Solved on Nov 27, 2023

Find the rate of change of the area of a square as its sides increase at 6 m/sec6 \mathrm{~m} / \mathrm{sec}, when the sides are 20 m20 \mathrm{~m} and 26 m26 \mathrm{~m} long.

STEP 1

Assumptions1. The sides of the square are increasing at a constant rate of 6 m/ sec6 \mathrm{~m} / \mathrm{~sec}. . We want to find the rate of change of the area of the square when the sides are 20 m20 \mathrm{~m} and 26 m26 \mathrm{~m} long.

STEP 2

The area of a square is given by the square of the length of its side. If we denote the side length by ss and the area by AA, we haveA=s2A = s^2

STEP 3

We are given that the sides of the square are increasing at a rate of 6 m/ sec6 \mathrm{~m} / \mathrm{~sec}. This is the derivative of the side length with respect to time, or dsdt\frac{ds}{dt}.
dsdt=6 m/sec\frac{ds}{dt} =6 \mathrm{~m/sec}

STEP 4

We want to find the rate of change of the area of the square, or dAdt\frac{dA}{dt}. We can find this by differentiating the equation for AA with respect to time.
dAdt=ddt(s2)\frac{dA}{dt} = \frac{d}{dt}(s^2)

STEP 5

Using the chain rule, the derivative of s2s^2 with respect to time is 2sdsdt2s \frac{ds}{dt}.
dAdt=2sdsdt\frac{dA}{dt} =2s \frac{ds}{dt}

STEP 6

Now we can substitute the given value of dsdt\frac{ds}{dt} into the equation.
dAdt=2s×6 m/sec\frac{dA}{dt} =2s \times6 \mathrm{~m/sec}

STEP 7

implify the equation.
dAdt=12s m2/sec\frac{dA}{dt} =12s \mathrm{~m^2/sec}

STEP 8

Now we can find the rate of change of the area when the sides are 20 m20 \mathrm{~m} long by substituting s=20 ms =20 \mathrm{~m} into the equation.
dAdt=12×20 m2/sec\frac{dA}{dt} =12 \times20 \mathrm{~m^2/sec}

STEP 9

Calculate the rate of change of the area.
dAdt=12×20 m2/sec=240 m2/sec\frac{dA}{dt} =12 \times20 \mathrm{~m^2/sec} =240 \mathrm{~m^2/sec}

STEP 10

Next, we can find the rate of change of the area when the sides are 26 m26 \mathrm{~m} long by substituting s=26 ms =26 \mathrm{~m} into the equation.
dAdt=12×26 m2/sec\frac{dA}{dt} =12 \times26 \mathrm{~m^2/sec}

STEP 11

Calculate the rate of change of the area.
dAdt=×26 m/sec=312 m/sec\frac{dA}{dt} = \times26 \mathrm{~m^/sec} =312 \mathrm{~m^/sec}a. The area of the square is changing at a rate of 240 m/sec240 \mathrm{~m^/sec} when the sides are 20 m20 \mathrm{~m} long. b. The area of the square is changing at a rate of 312 m/sec312 \mathrm{~m^/sec} when the sides are 26 m26 \mathrm{~m} long.

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