Solved on Nov 27, 2023

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

STEP 1

Assumptions1. The sides of the square are increasing at a constant rate of $6 \mathrm{~m} / \mathrm{~sec}$ . . We want to find the rate of change of the area of the square when the sides are $20 \mathrm{~m}$ and $26 \mathrm{~m}$ 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 $s$ and the area by $A$ , we have $A = s^2$

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

STEP 11

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

Was this helpful?

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

STEP 1

Assumptions1. The sides of the square are increasing at a constant rate of $6 \mathrm{~m} / \mathrm{~sec}$ . . We want to find the rate of change of the area of the square when the sides are $20 \mathrm{~m}$ and $26 \mathrm{~m}$ 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 $s$ and the area by $A$ , we have $A = s^2$

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

STEP 11

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

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

STEP 1

Assumptions1. The sides of the square are increasing at a constant rate of 6 m/ sec6 \mathrm{~m} / \mathrm{~sec}6 m/ sec. . We want to find the rate of change of the area of the square when the sides are 20 m20 \mathrm{~m}20 m and 26 m26 \mathrm{~m}26 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 sss and the area by AAA, we haveA=s2A = s^2A=s2

STEP 3

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

STEP 4

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

STEP 5

Using the chain rule, the derivative of s2s^2s2 with respect to time is 2sdsdt2s \frac{ds}{dt}2sdtds​.dAdt=2sdsdt\frac{dA}{dt} =2s \frac{ds}{dt}dtdA​=2sdtds​

STEP 6

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

STEP 7

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

STEP 8

Now we can find the rate of change of the area when the sides are 20 m20 \mathrm{~m}20 m long by substituting s=20 ms =20 \mathrm{~m}s=20 m into the equation.dAdt=12×20 m2/sec\frac{dA}{dt} =12 \times20 \mathrm{~m^2/sec}dtdA​=12×20 m2/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}dtdA​=12×20 m2/sec=240 m2/sec

STEP 10

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

STEP 11

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

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

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

The area of a square is given by the square of the length of its side. If we denote the side length by $s$ and the area by $A$ , we have $A = s^2$

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

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

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

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

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

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

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

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