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Chapter 5: Problem 11

A woman is cmptying her aquarium at a steady rate with a small pump. The water pumped to a 12 -in.-diameter cylindrical bucket, and its depth is increasing at the rate of 4.0 ir. per minute. Find the rate at which the aquarium water level is dropping if the aquarium measures 24 in. (wide) \(\times 36\) in. (long) \(x\) 18 in. (high)

Short Answer

Expert verified
The rate at which the aquarium water level is dropping is \( -144π \) cubic inches per minute.

Step by step solution

01

Identify Known Dimensions and Rates

The aquarium has dimensions 24 in. (wide) \(\times 36\) in. (long) \(\times 18\) in. (high), which form a rectangular prism. The volume of such a prism is given by \(V= lwh\), where \(l\), \(w\), and \(h\) are the length, width, and height respectively. Here, \(V = 24 \times 36 \times 18\). The cylindrical bucket is filling at a rate of 4.0 in. per minute. The volume is given by \(V= πr^2h\). The diameter of the bucket is 12 inches, meaning the radius is 6 inches, and \(h\) increases at a steady 4.0 inches per minute.
02

Set Up Differentiation

By the chain rule of differentiation, by differentiating the formula for the volume of the cylinder \(V= πr^2h\) with respect to time \(t\), we get \(dV/dt= π \times 2r \times (dr/dt) \times h + πr^2 \times (dh/dt)\). Substituting \(r=6\), \(dr/dt=0\) (since the bucket radius is constant), and \(dh/dt=4\) in, we find that \((dV/dt)_{bucket}= 144π\) cubic inches per minute.
03

Apply to the Aquarium

The rate at which the aquarium water level is dropping is equal to the rate at which the water is being pumped out, i.e., \((dV/dt)_{aquarium} = - (dV/dt)_{bucket}\). This is negative because the volume is decreasing. Therefore, \((dV/dt)_{aquarium}= -144π\) cubic inches per minute.

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