Worked examples
The question
Question: A 10 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 6 ft from the wall?
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 bottom slides away: x grows by 2 ft per second.
So dx/dt = 2.
dxdt = 2 ft/s
dx/dt means how fast x changes: the change each second.
dxdt: how fast x changes, per second
We want dy/dt at the moment when x = 6.
dydt = ?
2 Write the equation
The wall, the ground and the ladder make a right triangle.
Pythagoras: x squared plus y squared is 10 squared.
x2 + y2 = 102
Careful!
Do not put x = 6 in yet!
x changes with time.
Numbers go in only after the derivative.
Careful: not 62 + y² = 10² first ✗
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.
2x dxdt + 2y dydt = 0
4 Put the numbers in
First the missing side at that moment: y.
Pythagoras with x = 6.
62 + y2 = 102
10 squared is 100, 6 squared is 36.
Take away: 64.
y2 = 100 − 36 = 64
So y = 8.
A length is positive.
y = √64 = 8
Now put in x = 6, y = 8, and dx/dt = 2.
2(6)(2) + 2(8)dydt = 0
Multiply out.
24 + 16dydt = 0
5 Solve for the rate
Take away 24 on both sides.
16dydt = −24
To remove times 16, we divide by 16.
dydt = −2416
So dy/dt = −1.5 ft per second.
= −1.5 ft/s
The minus means y gets smaller.
The top slides down at 1.5 ft per second.
The top slides down at 1.5 ft/s.
6 Check
Let's check: a tiny moment later, 0.001 seconds.
x is 6.002.
Then y, from Pythagoras, is about 7.9985.
x = 6.002: y = √(10² − 6.002²) ≈ 7.9985
y changed by −0.0015 in 0.001 seconds.
That is about −1.5 ft per second.
Right!
change ÷ 0.001 ≈ −1.5
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.
▶ Watch it solved
The question
Question: A 13 ft ladder leans against a wall.
The top slides down at 3 ft/s.
How fast is the bottom sliding away when the top is 12 ft above the ground?
1 Draw and name
Draw it.
The ladder stays 13 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 3 ft per second.
Getting smaller means a minus rate: dy/dt = −3.
dydt = −3 ft/s
dx/dt means how fast x changes: the change each second.
dxdt: how fast x changes, per second
We want dx/dt at the moment when y = 12.
dxdt = ?
2 Write the equation
The wall, the ground and the ladder make a right triangle.
Pythagoras: x squared plus y squared is 13 squared.
x2 + y2 = 132
Careful!
Do not put y = 12 in yet!
y changes with time.
Numbers go in only after the derivative.
Careful: not 122x² + 122 = 13² first ✗
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.
13 squared is a number: 0.
2x dxdt + 2y dydt = 0
4 Put the numbers in
First the missing side at that moment: x.
Pythagoras with y = 12.
122 + x2 = 132
13 squared is 169, 12 squared is 144.
Take away: 25.
x2 = 169 − 144 = 25
So x = 5.
A length is positive.
x = √25 = 5
Now put in x = 5, y = 12, and dy/dt = −3.
2(5)dxdt + 2(12)(−3) = 0
Multiply out.
10dxdt − 72 = 0
5 Solve for the rate
Add 72 on both sides.
10dxdt = 72
To remove times 10, we divide by 10.
dxdt = 7210
So dx/dt = 7.2 ft per second.
= 7.2 ft/s
The bottom slides away at 7.2 ft per second.
The bottom slides away at 7.2 ft/s.
6 Check
Let's check: a tiny moment later, 0.001 seconds.
y is 11.997.
Then x, from Pythagoras, is about 5.00719.
y = 11.997: x = √(13² − 11.997²) ≈ 5.00719
x changed by 0.00719 in 0.001 seconds.
That is about 7.19 ft per second.
Right!
change ÷ 0.001 ≈ 7.19
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.
▶ Watch it solved
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.
dydt = −2 ft/s
dx/dt means how fast x changes: the change each second.
dxdt: how fast x changes, per second
We want dx/dt at the moment when y = 9.
dxdt = ?
2 Write the equation
The wall, the ground and the ladder make a right triangle.
Pythagoras: x squared plus y squared is 15 squared.
x2 + y2 = 152
Careful!
Do not put y = 9 in yet!
y changes with time.
Numbers go in only after the derivative.
Careful: not 92x² + 92 = 15² first ✗
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.
2x dxdt + 2y dydt = 0
4 Put the numbers in
First the missing side at that moment: x.
Pythagoras with y = 9.
92 + x2 = 152
15 squared is 225, 9 squared is 81.
Take away: 144.
x2 = 225 − 81 = 144
So x = 12.
A length is positive.
x = √144 = 12
Now put in x = 12, y = 9, and dy/dt = −2.
2(12)dxdt + 2(9)(−2) = 0
Multiply out.
24dxdt − 36 = 0
5 Solve for the rate
Add 36 on both sides.
24dxdt = 36
To remove times 24, we divide by 24.
dxdt = 3624
So dx/dt = 1.5 ft per second.
= 1.5 ft/s
The bottom slides away at 1.5 ft per second.
The bottom slides away at 1.5 ft/s.
6 Check
Let's check: a tiny moment later, 0.001 seconds.
y is 8.998.
Then x, from Pythagoras, is about 12.0015.
y = 8.998: x = √(15² − 8.998²) ≈ 12.0015
x changed by 0.0015 in 0.001 seconds.
That is about 1.5 ft per second.
Right!
change ÷ 0.001 ≈ 1.5
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.
▶ Watch it solved
