Question

1. a rectangular swimming pool 10 ft wide by 20 ft long and of uni - form depth is being filled with water.

a. if ( t ) is elapsed time, ( h ) is the height of the water, and ( v ) is the volume of the water, find equations relating ( v ) to ( h ) and ( dv/dt ) to ( dh/dt ).
b. at what rate is the volume of the water increasing if the water level is rising at ( \frac{1}{4} ) ft/min?
c. at what rate is the water level rising if the pool is filled at a rate of ( 10 mathrm{ft}^{3}/mathrm{min} )?

Explanation:

Step1: Find the equation relating \(V\) to \(h\)

The volume \(V\) of a rectangular prism (pool) is given by \(V=\text{length}\times\text{width}\times\text{height}\). Here, length \(l = 20\) ft, width \(w=10\) ft, and height \(h\) (of water). So, \(V = 20\times10\times h=200h\).

Step2: Differentiate \(V\) with respect to \(t\)

Differentiate \(V = 200h\) with respect to \(t\) using the chain - rule. \(\frac{dV}{dt}=\frac{d}{dt}(200h)\). Since \(200\) is a constant, \(\frac{dV}{dt}=200\frac{dh}{dt}\).

Step3: Solve part (b)

Given \(\frac{dh}{dt}=\frac{1}{4}\) ft/min. Substitute into \(\frac{dV}{dt}=200\frac{dh}{dt}\). Then \(\frac{dV}{dt}=200\times\frac{1}{4}=50\) ft³/min.

Step4: Solve part (c)

Given \(\frac{dV}{dt}=10\) ft³/min. Substitute into \(\frac{dV}{dt}=200\frac{dh}{dt}\). Then \(10 = 200\frac{dh}{dt}\), and \(\frac{dh}{dt}=\frac{10}{200}=\frac{1}{20}\) ft/min.

Answer:

a. \(V = 200h\), \(\frac{dV}{dt}=200\frac{dh}{dt}\)
b. \(50\) ft³/min
c. \(\frac{1}{20}\) ft/min