sinclairharry
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- Joined
- Mar 22, 2015
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- 17
The question is an inverted conical tank with a height 14 ft and radius 7ft is drained through a hole in the vertex at a rate of 8ft^3/second.
What is the change of rate of the water depth when the water is 2 ft ?
so dv/dt =-8ft^3/second
similar triangles show us that 7/14=r/h meaning r =1/2*h
Volume of a cone is v=1/3*pi*r^2*h
so dv/dt = 1/12pi(1/2h)^2*h * dh/dt
Sub in 2 as h
-8 = 1/12pi(2)^3 * dh/dt
dh/dt=-12/pi = -3.82
Is that correct ? Or have I done something wrong in the question stage?
What is the change of rate of the water depth when the water is 2 ft ?
so dv/dt =-8ft^3/second
similar triangles show us that 7/14=r/h meaning r =1/2*h
Volume of a cone is v=1/3*pi*r^2*h
so dv/dt = 1/12pi(1/2h)^2*h * dh/dt
Sub in 2 as h
-8 = 1/12pi(2)^3 * dh/dt
dh/dt=-12/pi = -3.82
Is that correct ? Or have I done something wrong in the question stage?