A reservoir is in the form og the frustum of a cone with upper base of radius 8ft and lower base radius of 4 ft and altitude of 10ft. The water in the reservoir is xft deep. If the level of the water is increasing at 4ft/min., how fast is the volume of water in the reservoir is increasing when its depth is 2ft.?

Answer: 100pi ft^3/min..

Help please, i need to see the solution...

100pi ft^3/min

Solution pls

To find the rate at which the volume of water in the reservoir is increasing, we need to find the derivative of the volume of water with respect to time.

Step 1: Find the formula for the volume of a frustum of a cone. The formula is:

V = (1/3)πh(r₁² + r₁r₂ + r₂²)

where V is the volume, h is the height (depth) of the frustum, r₁ is the radius of the upper base, and r₂ is the radius of the lower base.

Step 2: Determine the values given in the problem. We are given that the upper base radius (r₁) is 8 ft, the lower base radius (r₂) is 4 ft, and the water depth (h) is x ft.

Step 3: Substitute the values into the formula. The volume (V) becomes:

V = (1/3)πx(8² + 8*4 + 4²)
= (1/3)πx(64 + 32 + 16)
= (1/3)πx(112)
= (112/3)πx

Step 4: Take the derivative of the volume with respect to time (t), using the chain rule. The derivative of the volume (dV/dt) becomes:

dV/dt = (112/3)π(dx/dt)

Step 5: We are given that the water level is increasing at a rate of 4 ft/min, which means dx/dt = 4 ft/min. Substitute this into the derivative equation:

dV/dt = (112/3)π(4)
= (448/3)π
= 448π/3 ft³/min

Step 6: Finally, we need to find the rate at which the volume of water in the reservoir is increasing when the depth is 2 ft. Substitute x = 2 ft into the derivative equation:

dV/dt = 448π/3 ft³/min

Therefore, the rate at which the volume of water in the reservoir is increasing when its depth is 2 ft is approximately 100π ft³/min.

when the water is x feet deep, the surface has radius 4+(8-4)x/10 = 4+x/5

The volume of the cut-off part of the cone is 1/3 pi *16*10 = 160/3 pi

So, the volume of water is
v = 1/3 pi (4+x/5)^2 (x+10) - 160pi/3
= pi/75 x^3 + 2pi/3 x^2 + 32pi/3 x

dv/dt = (pi/25 x^2 + 4pi/3 x + 32pi/3) dx/dt

when x=2,
dv/dt = (4pi/25 + 8pi/3 + 32pi/3)(4)
= 4048/75 pi