Step by step
The question
Question: A 15 ft ladder leans against a wall.
The top slides down at 2 ft/s.
How fast is the bottom sliding away when the top is 9 ft above the ground?
1 Draw and name
Draw it.
The ladder stays 15 ft long.
The bottom's distance from the wall, x, and the top's height, y, change.
The top slides down: y gets smaller by 2 ft per second.
Getting smaller means a minus rate: dy/dt = −2.
dx/dt means how fast x changes: the change each second.
We want dx/dt at the moment when y = 9.
2 Write the equation
The wall, the ground and the ladder make a right triangle.
Pythagoras: x squared plus y squared is 15 squared.
Careful!
Do not put y = 9 in yet!
y 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.
15 squared is a number: 0.
4 Put the numbers in
First the missing side at that moment: x.
Pythagoras with y = 9.
15 squared is 225, 9 squared is 81.
Take away: 144.
So x = 12.
A length is positive.
Now put in x = 12, y = 9, and dy/dt = −2.
Multiply out.
5 Solve for the rate
Add 36 on both sides.
To remove times 24, we divide by 24.
So dx/dt = 1.5 ft per second.
The bottom slides away at 1.5 ft per second.
6 Check
Let's check: a tiny moment later, 0.001 seconds.
y is 8.998.
Then x, from Pythagoras, is about 12.0015.
x changed by 0.0015 in 0.001 seconds.
That is about 1.5 ft 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.