Maxima and minima problems

In a nutshell

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

Equations

DESCRIPTION	EQUATION
Turning point on a displacement-time graph.	$\dfrac{ds}{dt}=0$
Turning point on a velocity-time graph.	$\dfrac{dv}{dt}=0$

Variable definitions

QUANTITY NAME	SYMBOL	UNIT NAME	UNIT
$Displacement$	$s$	$Metre$	$m$
$Velocity$	$v$	$Metres\ per\ second$	$ms^{-1}$
$Acceleration$	$a$	$Metres\ per\ second\ squared$	$ms^{-2}$
$Time$	$t$	$Seconds$	$s$

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 $\underline{6.13\ m}$ .