View Full Version : Question On Related Time Rate (calculus)
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}