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Applications of Differentiation

Use derivatives to find velocity from position, and to solve simple maximum-value problems.

Motion: Velocity as a Derivative

Example 1 — finding velocity

A ball's position is s(t) = t^{2} + 3 t metres, t seconds after it is thrown. Find its velocity at t = 4 . 1 Differentiate: v(t) = s'(t) = 2 t + 3 2 Substitute t = 4 : v(4) = 2 (4) + 3 = 8 + 3 Answer: v(4) = 11 m /s

Some real problems ask for the largest (or smallest) possible value of a quantity. Write the quantity as a function, differentiate it, set the derivative to 0 to find the stationary point, then substitute back to get the actual maximum (or minimum) value.

Example 1 — maximum area

A farmer has 40 m of fencing to build a rectangular pen against an existing wall (so only 3 sides need fencing: two equal widths, w, and one length, l). Find the maximum possible area. 1 The fencing used is: l + 2 w = 40, so l = 40 - 2 w 2 Area = length \times width: A(w) = (40 - 2 w) \times w = 40 w - 2 w^{2} 3 Differentiate: A'(w) = 40 - 4 w 4 Set A'(w) = 0 : 40 - 4 w = 0, so w = 10 5 Find l: l = 40 - 2 (10) = 20 6 Find the maximum area: A = 20 \times 10 Answer: Maximum area = 200 m^{2} (when the pen is 20 m long and 10 m wide)

Practice

s(t) = t^{2} + 5 t . Find the velocity at t = 3 .

Your answer

Motion: Velocity as a Derivative

Example 1 — finding velocity

A ball's position is s(t) = t^{2} + 3 t metres, t seconds after it is thrown. Find its velocity at t = 4 . 1 Differentiate: v(t) = s'(t) = 2 t + 3 2 Substitute t = 4 : v(4) = 2 (4) + 3 = 8 + 3 Answer: v(4) = 11 m /s

Optimisation: Finding a Maximum Value

Some real problems ask for the largest (or smallest) possible value of a quantity. Write the quantity as a function, differentiate it, set the derivative to 0 to find the stationary point, then substitute back to get the actual maximum (or minimum) value.

Example 1 — maximum area

A farmer has 40 m of fencing to build a rectangular pen against an existing wall (so only 3 sides need fencing: two equal widths, w, and one length, l). Find the maximum possible area. 1 The fencing used is: l + 2 w = 40, so l = 40 - 2 w 2 Area = length \times width: A(w) = (40 - 2 w) \times w = 40 w - 2 w^{2} 3 Differentiate: A'(w) = 40 - 4 w 4 Set A'(w) = 0 : 40 - 4 w = 0, so w = 10 5 Find l: l = 40 - 2 (10) = 20 6 Find the maximum area: A = 20 \times 10 Answer: Maximum area = 200 m^{2} (when the pen is 20 m long and 10 m wide)

Practice

s(t) = t^{2} + 5 t . Find the velocity at t = 3 .

Your answer