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Science Forum Index » Mathematics Forum » Rate of Change - Derivative Question
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Message |
| Ryan |
 Posted: Thu May 01, 2008 4:48 pm |
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Guest
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Hi all -
I'm stuck on one of the steps to solving this problem. I've already
taken calculus, and I am trying to refresh my memory on a few
concepts. I'd really appreciate any help!...
Problem:
A water tank has a shape of a cone with height 2m and radius 2/3 m. If
water is pumped into the tank at a rate of 3m^3/hr, how fast is the
water level rising at the instant when the tank begins to overflow?
Steps:
* Let y = depth at time t (distance from the vertex of the cone to the
surface of the water)
* The dv/dt = 3m^3/hr b/c water is pumped into the tank at this
constant rate.
* The speed at which the water level is rising is the rate of change
of y with respect to t (dy/dt)
* We want to find dy/dt at the instant that y = 2 b/c that's when it
will start to overflow
* Cone Volume (V) = (pi/3)(r^2)y
* r/y = 2/3/2 or r = 1/3y
* so V = (pi/3)(y/3)^2y or (pi/27)y^3
*Take derviative of both sides of equation with respect to t
("implicit differentiation" b/c t isn't in the relationship):
dv/dt = (pi/9)(y^2)(dy/dt)
In words, I took the derivative of v with respect to y times
derivative of y with respect to t. It balances out: dv/dt = dv/dy*dy/
dt. I understand how the power rule is being used here, but I'm having
a hard time understanding the chain rule. Could someone explain in
words the inside function and outside function?
Thanks,
Ryan |
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