An isosceles right triangle with legs of length s has area A=(1)/(2)s^2. At the instant when s= sqrt( 32) centimeters, the area of the triangle is increasing at a rate of 12 squ…

Question

An isosceles right triangle with legs of length s has area A=
`(1)/(2)`s^2. At the instant when s= sqrt( 32) centimeters, the area of the triangle is increasing at a rate of 12 square centimeters per second. At what rate is the length of they hypotenuse of the triangle increasing, in centimeters per second, at that instant?

Answer 1

The rate at which the length of the hypotenuse is increasing at the given instant is 48sqrt(2)/c centimeters per second.

Step-by-step explanation:

To find the rate at which the length of the hypotenuse is increasing, we need to use the formula for the area of a triangle. In this case, the area of the triangle is given by A = 1/2s^2. We are given that the area is increasing at a rate of 12 square centimeters per second when s = sqrt(32) centimeters.

We can differentiate the equation A = 1/2s^2 with respect to time to find the rate of change of the area:

dA/dt = 2(1/2)s(ds/dt)

Substituting the given values, we have:

12 = 2(1/2)s(ds/dt)

12 = s(ds/dt)

Now we can find the rate at which the length of the hypotenuse is increasing by differentiating the Pythagorean theorem, which relates the lengths of the legs and the hypotenuse:

c^2 = a^2 + b^2

Taking the derivative with respect to time, we get:

2c(dc/dt) = 2a(da/dt) + 2b(db/dt)

Since this is an isosceles right triangle, a = b = s, and the derivative of s with respect to time is ds/dt. Substituting these values and the known values of s and ds/dt, we can solve for dc/dt:

2c(dc/dt) = 2s(ds/dt) + 2s(ds/dt)

2c(dc/dt) = 4s(ds/dt)

dc/dt = 2s(ds/dt)/c

Substituting the values, we have:

dc/dt = 2(sqrt(32))(12)/c

dc/dt = 48sqrt(2)/c centimeters per second

Answer 2

At the instant when s = √32, the length of the hypotenuse of the triangle is increasing at a rate of 1.5√2 centimeters per second.

At what rate is the length of the hypotenuse of the triangle increasing

From the question, we have the following parameters that can be used in our computation:

A = 1/2s²

Differentiate

So, we have

dA/ds = s

Given that

The area of the triangle is increasing at a rate of 12 square centimeters per second.

This means that

dA/dt = 12

This can also be represented as

12 = s * ds/dt

Make ds/dt the subject

ds/dt = 12/s

When the value of s is √32, we have

ds/dt = 12/√32

Evaluate

ds/dt = 12/4√2

ds/dt = 3/√2

Rationalize

ds/dt = 3√2/2

ds/dt = 1.5√2

Hence, the length of the hypotenuse of the triangle is increasing at a rate of 1.5√2 centimeters per second.