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# Calculus/Calculus - Related Rates

Question
The legs of an isosceles right triangle increase in length at a rate of 2m/s.

a) At what rate is the area of the triangle changing when the legs are 2 m long?
b) At what rate is the area of the triangle changing when the hypotenuse is 1 m long?
c) At what rate is the length of the hypotenuse changing?

Let the leg of an isosceles triangle be x.
Then the hypotenuse of this triangle is given by sqrt ( x^2 + x^2) = sqrt(2) * x

Area of triangle A= 0.5 * x* sqrt(2) * x

= 1/ (sqrt 2) * x^2

dA/dx = 2 / (sqrt 2) * x  = sqrt(2) * x

(a)    dA/dt = dA/dx *dx/dt

= sqrt(2) * x * (2)        (note that dx/dt = 2m/s)

When x=2, dA/dt =  sqrt(2) * 2 * (2) = 4 *sqrt (2)  m^2 /s  (shown)

(b) When the hypotenuse is 1m long, each leg x= 1/sqrt(2)  m

Therefore,  dA/dt =  sqrt(2) * 1/sqrt(2) * (2) = 2  m^2 /s   (shown)

(c) Since the leg is changing at a rate of 2m/s,
the hypotenuse changes at a rate of  sqrt (2)* dx/dt = 2* sqrt (2) m/s  (shown)

Hope this helps. Peace.

Calculus

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