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🌌 Calculus Kingdom · Calculus

Applications of Derivatives

Use derivatives to find tangent lines, velocities, and maximum and minimum values.

In short

  • A tangent line needs both a point, f(a), and a slope, f'(a).
  • Velocity is the derivative of position and acceleration is the derivative of velocity.
  • The sign of f' says whether the function rises or falls; f' = 0 marks a stationary point.
  • The second derivative classifies it: positive means a minimum, negative means a maximum.

The tangent line

A tangent is the straight line that touches a curve at a point and matches its direction there. Building its equation needs exactly two things, and the derivative supplies only one of them.

  • the point: substitute into f to get the height, giving (a, f(a))
  • the slope: substitute into f' to get the steepness, m = f'(a)

Then use point-slope form: y - f(a) = f'(a)(x - a).

For f(x) = x2 - 3x at x = 4:

f(4) = 16 - 12 = 4, so the point is (4, 4) f'(x) = 2x - 3, so f'(4) = 5 y - 4 = 5(x - 4) y = 5x - 16

Confusing f(a) with f'(a) is the classic error here. The first is a height, the second is a slope, and the tangent needs both.

Motion: position, velocity, acceleration

If s(t) is position at time t, then

velocity v(t) = s'(t) acceleration a(t) = v'(t) = s''(t)

Each derivative asks "how fast is the thing below it changing?"

For s(t) = t3 - 6t2 + 9t metres:

v(t) = 3t2 - 12t + 9 a(t) = 6t - 12

At t = 1 the velocity is 0 (the object is momentarily at rest, about to turn round) and the acceleration is -6.

The signs carry meaning. Positive velocity means moving forwards, negative means backwards, and zero means momentarily at rest. Positive acceleration means speeding up in the positive direction. A negative velocity with a negative acceleration means going backwards and getting faster.

Increasing, decreasing and stationary points

The derivative is a slope, so its sign describes the shape of the curve:

f'(x) > 0 -> f is increasing (rising left to right) f'(x) < 0 -> f is decreasing f'(x) = 0 -> a stationary point, where the tangent is horizontal

Finding stationary points means solving f'(x) = 0. For f(x) = x3 - 3x, f'(x) = 3x2 - 3 = 0 gives x = 1 and x = -1.

To classify them, use the second derivative:

  • f''(x) > 0 — the curve is concave up (holds water), so the point is a minimum
  • f''(x) < 0 — the curve is concave down, so the point is a maximum

Here f''(x) = 6x, so at x = 1 it is positive (a minimum) and at x = -1 it is negative (a maximum).

The alternative test is to check the sign of f' just left and just right of the point. Either method works; the point is that finding a stationary point is not the same as knowing what kind it is.

Optimization

An optimization problem asks for the biggest or smallest value something can take. The derivative finds it, but the setting up is most of the work.

The routine:

1. Name the variables and write down what is to be maximised or minimised. 2. Use the constraint in the problem to express it in one variable. 3. Differentiate, set the derivative to zero, and solve. 4. Confirm it is the right kind of point, and answer the question actually asked.

"You have 40 m of fencing for three sides of a rectangular pen, using a wall as the fourth side. What is the largest area?"

Let x be each side perpendicular to the wall, y the side along it. Constraint: 2x + y = 40, so y = 40 - 2x. Area: A(x) = x(40 - 2x) = 40x - 2x2. A'(x) = 40 - 4x = 0, so x = 10. A''(x) = -4 < 0, so it is a maximum. y = 40 - 20 = 20, giving an area of 200 m2.

Notice that the answer is not a square. With one side free, the best pen is twice as wide as it is deep. Step 4 matters too: if the question asks for the area, do not hand back a side length.

Worked examples

Example 1

Find the equation of the tangent to f(x) = x2 + 2x at x = 3.

  1. Height: f(3) = 9 + 6 = 15, so the tangent touches at (3, 15).
  2. Slope: f'(x) = 2x + 2, so f'(3) = 8.
  3. Point-slope form: y - 15 = 8(x - 3).
  4. Rearranged: y = 8x - 9.

Example 2

A stone thrown upward has height s(t) = 30t - 5t2 metres. When is it at its highest, and how high is that?

  1. At the highest point the stone is momentarily at rest, so solve s'(t) = 0.
  2. s'(t) = 30 - 10t, so 30 - 10t = 0 gives t = 3 seconds.
  3. s''(t) = -10 < 0, confirming a maximum.
  4. s(3) = 90 - 45 = 45 metres.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

f(x) = x2 + 2x - 3 Find the equation of the tangent line to the curve at x = 3. Give the right-hand side as an expression in x (for example 4x - 3).

Answer: 8*x + (-12)

  1. f(3) = 1(3)2 + 2(3) - 3 = 12
  2. f'(x) = 2x + 2
  3. f'(3) = 8, so the tangent has slope 8.
  4. y - 12 = 8(x - 3)
  5. y = 8x - 12

Problem 2

Difficulty 3 of 5

Noor's cart moves so that its position after t seconds is s(t) = 3t3 - 6t2 + 8t + 9, measured in metres. Find its velocity at t = 4 seconds.

Answer: 104 m/s

  1. s(t) = 3t3 - 6t2 + 8t + 9
  2. v(t) = s'(t) = 9t2 - 12t + 8
  3. At t = 4: 104 m/s

Problem 3

Difficulty 4 of 5

f(x) = x3 - 15x2 + 27x This curve has two stationary points (places where the slope is zero). What is the SMALLER of the two x-values?

Answer: 1

  1. f'(x) = 3x2 - 30x + 27
  2. 3x2 - 30x + 27 = 3(x - 1)(x - 9) = 0
  3. x = 1 or x = 9
  4. The smaller value is x = 1.

Common mistakes

  • Using f(a) as the slope of the tangent, or f'(a) as the point.
  • Finding a stationary point and never checking whether it is a maximum or a minimum.
  • Optimising without first using the constraint to reduce to one variable.
  • Answering with the x-value when the question asked for the maximum value itself.

What you should be able to do

  • Write the equation of a tangent line at a point.
  • Relate position, velocity and acceleration.
  • Find critical points and classify maxima and minima.
  • Solve a simple optimisation problem.

Where this fits in the curriculum

Common Core

  • CHA-3.A

    AP Calculus AB, Unit 4 — Interpret the meaning of a derivative in context.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

  • CHA-3.B

    AP Calculus AB, Unit 4 — Calculate rates of change in applied contexts, including position, velocity and acceleration.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

  • FUN-4.A

    AP Calculus AB, Unit 5 — Justify conclusions about the behaviour of a function based on the behaviour of its derivatives.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

  • FUN-4.B

    AP Calculus AB, Unit 5 — Calculate minimum and maximum values in applied contexts or analysis of functions.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

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