A) A water trough is 6 m long and has a cross-section in the shape of an isoscel

Question

A) A water trough is 6 m long and has a cross-section in the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 90 cm wide at the top, and has height 60 cm. If the trough is being filled with water at the rate of 0.1 m3/min how fast is the water level rising when the water is 10 cm deep?

B)A trough is 16 ft long and its ends have the shape of isosceles triangles that are 5 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 15 ft3/min, how fast is the water level rising when the water is 4 inches deep?

Explanation / Answer

Consider that at a general time t min, the depth of the water in the trough is h cm.

Let a volume dV cm^3 of water fall into the trough in dt minutes after that, causing a change in depth of water = dh cm.

Then dV = A(h)*dh ...........(1)
where A(h) is the area of the water surface as a function of the depth of water (h).

Now we need to find A(h).

Draw the trapezoidal cross-section of the trough.
Draw a line parallel to the base between the base and the top edges at a distance h from the base (this marks the water level as seen from the side). Using some properties of similar triangles, we find that the length of this line is (h+20) cm.
Now the area of the water surface = length of this line * length of the trough
= 600 * (h+20) cm^2
So
A(h) = 600 * (h+20) cm^2

Substituting in (1):

dV = 600 * (h+20) dh cm^3

Divide thoughout by dt

dV/dt = 600 * (h+20) dh/dt

dV/dt is the rate at which trough is being filled = 0.3m^3/min = 300000 cm^3/min

Rate at which water level is rising = dh/dt = 300000 / (600*(h+20)) cm/min
= 5/(h+20) m/min

Put h=30

dh/dt = 0.1 m/min

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