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# A swimming pool is 20 $\mathrm{ft}$ wide, 40 $\mathrm{ft}$ long, 3 $\mathrm{ft}$ deep at theshallow end, and 9 ft deep at its deepest point. A cross-section is shown in the figure. If the pool is being filled at a rate of 0.8 $\mathrm{ft}^{3} / \mathrm{min}$ , how fast is the water level rising whenthe depth at the deepest point is 5 $\mathrm{ft}$ ?

## The water level is Increasing at the rate of $1.32 \times 10^{-3} \mathrm{ft} / \mathrm{min}$

Derivatives

Differentiation

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### Video Transcript

to figure out how fast the water level is rising. The first thing you want to do is write out our differential equation. You know, the volume of the swimming pool is the area of the cross section eight times the width of the pool W therefore we know differentiating we have DV over d t is w times d A over DT. This is our equation. Therefore, we now know that a is H and B is a h over three because remember, we had 16. Overby is six over h two. Just simplifying this therefore we know now, but we can write out our equation as 11 h squared over six plus 12 h which means D A over GT's. Now this is differentiating. There's gonna be 11 h plus 12 in the 11 is divided by three Then this is all times de age divide by DT. So now, given what we have in the problem, we can plug in 0.8 feet. You per minute is our devi over. DT is 20 times 11 times five over three because 12 times d h over DT, which gives us d h over DT is 1.32 times 10 to the negative three feet per minute. And because the answer is positive, we know the rate is increasing, therefore the water level's increasing.

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