Step by step
The question
Question: A 5 m ladder leans against a wall.
The bottom slides away at 1 m/s.
How fast is the top sliding down when the bottom is 3 m from the wall?
1 Draw and name
Draw it.
The ladder stays 5 m long.
The bottom's distance from the wall, x, and the top's height, y, change.
The bottom slides away: x grows by 1 m per second.
So dx/dt = 1.
dx/dt means how fast x changes: the change each second.
We want dy/dt at the moment when x = 3.
2 Write the equation
The wall, the ground and the ladder make a right triangle.
Pythagoras: x squared plus y squared is 5 squared.
Careful!
Do not put x = 3 in yet!
x changes with time.
Numbers go in only after the derivative.
3 Take the derivative
Both x and y change with time.
The chain rule: x squared gives 2x times dx/dt.
y squared gives 2y times dy/dt.
5 squared is a number: 0.
4 Put the numbers in
First the missing side at that moment: y.
Pythagoras with x = 3.
5 squared is 25, 3 squared is 9.
Take away: 16.
So y = 4.
A length is positive.
Now put in x = 3, y = 4, and dx/dt = 1.
Multiply out.
5 Solve for the rate
Take away 6 on both sides.
To remove times 8, we divide by 8.
So dy/dt = −0.75 m per second.
The minus means y gets smaller.
The top slides down at 0.75 m per second.
6 Check
Let's check: a tiny moment later, 0.001 seconds.
x is 3.001.
Then y, from Pythagoras, is about 3.99925.
y changed by −0.00075 in 0.001 seconds.
That is about −0.75 m per second.
Right!
Let's look back.
First: draw and name.
Then: write the equation.
Then: take the derivative.
Then: put the numbers in.
Then: solve for the rate.
Last: check.