Applications of Differentiation · Hook

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Ever wondered how to get the biggest goat pen from the same fencing?

Mallam Musa has 40 m of fencing and a long wall to build against. Some shapes enclose far more ground than others. Differentiation tells him the exact best size — no guessing.

Revision notes

📒 Applications of Differentiation: lesson notes

Use dy/dx to find speeds, turning points and the biggest or smallest value of a quantity.

Mallam Musa has 40 m of fencing and a long wall to build against. Some shapes enclose far more ground than others. Differentiation tells him the exact best size — no guessing.

Rates of change and velocity

dy/dx tells you how fast y changes as x changes. Substitute a value of x to get the rate at that point.

If s is distance and t is time, the velocity is v = ds/dt, and the acceleration is a = dv/dt.

For s = 3t2 + 2t: v = 6t + 2, so at t = 4 the velocity is 6 × 4 + 2 = 26 m/s, and a = 6 m/s2.

🔑 Key idea: Velocity = ds/dt; acceleration = dv/dt.

Turning points

At the top of a hill or the bottom of a valley the tangent is flat, so the gradient is 0.

To find a turning point, solve dy/dx = 0 for x, then put that x into y.

For y = x2 − 6x + 5: dy/dx = 2x − 6 = 0 gives x = 3, and y = 9 − 18 + 5 = −4. The turning point is (3, −4).

🔑 Key idea: Turning point: dy/dx = 0.

Maximum or minimum?

Differentiate again. If d2y/dx2 > 0 the point is a minimum (∪ shape); if d2y/dx2 < 0 it is a maximum (∩ shape).

For y = x2 − 6x + 5, d2y/dx2 = 2 > 0, so (3, −4) is a minimum.

In word problems, write the quantity (area, profit, cost) in terms of one variable, differentiate, and set the derivative to 0.

🔑 Key idea: d2y/dx2 > 0 → minimum; d2y/dx2 < 0 → maximum.

Worked example

Find the y-coordinate of the turning point of the curve y = x2 − 2x + 5.

  1. Step 1: Differentiate

    dy/dx = 2x − 2

  2. Step 2: Set dy/dx = 0

    2x − 2 = 0

    2x = 2

    x = 2 ÷ 2 = 1

    Why? At a turning point the tangent is flat, so the gradient is 0.

  3. Step 3: Substitute x = 1 into y

    y = 1 × 12 + (−2) × 1 + 5

    = 4

Answer: y = 4

Summary

  • Substitute into dy/dx to get the gradient or rate at a point.
  • v = ds/dt and a = dv/dt.
  • At a turning point dy/dx = 0; put that x back into y for the y-coordinate.
  • Second derivative positive → minimum; negative → maximum.
  • Word problems: one-variable formula → differentiate → set to 0.

Ready to check you've got it? Scroll up and work through the lesson: the interactive visuals, practice questions and mastery check are all there.