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?
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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)`
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