# The area of a right triangle is decreasing at 2 cm^2/s. If the hypotenuse is increasing at 1 cm/s, find the rate of change of the ratio of the sides You should notice that the problem involves related rates.

You need to come up with the notations: A for area, H for hypotenuse, x for leg 1 and y for leg 2.

You need to remember the formula of area of right triangle such taht:

`A = (xy)/2`

Differentiating both sides with...

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You should notice that the problem involves related rates.

You need to come up with the notations: A for area, H for hypotenuse, x for leg 1 and y for leg 2.

You need to remember the formula of area of right triangle such taht:

`A = (xy)/2`

Differentiating both sides with respect to t yields:

`(dA)/(dt) = (1/2)*(x(dy)/(dt) + y(dx)/(dt))`

The problem provides the information that the area of right triangle is decreasing at 2 `cm^2/s`  such  that:

`(dA)/(dt) = -2`

`-2 = (1/2)*(x(dy)/(dt) + y(dx)/(dt)) =gt -4 = (x(dy)/(dt) + y(dx)/(dt)) =gt -4 - x(dy)/(dt) = y(dx)/(dt)`

You need to use Pythagorean theorem such that:

`H^2 = x^2 + y^2`

Differentiating both sides yields:

`2H(dH)/(dt) = 2x(dx)/(dt) + 2y(dy)/(dt)`

The problem provides the information that the hypotenuseis increasing at 1 `cm/s`  such that:

`2H*1 = 2x(dx)/(dt) + 2y(dy)/(dt)`

`H = x(dx)/(dt) + y(dy)/(dt) =gt y(dy)/(dt) = H - x(dx)/(dt) `

The ratio of sides is `x/y` , hence, differentiating with respect to t yields:

`(x(dx)/(dt)*y - x*y*(dy)/(dt))/(y^2)`

`(x(dx)/(dt)*y - x*(H - x(dx)/(dt)))/(y^2)`

`(x(dx)/(dt)*(y + x)- x*H)/(y^2)`

Hence, evaluating the rate of change of ratio of sides yields `(d(x/y))/(dt) = (x(dx)/(dt)*(y + x) - x*H)/(y^2)`.

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