Skip to lesson

Math · Calculus

Chapter 5: Applications of the Derivative

Rates of Change and Related Rates

The derivative as a speed.

Lesson
1
Time
About 22 minutes
0 of 8 done

Step 1: Let's Learn

Read it, or press Listen and follow the words.

If s(t) is position, then s′(t) is velocity: how fast the position is changing.

And again

The derivative of velocity is acceleration. Differentiating twice is meaningful, not just possible.

Related rates

When two quantities are linked, their rates are linked too. Differentiate the relationship with respect to time.

The method

Write the relationship, differentiate both sides with respect to t, then substitute the known values.

The derivative as a rate

If s(t) is position, s′(t) is velocity and s″(t) is acceleration. The derivative measures how fast one quantity changes with respect to another, which is why it appears throughout science.

The units say what it measures

A derivative's units are the output units divided by the input units — metres per second, dollars per item, degrees per minute. Reading the units is the quickest way to interpret a derivative in context.

The sign carries meaning

A positive derivative means the quantity is increasing, negative that it is decreasing, zero that it is momentarily steady. In applied problems the sign is often the answer being asked for.

Marginal quantities in economics

Marginal cost is the derivative of total cost — roughly the cost of one more unit. Economics adopted calculus for exactly this, and "marginal" is the economist's word for "derivative".

Step 2: Try It Yourself

Tap and try it out.

If this is position against time, the tangent slope is the velocity.
-8-8-6-6-4-4-2-222446688
y = 1x² + 0x + 0
  • Point(2, 4)
  • Slope of the tangent4

Step 3: Watch an Example

One step at a time.

Watch Priya Solve a Related Rate

A square’s sides grow at 2 cm per second. How fast is its area growing when the side is 5 cm?

  1. Step 1

    The relationship is A = s².

Step 4: Your Turn

Practice makes it stick.

The Velocity

Problem 1 of 2

s(t) = t², so v(t) = 2t. What is the velocity at t = 6?

The Square

Problem 2 of 2

A square with side 4 growing at 3 cm/s. How fast is the area growing?

cm² per second

Rates

1 of 4

v(t) = 2t. Velocity at t = 9?

2 of 4

A square with side 6 growing at 2 cm/s. Area rate?

3 of 4

A square with side 10 growing at 1 cm/s. Area rate?

4 of 4

What is the derivative of velocity?

Step 5: Quick Check

Show what you know.

Question 1 of 2

A square with side 7 growing at 2 cm/s. Area rate?

Question 2 of 2

What does a derivative measure in a real situation?

What You Learned

  • The derivative of position is velocity; of velocity, acceleration.
  • Linked quantities have linked rates.
  • Differentiate the relationship with respect to time, then substitute.