Step by step
The question
Question: A 10 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 8 ft above the ground?
1 Draw and name
Draw it.
The ladder stays 10 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 = 8.
2 Write the equation
The wall, the ground and the ladder make a right triangle.
Pythagoras: x squared plus y squared is 10 squared.
Careful!
Do not put y = 8 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.
10 squared is a number: 0.
4 Put the numbers in
First the missing side at that moment: x.
Pythagoras with y = 8.
10 squared is 100, 8 squared is 64.
Take away: 36.
So x = 6.
A length is positive.
Now put in x = 6, y = 8, and dy/dt = −2.
Multiply out.
5 Solve for the rate
Add 32 on both sides.
To remove times 12, we divide by 12.
A calculator: about 2.67 ft per second.
The bottom slides away at 2.67 ft per second.
6 Check
Let's check: a tiny moment later, 0.001 seconds.
y is 7.998.
Then x, from Pythagoras, is about 6.00267.
x changed by 0.00267 in 0.001 seconds.
That is about 2.67 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.