1. ## Related Rates

problem: A 5 meters ladder is leaning against a vertical wall with the lower end on the floor. If the lower end is pulled away from the wall at a rate of 1 meter/sec, how fast is the upper end of the ladder moving along the wall when the lower end is 3 meters from the wall?

2. Originally Posted by cazimi
problem: A 5 meters ladder is leaning against a vertical wall with the lower end on the floor. If the lower end is pulled away from the wall at a rate of 1 meter/sec, how fast is the upper end of the ladder moving along the wall when the lower end is 3 meters from the wall?
First let's modify this diagram a bit, see below:

So when faced with a related rates problem, you want to begin by drawing a diagram and filling in what you know and don't know. Then you try to figure out a formula that connects what you know to what you don't, hopefully, what you don't know will be the only unknown in this formula. Here goes:

The first formula that should come into your head for this is Pythagoras' formula, and it is the one we'll be using:

By Pythagoras:

5^2 = x^2 + y^2 .........now we differentiate implicitly with respect to time
=> 0 = 2x dx/dt + 2y dy/dt .........now solve for dy/dt
=> dy/dt = (-2x dx/dt)/2y

now we know dx/dt, the question told us. we also know we want x to be 3, but what is y? well, we go back to the faithful Pythagoras' formula to find out what y is when x = 3

by pythagoras:
y^2 = 5^2 - x^2
=> y^2 = 5^2 - 3^2
=> y^2 = 25 - 9 = 16
=> y = 4 ....now we know everything, let's plug all these values into our formula.

dy/dt = (-2x dx/dt)/2y
=> when x = 3, dx/dt = 1, y = 4, we get:
dy/dt = (-2(3) (1))/(2(4))
=> dy/dt = -6/8
=> dy/dt = -3/4 m/s
this should be a negative rate, since the length of y is decreasing (see diagram below to see what i call x and y)