- #1
ilovemynny
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Homework Statement
A water tank the shape of an inverted circular cone with a base radius of 2m and height of 4m. if water is being pumped into the tank at a rate of 2m^3/min, find the rate at which the water level is rising when the water is 3m deep.
dv/dt = 2m^3/min
h = 3m
r = h/2
I know how to find dh/dt and the answer is dh/dt = 8/9pi m/min
but the next part asks for the rate of change of the radius (dr/dt) at that moment (when h = 3m) and i don't know how to get get this
Homework Equations
V = (pi/3)(r^2)h
The Attempt at a Solution
i was thinking that i would find the derivative of the volume formula so:
V = (pi/3) * r^2 * h
dV/dt = (pi/3) * [r^2 * (dh/dt) + 2rh * (dr/dt)]
and plug in the values:
2m^3/min = (pi/3) * [2^2 * (8/9pi m/min) + 2(2)(3) * (dr/dt)]
and find dr/dt from there
is this right or am i completely wrong?
if i am wrong can someone please help me? I don't know if I'm even using the right variables.