An inverted conical water tank with a height of 10 ft and a radius of 5 ft is drained through a hole in the vertex at a rate of 5 ft^3/s. What is the rate of change of the water depth when the water is 4 ft? (Hint: Use similar triangles.)
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Let h = height of water at time t
Let r = radius at top of water at time t
By similar triangles
r/h = 5/10
r = h/2
V = 1/3 πr²h
V = 1/3 π (h/2)² h
V = 1/12 π h³
Differentiate both sides with respect to t:
dV/dt = 1/4 π h² dh/dt
Now we plug in known values:
Water is draining at a rate of 5 ft³/s -----> dV/dt = −5
Water depth is 4 ft -----> h = 4
−5 = 1/4 π 4² dh/dt
−5 = 4π dh/dt
dh/dt = −5/(4π) = −0.397887358
Depth is decreasing (since dh/dt < 0) at a rate of about 0.4 ft/s
Let r = radius at top of water at time t
By similar triangles
r/h = 5/10
r = h/2
V = 1/3 πr²h
V = 1/3 π (h/2)² h
V = 1/12 π h³
Differentiate both sides with respect to t:
dV/dt = 1/4 π h² dh/dt
Now we plug in known values:
Water is draining at a rate of 5 ft³/s -----> dV/dt = −5
Water depth is 4 ft -----> h = 4
−5 = 1/4 π 4² dh/dt
−5 = 4π dh/dt
dh/dt = −5/(4π) = −0.397887358
Depth is decreasing (since dh/dt < 0) at a rate of about 0.4 ft/s