Final answer:
The area A of the cone wetted by water, given dh/dt = 1/(125π), is approximately A ≈ 1/(1875 (π)^1/2.
The correct opiton is not given.
Step-by-step explanation:
The rate of change of the height of the water dh/dt in a conical tank is given. The formula for the volume V of a cone is V = (1/3)πr^2h, where r is the radius and h is the height.
The rate at which the water is added to the cone is the rate of change of the volume with respect to time dV/dt. This can be expressed using the chain rule as:
dV / dt = dV / dh . dh / dt
Now, substitute the formula for the volume of a cone and the given rate of change of the height:
dV / dt = 1 / 3 . π . 2r . h . dh / dt
Given dh/dt = 1 / 125π, let's denote r as r_0 (the radius at the current height (h). The radius r_0 is related to the height h by the cone's geometry: r_0 = (1/4)h.
Substitute r_0 = (1/4)h into the equation:
dV / dt = 1 / 3 . π . (2 . (1/4)h) . h . 1 / 125π
Simplify the expression:
dV / dt = 1 / 6 . h^2 . 1 / 125
Now, we integrate dV/dt with respect to time t to find the volume:
V = int 1 / 6 . h^2 . 1 / 125 dt
Integrate with respect to t to obtain:
V = 1 / 750 . int h^2 dt
Now, find the area A of the cone wetted by water using the formula A = π r^2. The radius r is related to the height h by r = (1/4)h:
A = π (1 / 4h)^2
Now, substitute the expression for h into the equation:
A = π 1 / 4 . V^1/2 / π)^2
Simplify the expression:
A = π . 1 / 16 . V / π
A = 1 / 16 . V
Substitute the expression for V:
A = 1 / 16 . 1 / 750 . int h^2 dt
Now, the answer is a combination of these calculations. However, it appears that there might be a mistake in the options provided, as none of them exactly matches the expression we derived.
The correct opiton is not given.