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a trough is 12 feet long and 3 feet across the top (see figure). its ends are isosceles triangles with altitudes of 3 feet. if water is being pumped into the trough at 2 cubic feet per minute, how fast is the water level rising (in feet per minute) when \displaystyle hh is 1 foot deep?

Sagot :

If water is being pumped into the trough at 2 cubic feet per minute, then the water level rising (in feet per minute) when h is 1.6 foot deep is 0.1 feet cubic per minute.

Suppose that we have a function y = f(x), and wish to find the rate of change of y with respect to a third variable t, where we do not have y expressed explicitly in terms of t. We use the chain rule in differentiation to find this rate of change in a process called implicit differentiation.

                     dy/dt = dy/dx × dx/dt

If base is equal to the height, by similar triangles:

Volume v = base x height x length/2

                = bhl/2

                = h²12/2

                = 6h².

Therefore,  

dv/dt = 12hdh/dt

         = 2 ft³/min

 Now,

dh/dt = 2/(12h)

when h = 1.6,

dh/dt = 2/(12 × 1.6) ≈ 0.1 ft³/min

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