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3.4 Calculus

Syllabus
2017
Topic
3.4
Level
Higher

Learning objectives

3.4HA Variable rates of changeUnderstand the concept of a variable rate of change.3.4HB Differentiate integer powers of xdifferentiate integer powers of x3.4HC Gradients, rates of change, stationary points, turningdetermine gradients, rates of change, stationary points, turning points (maxima and minima) by differentiation and relate these to graphs Find the coordinates of the maximum and minimum points3.4HD Distinguish between maxima and minima by consideringdistinguish between maxima and minima by considering the general shape of the graph only3.4HE Calculus in kinematics and practical problemsapply calculus to linear kinematics and other simple practical problems

Understand a variable rate of change

A rate of change compares how quickly one quantity changes with another. A straight line has constant gradient; a curve usually has a gradient that varies from point to point.

Situation Rate of change
displacement against time velocity
velocity against time acceleration
curve y=f(x)y=f(x) gradient dy/dxdy/dx
tangent at a point instantaneous rate there

For a very small change in xx, the corresponding change in yy is approximately (dy/dx)Δx(dy/dx)\Delta x. Differentiation gives this local gradient exactly for an algebraic function.

An average rate over an interval is the gradient of a chord; an instantaneous rate is the gradient of a tangent at one point.

Differentiate integer powers of x

Use the power rule: d(axn)/dx=anxn−1d(ax^n)/dx=anx^{n-1} for an integer power nn. Differentiate every term separately, and constants differentiate to zero.

Term Derivative
5x35x^3 15x215x^2
−2x-2x −2-2
77 00
4/x2=4x−24/x^2=4x^{-2} −8x−3-8x^{-3}

If y=4x3+5x2+2x−7y=4x^3+5x^2+2x-7, then dy/dx=12x2+10x+2dy/dx=12x^2+10x+2.

The exponent decreases by 1 after multiplying by the old exponent. Do not merely reduce the exponent or differentiate a constant as 1.

Use derivatives to find gradients and turning points

The derivative gives the gradient function. Substitute an xx-value for a gradient at a point; solve dy/dx=0dy/dx=0 to find stationary points.

Task Method
gradient at x=ax=a find dy/dxdy/dx, then substitute aa
stationary xx-values solve dy/dx=0dy/dx=0
stationary coordinates substitute each xx into the original yy
negative gradient solve dy/dx<0dy/dx<0

For y=x3−x2−8x+12y=x^3-x^2-8x+12, dy/dx=3x2−2x−8=(3x+4)(x−2)dy/dx=3x^2-2x-8=(3x+4)(x-2). Thus stationary points occur at x=−4/3x=-4/3 and x=2x=2.

Solving dy/dx=0dy/dx=0 gives only the xx-coordinates. Use the original function—not the derivative—to find the corresponding yy-coordinates.

Classify maxima and minima from graph shape

A stationary point is a local maximum when the graph rises then falls, and a local minimum when it falls then rises.

Gradient change through point Classification
positive to negative local maximum
negative to positive local minimum
no sign change stationary point of inflection

Sketch or inspect the general shape around each stationary point. The curve's direction on both sides decides whether the point is a maximum or minimum.

‘Maximum’ or ‘minimum’ may mean local unless a domain is given. Endpoints must also be checked when finding an absolute extreme on a closed interval.

A horizontal tangent alone does not prove maximum or minimum; a stationary point can be an inflection if the gradient keeps the same sign.

Apply calculus to motion and practical models

For displacement s(t)s(t), velocity is v=ds/dtv=ds/dt and acceleration is a=dv/dt=d2s/dt2a=dv/dt=d^2s/dt^2. Keep units consistent: metres, seconds, m/s and m/s².

Request Equation to use
instantaneously at rest set v=0v=0
given acceleration set aa to that value
displacement at a time substitute into s(t)s(t)
optimise a quantity differentiate model, solve derivative =0=0, classify

If s=2t3−5t2+6t−5s=2t^3-5t^2+6t-5, then v=6t2−10t+6v=6t^2-10t+6 and a=12t−10a=12t-10. Setting a=5a=5 gives t=1.25t=1.25 s.

Reject solutions outside the stated physical domain, such as negative time, impossible length or a value making a denominator zero.

Velocity can be negative because it includes direction; speed is ∣v∣|v| and cannot be negative.