Cogito
Calculus · Chapter 5 · Lesson 1
Rates of Change and Related Rates
The derivative as a speed.
10 problems · about 22 minutes · CHA-3.A, CHA-3.B
What this lesson teaches
The student interprets the derivative as a rate of change and solves related rates problems.
- The derivative of position is velocity; of velocity, acceleration.
- Linked quantities have linked rates.
- Differentiate the relationship with respect to time, then substitute.
Warm-Up Review
From earlier lessons. Loosen up before the new work.
2 problemsReview — The Chain Rule: For (6x + 1)³, what is the inner derivative?
Answer 6
Why 6.
Review — The Product and Quotient Rules: y = x² (3x + 1), so y′ = 9x² + 2x. What is y′ at x = 2?
Answer 40
Why 40.
Warm Up
Straightforward practice. Get the method working first.
4 problemsA square with side 7 growing at 2 cm/s. Area rate?
Answer 28
Why 28 cm²/s.
What does a derivative measure in a real situation?
Answer How fast one quantity changes as another does.
Why A rate of change.
v(t) = 2t. Velocity at t = 9?
Answer 18
Why 2 × 9.
A square with side 6 growing at 2 cm/s. Area rate?
Answer 24
Why 2 × 6 × 2.
Build It Up
The same ideas with more to keep track of.
2 problemsA square with side 10 growing at 1 cm/s. Area rate?
Answer 20
Why 2 × 10 × 1.
What is the derivative of velocity?
Answer Acceleration.
Why Differentiating again.
Stretch Yourself
Mixed problems. Work out what kind of question it is before you start.
2 problemsThe Velocity: s(t) = t², so v(t) = 2t. What is the velocity at t = 6?
Answer 12
Why 12.
The Square: A square with side 4 growing at 3 cm/s. How fast is the area growing?
Answer 24 cm² per second
Why 24 cm²/s.