related rate problem?

Question

Water is released from a conical tank 50 inches tall and 30 inches in radius, and falls into a rectangular tank whose base has an area of 400 square inches. The rate of release is controlled so that when the height of the water in the conical tank is x inches, the height is decreasing at the rate of 50-x inches per minute. How fast is the water level in the rectangular tank rising when the height of the water in the conical tank is 10 inches

Answer 1

V cone = ⅓πr²h
h/r = 50/30 so h = 5r/3 and r = 3h/5 so when h = x, r = 3x/5
Thus V = ⅓π(3x/5)²x
= 3/25 * πx³
When h = x, dx/dt = (50 - x)
So dV/dt = dV/dx * dx/dt
= 9/25 * πx² * (50 - x)

For the rectangular tank

V = Ah
= 400h
dV/dt = dV/dh * dh/dt
9/25 * πx² * (50 - x) = 400 * dh/dt

So dh/dt = 9/10000 * πx² * (50 - x)
when x = 10
dh/dt = 9/10000 * π * 10² * (50 - 10)
= 9/10000 * π *100 * 40
= 18π/5 in/min

Answer 2

Assume the vertex of the cone is at the bottom.
Then r = 30x/50 in.
V = (1/3)πr²x
V = (1/3)π(3x/5)²x
V = 3πx³/25
V2 = 400h
dV2/dt = (dV2/dh)(dh/dt
dV2/dt = 400 dh/dt = -dV/dt
dx/dt = x - 50
dV/dt = (dV/dx)(dx/dt
dV/dt = 3πx²(x - 50) = -400dh/dt
dh/dt = 3πx²(50 - x)/(25*400)
dh/dt = 300π(40)/(25*400) = @ x = 10
dh/dt = 6π/5
dh/dt = 3.77 in/min

Answer 3

related rates