A feeding trough full of water is 5 ft long and its ends are isosceles triangles having a base and height of 3 ft. Water leaks out of the tank at a rate of 5 (ft)^3/min. How fast is the water level falling when the water in the tank is 6 in. deep?

Question

A feeding trough full of water is 5 ft long and its ends are isosceles triangles having a base and height of 3 ft.  Water leaks out of the tank at a rate of 5 (ft)^3/min.  How fast is the water level falling when the water in the tank is 6 in. deep?

Ezra Adams · Accepted Answer

falling at a rate of 2ft /min

If you consider the generalised position when the depth of the water in the trough is at x where #x = x(t). Due to the similar triangles, the "width" of the water channel will also be x, as indicated in the drawing

We can say that the volume in the tank at that time, #V(t)#, is:

#V = 1/2 * x * x * l# where l is the length of the trough [ie 5 ft]

so #V = 5/2 x^2#

taking the derivative wrt time

#dot V = 5/2* 2 x * dot x = 5 x dot x# [where the dot is d/dt]

so # dot x = (dot V) / (5 x)#

we know that #dot V = - 5 (ft^3) /min# and we are asked about #dot x# at #x = 1/2 ft#

so so # dot x = (- 5 (ft^3) /min) / (5 ft * 1/2 ft) = - 2 (ft) /min#

so it's falling at a rate of 2ft /min at that moment

Isabella Ackerman · Answer

undefined To find the rate at which the water level is falling, you need to use related rates. The volume of water in the trough can be calculated as the product of its length and the area of its cross-section, which is a triangle.

Given:
Length of trough (L) = 5 ft
Base of triangle (b) = 3 ft
Height of triangle (h) = 3 ft
Rate of water leaking (dV/dt) = 5 (ft)^3/min

Volume of water in trough (V) = (1/2) * base * height * length

Using the formula for the volume, differentiate both sides with respect to time (t) to find the rate at which the volume is changing:

dV/dt = (1/2) * (dh/dt) * b * L + (1/2) * h * db/dt * L + (1/2) * h * b * dL/dt

At the given condition, when the depth of water (h) is 6 inches (0.5 ft), and since the trough is full, we can assume that:

h = 0.5 ft

Now, plug in the given values and solve for dh/dt:

5 = (1/2) * (dh/dt) * 3 * 5 + (1/2) * 3 * 3 * db/dt + (1/2) * 3 * 3 * 5

Solve for db/dt:

5 = (15/2) * (dh/dt) + (9/2) * db/dt + 22.5

db/dt = (5 - (15/2) * (dh/dt) - 22.5) / (9/2)

Now, substitute the value of h = 0.5 ft and solve for dh/dt:

5 = (15/2) * (dh/dt) + (9/2) * (db/dt) + 22.5

dh/dt = (5 - (9/2) * (db/dt) - 22.5) / (15/2)

Calculate db/dt using the value of dh/dt obtained from the second equation.

Finally, find the value of dh/dt when h = 0.5 ft.

This will give you the rate at which the water level is falling when the water in the trough is 6 inches deep.