Step by step
The question
Question: A 17 ft ladder leans against a wall.
The bottom slides away at 2 ft/s.
How fast is the top sliding down when the bottom is 8 ft from the wall?
1 Draw and name
Draw it.
The ladder stays 17 ft long.
The bottom's distance from the wall, x, and the top's height, y, change.
The bottom slides away: x grows by 2 ft per second.
So dx/dt = 2.
dx/dt means how fast x changes: the change each second.
We want dy/dt at the moment when x = 8.
2 Write the equation
The wall, the ground and the ladder make a right triangle.
Pythagoras: x squared plus y squared is 17 squared.
Careful!
Do not put x = 8 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.
17 squared is a number: 0.
4 Put the numbers in
First the missing side at that moment: y.
Pythagoras with x = 8.
17 squared is 289, 8 squared is 64.
Take away: 225.
So y = 15.
A length is positive.
Now put in x = 8, y = 15, and dx/dt = 2.
Multiply out.
5 Solve for the rate
Take away 32 on both sides.
To remove times 30, we divide by 30.
A calculator: about −1.07 ft per second.
The minus means y gets smaller.
The top slides down at 1.07 ft per second.
6 Check
Let's check: a tiny moment later, 0.001 seconds.
x is 8.002.
Then y, from Pythagoras, is about 14.99893.
y changed by −0.00107 in 0.001 seconds.
That is about −1.07 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.