Applications of Differential Calculus (Grade 12 NSC Matric Mathematics): Flashcards

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Applications of Differential Calculus
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Optimisation problems

Problems finding maximum or minimum values

Stationary points

Points where rate of change is zero

How to find stationary points

Set f′(x)=0f'(x) = 0 and solve

If f′′(a)<0f''(a) < 0, what type of point is x=ax = a?

Local maximum

If f′′(a)>0f''(a) > 0, what type of point is x=ax = a?

Local minimum

Variables needed before differentiating in optimisation

Only one variable

Instantaneous rate of change equals

The derivative f′(x)f'(x)

Velocity formula (if ss is distance)

v(t)=dsdt=s′(t)v(t) = \frac{ds}{dt} = s'(t)

Acceleration formula (if vv is velocity)

a(t)=dvdt=v′(t)a(t) = \frac{dv}{dt} = v'(t)

Acceleration in terms of distance ss

a(t)=s′′(t)a(t) = s''(t) (second derivative)

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