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Shahdood · Dec 3, 2011, 11:28 AM

If the area of an isosceles right angle triangle equals (A)cm2 anf if the rate of change of its area is 32cm/sec at the instant at which the length of each of the equal legs is 8 cm. find (at the instant)the time rate of change of the perimeter (p) of the triangle

Unknown008 · Dec 13, 2011, 05:11 AM

Write down your formulae:

$A = \frac{l^2}{2}$

l is the length of one side the 'leg'.

P = 2l + h

h is the hypotenuse and is obtained through Pythagoras' Theorem;

$h^2 = l^2 + l^2$

$P = 2l + \sqrt{2}l = l(2+\sqrt2)$

You are given $\frac{dA}{dt} = 32$

At that time, $l = 8$

You must first the rate at which the length is changing and from there you find the rate at which the perimeter is changing.

You can find dl/dt from the chain rule:

$\frac{dA}{dt} = \frac{dA}{dl} \times \frac{dl}{dt}$

Then you can get the rate of change of perimeter:

$\frac{dP}{dt} = \frac{dP}{dl} \times \frac{dl}{dt}$