Stationary Points 02

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Michael Proudman

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Michael Proudman 4 years, 10 months ago

Published this.

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Michael Proudman 4 years, 10 months ago

Created this as a copy of Stationary Points (non random) 01.

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Question statement

Points on the curve $y = f(x)$ at which $\frac{dy}{dx}=0$ are called stationary points of the function.

If $x_0$ locates a stationary point of $f(x)$ , so that $\frac{dy}{dx} \large \vert_{x_0}=0$ then the stationary point is:

a local minimum if $\frac{d^2y}{dx^2} \Large \vert_{x_0}>0$

a local maximum if $\frac{d^2y}{dx^2} \Large \vert_{x_0}<0$

inconclusive if $\frac{d^2y}{dx^2} \Large \vert_{x_0}=0$

Give any introductory information the student needs.

			Name	Type	Generated Value

CF1	integer	9
CF2	integer	17
C	integer	2
t	decimal	dec("0.94")
h	number	11.03

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(2..9)

JME function reference

Description

x^2 co-efficient

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `integer`

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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3. Gap Gap 2. Number entry
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4. Gap Gap 3. Number entry
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Part b) Gap-fill

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

A ball is thrown in the air. Its height (in metres) at any time $t$ is given by:

$h=\simplify{ {C}+{CF2}t-{CF1}t^{2} }$

What is its maximum height?

The ball will reach its highest at the stationary (turning) point.

First differentiate the function:

$\large \frac{dh}{dt}=$ Gap 0.

At the stationary point, this gradient will be zero.

Enter the equation that shows this:

Gap 1.

Solve this equation to find the time when this happens:

$t=$ Gap 2. seconds

Substitute this values, into the original function to find the height reached at that time:

$h=$ Gap 3. m

Advice

We are asked to find the greatest height reached by a thrown ball whose height can be modelled by the function $h=\simplify{ {C}+{CF2}t-{CF1}t^{2} }$

You don't need to produce a graph to do this but right now, it may help us to visualise what it is that we are trying to do. Here is a graph of this function with the highest point indicated:

It is easy to see that the highest point is where the graph "turns" and the gradient changes from positive to negative. That is the point that we are looking for.

At this point the gradient will be zero. The gradient is given by the derivative of the function $\frac{dh}{dt}$ .

Differentiate the function:

$\frac{dh}{dt}=\simplify{{CF2}-{CF1}*2*t}$

The stationary points will be where this is zero, so:

$\simplify{{CF2}-{CF1}*2*t}=0$

If we now solve this equation it will give us the time to reach that height.

$\simplify{{CF2}-{CF1}*2*t}=0$

$\var{CF2}=\simplify{{CF1}*2*t}$

$\frac{\var{CF2}}{\simplify{{CF1}*2}}=t$

$t=\var{t}$ seconds

Now that we know the time it takes for the ball to reach its highest point, we are ask to calculate how high it reaches.

We simply need to use the original equation that models the height of the ball and substitute $t=\var{t}$

$h=\simplify{ {C}+{CF2}t-{CF1}t^{2} }$

becomes:

$h= \var{C}+\var{CF2}(\var{t})-\var{CF1}(\var{t})^{2}$

$h=\var{h}$ m

In this example it seems logically obvious that the stationary point is a maximum. After all, balls fly up and then fall down, they don't fly down and then float up again.

However, it is always good practice to check and it will help to stop you making silly mistakes - and it is quick and easy to do.

To find out, we need the second derivative $\frac{d^2h}{dt^2}$ .

Here is the original function:

$h=\simplify{ {C}+{CF2}t-{CF1}t^{2} }$

and its first derivative, that we worked out earlier:

$\frac{dh}{dt}=\simplify{{CF2}-{CF1}*2*t}$

Differentiate a second time to give:

$\frac{d^2h}{dt^2}=\simplify{-{CF1}*2}$

Which is negative, confirming that this point is a local maximum.

Give a worked solution to the whole question.

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Taxonomy: mathcentre

Taxonomy: Kind of activity

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Checkpoint description

Michael Proudman 4 years, 10 months ago

Michael Proudman 4 years, 10 months ago

Points on the curve $y = f(x)$ at which $\frac{dy}{dx}=0$ are called stationary points of the function.

Error in variable testing condition

Error

Warning

Generated value: `integer`

Parts

Gap-fill

A ball is thrown in the air. Its height (in metres) at any time $t$ is given by:

$h=\simplify{ {C}+{CF2}t-{CF1}t^{2} }$

What is its maximum height?

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Stationary Points 02

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Michael Proudman 4 years, 10 months ago

Michael Proudman 4 years, 10 months ago

Points on the curve y=f(x)y = f(x) at which dydx=0\frac{dy}{dx}=0 are called stationary points of the function.

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

A ball is thrown in the air. Its height (in metres) at any time tt is given by:

h={{C}+{CF2}t−{CF1}t{2}}h=\simplify{ {C}+{CF2}t-{CF1}t^{2} }

What is its maximum height?

Points on the curve $y = f(x)$ at which $\frac{dy}{dx}=0$ are called stationary points of the function.

Generated value: `integer`

A ball is thrown in the air. Its height (in metres) at any time $t$ is given by:

$h=\simplify{ {C}+{CF2}t-{CF1}t^{2} }$