A trough built for your pet pigs has an equilateral triangle cross-section which is 1m across the top. The trough is 500cm long. Your pet pigs...
are consuming their food at a rate of 120cm^3/s and the cylindrical automatic food dispenser is adding food at a rate of 0.36m^3/hr. How fast is the food level falling at the instant the food in the trough is 40 cm deep?
Volume of the prism equals the area of the base times the length. The length is given as 500cm.
Consider a cross-section. ` `Let `h` represent the depth of the food in cm, and `2x` the distance across the face in cm. The area is `1/2(2x)(h)=xh` .
(1)The volume is `V=500xh` .
(2) From the triangle, we have `x=h/sqrt(3)`
(3) So `V=(500)/sqrt(3)h^2`
Food is being consumed at the rate of `120 (cm)^3/s` , and replenished at the rate of `.36m^3/(hr)` . Converting to `(cm)/s` we get `100 (cm)^3/s` .
Thus `(dV)/(dt)=100 (cm)^3/s - 120 (cm)^3/s=-20 (cm)^3/s`
(4) Differentiating with respect to `t` yields:
`(dV)/(dt)=(1000)/sqrt(3)h (dh)/(dt)` . Given that `h=40cm` and the computed `(dV)/(dt)=-20 (cm)^3/s` we have:
`(dh)/(dt)=(-sqrt(3))/2000 ~~-.000866 (cm)/s ~~ -3.12 (cm)/(hr)`
Thus the rate that the food is falling is approximately `3.12 (cm)/(hr)`