165k views
3 votes
The legs of an isosceles right triangle increase in length at a rate of 3 m divided by s3 m/s. a. At what rate is the area of the triangle changing when the legs are 22 m​ long? b. At what rate is the area of the triangle changing when the hypotenuse is 66 m​ long? c. At what rate is the length of the hypotenuse​ changing?

1 Answer

2 votes

Answer:

dA/dt = 66 [m²/s]

dA/dt = 15,56 [m²/s]

dL/dt = 2 [m/s]

Explanation:

In an isosceles right triangle, we have:

L² = x² + y² (1)

Where L is the hypotenuse and x and y are the legs

And we also know that area of the triangle is

A = (1/2)* b*h

Where b is base of the triangle (in this case one leg) and h the height ( the other leg) then

A = (1/2)*x*y

in this case x=y then

A = (1/2)*x²

Differentiating in relation to time, at both sides of the equation let us get:

dA/dt = (1/2)*2*x*dx/dt

dA/dt = x*dx/dt

a)

When the legs are 22 meters long, and legs are increasing in length at a rate of 3 m/s we have:

dA/dt = x*dx/dt ⇒ dA/dt = = 22*3 [m²/s]

dA/dt = 66 [m²/s]

b) L² = x² + y² since x=y

L² = 2*x² ⇒ x² = L²/2

And A = (1/2)*x²

Differentiating in relation to time, at both sides of the equation let us get:

dA/dt = (1/2)* 2*x*dx/dt

When hypotenuse is 66 m long x² =(1/2)*L² x² = (1/2)*(66)²

Then x = 66/√2

dA/dt = (1/2)* 2* 66/√2 * 3

dA/dt = 22/√2 [m²/s ]

dA/dt = 15,56 [m²/s]

c)

L²= 2*x²

2*L*dL/dt = 4 * x*dx/dt

2*66* dL/dt = 4 * 22*3

dL/dt = 2 [m/s]

User Tena
by
7.4k points