Differentiation, one idea at a timeMathematics Methods, Unit 3 & 4

Before the method: the idea behind it

Putting derivatives to work

Everything so far has been about finding derivatives. This lesson is about what they’re for. Three jobs come up again and again: describing motion, finding the best possible value of something, and estimating a change without redoing a whole calculation. Each is a different question, and the derivative answers all three.

Part one

Motion is a derivative, twice over

Imagine a robot on a straight rail, and a sensor that logs its position every instant.

Plot position against time and you get a curve. The gradient of that curve at any moment is how fast the position is changing: the velocity. Wherever the curve flattens out, the velocity is zero and the robot is momentarily at rest, usually because it’s about to reverse. Differentiate again and you get the rate at which velocity changes: the acceleration. Positive acceleration doesn’t mean “getting faster”; it means velocity is increasing, which for something moving backwards actually means slowing down.

position–time graph: s(t) = t³ − 6t² + 9t 0 2 4 t=0 1 2 3 4 the robot on its rail, position s in metres 0 m 2 m 4 m

Drag the slider or press play. The blue dot moves along the graph while the red robot moves along its rail. The teal arrow is the velocity: its direction is the direction of travel, its length is the speed.

TIME
t = 0.00 s
POSITION
s = 0.00 m
VELOCITY, s′(t)
v = 9.00 m/s
moving right
ACCELERATION, s″(t)
a = −12.00 m/s²
slowing down

Watch the two moments the graph turns: at t = 1 and t = 3 the velocity hits zero, the arrow vanishes, and the robot reverses. Between them the arrow points left. And notice the sign game between velocity and acceleration: when they have the same sign the robot is speeding up, when they have opposite signs it is slowing down, regardless of which direction it’s actually going.

\[ v(t) = \dfrac{ds}{dt}, \qquad a(t) = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2} \]

Part two

The best of all possible values

A farmer has 100 metres of fencing and a long straight wall. What shape of paddock gives the most grass?

Push the paddock very narrow and it becomes a long, useless strip. Push it very wide and the fencing runs out before there’s any depth. Somewhere between those two hopeless extremes is a best shape, and until now you’d have had to guess it by trial and error. The derivative removes the guessing: wherever the area is at its greatest, the area’s rate of change with respect to the shape is exactly zero. That single fact turns “what’s the best?” into an equation you can solve.

Every optimisation question, whatever it’s dressed up as, follows the same four moves.

Write the quantity you care about as a function of one variable. Use the constraint in the question (fixed fencing, fixed volume, fixed sum) to eliminate any second variable.

Differentiate it and solve for where the derivative is zero. That gives the candidate values.

Use the second derivative to confirm it’s a maximum or a minimum. Negative means maximum, positive means minimum.

Check the answer makes sense and answer the actual question. A length can’t be negative, a box can’t have zero base. Then give the quantity that was asked for, with units.

For the paddock: with x metres of fence on each of the two sides running out from the wall, the third side gets 100 − 2x, so the area is A = x(100 − 2x). Its derivative is 100 − 4x, which is zero at x = 25. The second derivative is −4, so that’s a maximum. The paddock is 25 m by 50 m and holds 1250 m² of grass, and no other rectangle with that fencing can beat it.

Part three

Good enough, fast

A machinist measures a steel ball bearing’s radius as 6 mm. The gauge is only accurate to a twentieth of a millimetre. How wrong might the volume be?

You could recompute the volume at 6.05 mm and subtract. But there’s a faster idea that also explains what a derivative really is. Very close to any point, a curve is almost indistinguishable from its tangent line. So a small change in x produces a change in y of roughly gradient × step. That’s the increments formula, and it works for any smooth function, in any context, for any small enough change.

\[ \delta y \approx \dfrac{dy}{dx}\times\delta x \]

The symbol δ (delta) just means “a small change in.” Read the formula as: the change in the output is about the gradient times the change in the input. It’s an approximation, but for small steps it’s a remarkably good one, and it’s often all an engineer needs.

Where this shows up in real life

Crash investigation

Investigators reconstruct a vehicle’s motion backwards from skid marks and damage, working out velocity and acceleration from position over time. The mathematics of a stopping car is exactly Part one, with a real speed limit riding on the answer.

Why cans are shaped the way they are

A drink company wants a can that holds a fixed volume using the least metal. That’s an optimisation problem, solved with a derivative, and it’s why cans of a given volume tend to have similar proportions wherever you buy them.

Before you start

Which tool for which question

The question says…Reach for
velocity, acceleration, at rest, changes directionDifferentiate position: v = s′, a = s″
greatest, least, maximum, minimum, best, optimalSet the derivative to zero, confirm with the second derivative
estimate the change, approximate error, small increaseIncrements formula: δy ≈ (dy/dx) × δx

Next, the working method for all three, then practice from straightforward up to exam-style problems.