Related Rates

One thing changes, so another must too.

12 problems · about 23 minutes · CHA-3.E

What this lesson teaches

The student solves related rates problems by differentiating a relationship with respect to time.

Straightforward practice. Get the method working first.

5 problems

dV/dt = 3s² · ds/dt. With s = 3 and ds/dt = 2, what is dV/dt?
Answer 54
Why 54.
Why must substitution wait until after differentiating?
Answer A substituted value becomes a constant, and constants have zero derivative.
Why Early substitution freezes a moving quantity.
dV/dt = 3s² · ds/dt. With s = 4 and ds/dt = 2, what is dV/dt?
Answer 96
Why 3 × 16 × 2.
dA/dt = 2πr · dr/dt. With r = 5 and dr/dt = 2, what is dA/dt divided by π?
Answer 20
Why 2 × 5 × 2.
A tank drains. Is dV/dt positive or negative? 1 positive, 2 negative.
Answer 2
Why The volume shrinks.

The same ideas with more to keep track of.

3 problems

A square’s area is A = s². dA/dt = 2s · ds/dt. With s = 6 and ds/dt = 3, what is dA/dt?
Answer 36
Why 2 × 6 × 3.
Should a changing quantity be substituted before differentiating? 1 for yes, 0 for no.
Answer 0
Why It would become a constant.
dV/dt = 3s² · ds/dt. With s = 1 and ds/dt = 5, what is dV/dt?
Answer 15
Why 3 × 1 × 5.

Mixed problems. Work out what kind of question it is before you start.

4 problems

Put the related rates method in order.
Answer 1. Draw the situation and label what changes. 2. Write an equation relating the quantities. 3. Differentiate both sides with respect to time. 4. Substitute the known values and solve.
Why Substitution is always the last step.
dA/dt = 2s · ds/dt. With s = 10 and ds/dt = 1, what is dA/dt?
Answer 20
Why 2 × 10 × 1.
The Cube: dV/dt = 3s² · ds/dt. With s = 2 and ds/dt = 1, what is dV/dt?
Answer 12
Why 12.
The Circle: A circle’s area is A = πr². dA/dt = 2πr · dr/dt. With r = 3 and dr/dt = 1, what is dA/dt divided by π?
Answer 6
Why 6π, so 6 after dividing.