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

5. a rectangular swimming pool 10 ft wide by 20 ft long and of uni - form depth is being filled with water.\na. 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 ).\nb. at what rate is the volume of the water increasing if the water level is rising at ( \frac{1}{4} ) ft/min?\nc. at what rate is the water level rising if the pool is filled at a rate of ( 10 mathrm{ft}^{3}/mathrm{min} )?

5. a rectangular swimming pool 10 ft wide by 20 ft long and of uni - form depth is being filled with water.\na. 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 ).\nb. at what rate is the volume of the water increasing if the water level is rising at ( \frac{1}{4} ) ft/min?\nc. at what rate is the water level rising if the pool is filled at a rate of ( 10 mathrm{ft}^{3}/mathrm{min} )?

Answer

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