Math

QuestionModel the depth of water in a pool draining at 20%20\% per hour from an initial depth of 12 feet. Find D(t)D(t) for t=1,2,3t=1, 2, 3 hours.

Studdy Solution

STEP 1

Assumptions1. The initial depth of the water is12 feet. . The rate of decrease is20% per hour.
3. The rate of decrease is constant.
4. The depth of the water decreases exponentially, not linearly.

STEP 2

First, we need to understand that the depth of the water decreases at a constant percent rate, which means it decreases exponentially. The general formula for exponential decay is(t)=D0×(1r)t(t) = D0 \times (1 - r)^twhere- (t)(t) is the depth of the water at time tt, - 00 is the initial depth of the water, - rr is the rate of decrease, - tt is the time in hours.

STEP 3

Now, plug in the given values for the initial depth of the water and the rate of decrease to create the equation.
(t)=12×(10.20)t(t) =12 \times (1 -0.20)^t

STEP 4

Now we can use this equation to find the depth of the water at each given time.
For t=1t=1 hr(1)=12×(10.20)1(1) =12 \times (1 -0.20)^1

STEP 5

Calculate the depth of the water at t=1t=1 hr.
(1)=12×0.80=9.feet(1) =12 \times0.80 =9.\, feet

STEP 6

For t=2t=2 hr(2)=12×(10.20)2(2) =12 \times (1 -0.20)^2

STEP 7

Calculate the depth of the water at t=2t=2 hr.
(2)=12×0.802=7.68feet(2) =12 \times0.80^2 =7.68\, feet

STEP 8

For t=3t=3 hr(3)=12×(10.20)3(3) =12 \times (1 -0.20)^3

STEP 9

Calculate the depth of the water at t=3t=3 hr.
(3)=12×.803=6.144feet(3) =12 \times.80^3 =6.144\, feetSo, the depth of the water in the pool after hour is9.6 feet, after2 hours is7.68 feet, and after3 hours is6.144 feet.

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