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?

dA/dt = 66 [m²/s]

dA/dt = 15,56 [m²/s]

dL/dt  =  2 [m/s]

Step-by-step 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]

ian gonna lie that is way to much

maybe i can