2(dr/dt) = 6 cm^3/min
[FONT=&quot]etc.

Please double-check my work for accuracy.
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[HallsofIvy]

Your mistake is that you differentiated with respect to r, treating h as if it were independent of r, then replaced h by r/2 (so h is NOT indpendent of r).

Either
use the chain rule: \(\displaystyle \frac{dV}{dt}= \frac{1}{3}\pi (2r\frac{dr}{dt}h+ r^2\frac{dh}{dt})\).
You are given that r= 2h so h= r/2 and \(\displaystyle \frac{dh}{dt}= \frac{1}{2}\frac{dr}{dt}\).
Or (simpler)
replace h by r/2 to begin with: \(\displaystyle V= \frac{1}{3}\pi r^2h= \frac{1}{6}\pi r^3\)
so that \(\displaystyle \frac{dV}{dt}= \frac{1}{2}\pi r^2 \frac{dr}{dt}\).

 

[HWilliams44]

Thank you all so much! ^.^