Maxima and minima problems

​​In a nutshell

Using derivatives, you can determine maximum and minimum values of displacement and velocity. 

Equations

Variable definitions

Displacement

For maximum and mimimum displacement, you can work out the turning points of the curve on a displacement-time graph. At a turning point, the gradient is $$0$$. Therefore, the turning point can be worked out by:

$$\boxed{\dfrac{ds}{dt} = 0}$$​​

To work out if the turning point is a maximum or a minimum, you use the second derivative.

For a minimum point:

​$$\dfrac{d^2s}{dt^2}>0$$​​

For a maximum point:

​$$\dfrac{d^2s}{dt^2}<0$$​​

​

Velocity

For maximum and minimum velocity, you can work out the turning points of the curve on a velocity-time graph. The turning point can be worked out by:

​$$\boxed{\dfrac{dv}{dt}=0}$$​​

For a minimum point:

​$$\dfrac{d^2v}{dt^2}>0$$​​

For a maximum point:

​$$\dfrac{d^2v}{dt^2}<0$$​​

Example

A boy throws a ball up into the air and catches it again. The displacement of the ball at time $$t$$ from his hand is modelled by $$s=5t+3-2t^2$$, $$0\leq t \leq3$$. Find the maximum distance the ball travels from his hand.

For the maximum distance you have to find the turning point.

$$s=-2t^2+5t+3$$​​

Differentiate:

​$$\dfrac{ds}{dt} = -4t+5$$​​

​

At the maximum point, the gradient equals $$0$$:​

$$-4t+5=0$$

$$t=\dfrac54=1.25\ s$$​​​

Substitute this time into the formula for displacement to get the maximum distance:

$$s=5(1.25)+3-2(1.25^2) = 6.25+3 - 2(1.5625) = 6.125$$​​

The maximum distance the ball travels from his hand is $$\underline{6.13\ m}$$.