Related rates problems share the same idea, which is to compute the rate of change of one quantity in terms of the rate of change of the other quantity. So basically, all you need to do is to figure out an equation that can relate the known quantities and the unknown quantities, and then use the rules in calculus you learned to do differtiation on both sides of the equation with resect to time. (Mostly all the questions you see will be about time.)

Let's do a practice. Here's the question:

A 20-feet ladder leaning against a wall is sliding down in a way that the angle of inclination is decreasing at a rate of 0.1 rad/min. So how fast is the area enclosed by the ladder, the wall and the ground going decreasing when the inclination angle is 30 degree?

Solution:

let's draw a picture to show the given and unknown:

From algebra, we know the area of a triangle equals to A=2*x*y. That is the relation/equation we are looking for. ("A" represents area.)

However, x and y are unknown, so we have to convert it to something related to the angle a.

We know sin a = y/20 , cos a = x/20,

so we get y=20 sin a, x=20 cos a

So we can plug in x and y into the equation and get:

A=(1/2) x * y

A=(1/2) 20cos a * 20 sin a

By simplify the equation we get:

A=200 sina cosa

then since 2 sin a * cos a = sin 2a ( in algebra this is needed to be memorized.)

We get A= 100 * (2 sin a * cos a)

So A = 100 sin 2a

Now it's time to do differtiation on both sides of the equation:

dA/dt = 100 (cos 2a) * 2 * da/dt

dA/dt = 200 (cos 2a) * da/dt

Since we are asking the related rate of the area when the angle is 30 degree, so we put in 30, and also the changing rate of the angle is -0.1 rad/min, so we put it in as well: (the changing rate of angle is negative because it is moving out.)

dA/dt = 200 (cos (2*30)) (-0.1)

dA/dt = 200 (cos 60) X (-0.1)

dA/dt = 200 * 0.5 * (-0.1)

dA/dt = -10 ft^{2}/sec

So the area enclosed by the ladder, the wall and the ground going decreasing when the inclination angle is 30 degree is 10 ft^{2}/sec away from the wall.

**Subject :**Math**Topic :**Calculus-
**Posted By :**Jason