Step by step
The question
Question: A 25 ft ladder leans against a wall.
The bottom slides away at 3 ft/s.
How fast is the top sliding down when the bottom is 7 ft from the wall?
1 Draw and name
Draw it.
The ladder stays 25 ft long.
The bottom's distance from the wall, x, and the top's height, y, change.
The bottom slides away: x grows by 3 ft per second.
So dx/dt = 3.
dx/dt means how fast x changes: the change each second.
We want dy/dt at the moment when x = 7.
2 Write the equation
The wall, the ground and the ladder make a right triangle.
Pythagoras: x squared plus y squared is 25 squared.
Careful!
Do not put x = 7 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.
25 squared is a number: 0.
4 Put the numbers in
First the missing side at that moment: y.
Pythagoras with x = 7.
25 squared is 625, 7 squared is 49.
Take away: 576.
So y = 24.
A length is positive.
Now put in x = 7, y = 24, and dx/dt = 3.
Multiply out.
5 Solve for the rate
Take away 42 on both sides.
To remove times 48, we divide by 48.
So dy/dt = −0.875 ft per second.
The minus means y gets smaller.
The top slides down at 0.875 ft per second.
6 Check
Let's check: a tiny moment later, 0.001 seconds.
x is 7.003.
Then y, from Pythagoras, is about 23.99912.
y changed by −0.00088 in 0.001 seconds.
That is about −0.88 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.