Question

A trough is 16 ft long and its ends have the shape of isosceles triangles that are 4 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 5 inches deep?

Answer 1

Answer:

$\frac{dh}{dt} = 0.562$

Explanation:

given data:

rate of filling  = 15 ft3/min

determine dh\dt  when height is water is 5 inch

as from the similar triangle

$\frac{4}{1} = \frac{b}{h}$

b =4 h

we know that

volume = (area of base of trough){height of trough)

           $= \frac{1}{2}* (base of triangle of water* height of triangle of water)(16)$

           $= \frac{1}{2}*4h*h*16$

            $= 32h^{2}$

Differentiate w.r.t to t

$v(t) =32h^{2}$

$\frac{d}{dt} v(t) = \frac{d}{dt} 32h^{2}$

$\frac{dv}{dt} = 64h\frac{dh}{dt}$

we know h = 5 inch = 0.4166 ft

$\frac{dv}{dt} = 15$

$15 = 64*0.4166\frac{dh}{dt}$

$\frac{dh}{dt} = 0.562$