Related rates Calculus?

Question

a 3 meter ladder stands against a high wall.The foot of the ladder moves outward at a speed of .1 m/s when the foot is 2 meter from the wall.how fast is the area formed by the wall,ground,and ladder changing at this time?

Answer 1

Since the ladder against the wall formed a triangle we can use Pythagorean theorem having 3m as a constant number.

x^2 + y^2=3
then we differenciate in terms of t (time) with chain rule.

2x (dx/dt) + 2y (dy/dt) = 0

and then with some algebra we can put it as

(dt/dy) = -x/y (dx/dt)

We also know that when x=2, dx/dt=0.1 m/s and by completing the Pythagoras triangle y=√5
If we plug in those numbres to the equation (dt/dy) = -x/y (dx/dt)

(dt/dy) = -2/10√5


We also know that the area equals xy/2 so

A=xy/2 so we differentiate with chain rule

dA/dt = [y (dx/dt) - x (dy/dt)]/ y^2

and just plug in the numbers that you got before.

dA/dt = [ (√5)(.1) - (2)(-1/5√5)]/5

which is equivalent to .0805 m^2/s

Answer 2

I don't have a complete answer, but maybe this will help. It's been a long time since I did these.

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The Pythagorean Theorem says that:

d² + h² = 3²

Where d is the distance from the wall and h is the height above the wall where the ladder touches.

So, when d = 2,

2² + h² = 9
h = √5

The formula for the area, expressed in terms of d and h, is:

A = 1/2hd

When you differentiate this, you get:

A' = 1/2hd' + 1/2dh'

Now you know h=√5, d=2, and d' = 0.1.

I'm not sure what h' is or how to find it, unfortunately. Probably can be computed using trig functions somehow, but I can't remember.

Answer 3

You need to find dA/dt, the change in area wrt time. To do this, you need to write a function of area.

Area = 0.5 base * height

For this function, base is just x, and height is √(3^2-x^2), with help from Pythagorus. So,

A= 0.5*x*√(3^2-x^2)

This is a relation of the variables A, area, and x, distance from wall. Both of these variable are functions of time, ie, they change with time. So, we can differentiate them wrt time.

dA/dt = 0.5 *( x* (1/(2√(3^2-x^2))*(-2x*dx/dt)+√(3^2-x^2) * 0.5)

Now you have three "variable", dA/dt, x, and dx/dt. If you know two of them you can solve for the third. In this case you know dx/dt (0.1 m/s) and x (2 m), so you can solve for dA/dt

All related rates problems are exactly like this: get a relationship of the variables, then implicitly differentiate wrt time to get a relationship of their rates. Then solve for whichever you like, given all but one value. Sometime you will need to work a little to get the values.