Simon's copy of Functions of two variables: Stationary points 3

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Simon Thomas 6 years, 1 month ago

Created this as a copy of Functions of two variables: Stationary points 3.

Name	Status	Author	Last Modified
Functions of two variables: Stationary points 3	Ready to use	Bill Foster	01/06/2016 09:47
Morten's copy of Functions of two variables: Stationary points 3	Ready to use	Morten Brekke	01/06/2016 09:47
vijay's copy of Functions of two variables: Stationary points 3	Ready to use	vijay t	01/06/2016 09:47
Simon's copy of Functions of two variables: Stationary points 3	draft	Simon Thomas	14/03/2019 14:12

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Question statement

In the following question find the $(x,y)$ coordinates of the single stationary point $(a,b) \in D$ of the continuous function $f: D \rightarrow \mathbb{R}$ :

$f(x,y) = \simplify[std]{{a} + {b}*e^(-(x-{c})^2-(y-{d})^2)}$

where

$D = \{(x,y): \simplify[std]{(x-{c})^2+(y-{d})^2}\} \le \var{r}$

That is, $D$ is a disk of radius $\simplify[std]{sqrt({r})}$ and centre $(\var{c},\var{d})$ .

Input both cooordinates as fractions or integers and not decimals.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	7
c	integer	2
b	integer	7
d	integer	-9
s3	integer	1
s2	integer	1
s1	integer	1
s5	integer	1
s4	integer	-1
r	integer	2

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Value

xxxxxxxxxx
 
s1*random(1..9)

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Description

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Generated value: `integer`

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Parts

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1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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3. Gap Gap 2. Number entry
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1. Step Information only

Gap-fill

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

$x$ –coordinate, $a=$ Gap 0.

$y$ –coordinate, $b=$ Gap 1.

Input the value of $f(x,y)$ at $(a,b)$ :

$f(a,b)=$ Gap 2.

If you want some help, click on Show steps. You will not lose any marks if you do so.

Advice

The $(x,y)$ coordinates of the stationary point of a function of 2 variables $f(x,y)$ are given by solving
the following 2 equations for $x$ and $y$

$\begin{eqnarray*} \partial f \over \partial x &=&0\\ \\ \partial f \over \partial y &=&0 \end{eqnarray*}$

In this case you get two equations to solve for $x$ and $y$

$\begin{eqnarray*} \simplify[std]{{-2*b}*(x-{c})*e^(-(x-{c})^2-(y-{d})^2)}&=&0\\ \\ \simplify[std]{{-2*b}*(y-{d})*e^(-(x-{c})^2-(y-{d})^2)}&=&0 \end{eqnarray*}$
We can cancel off the term

$\simplify[std]{e^(-(x-{c})^2-(y-{d})^2)}$ in both equations as

$\simplify[std]{e^(-(x-{c})^2-(y-{d})^2)} \neq 0,\;\forall x,\;y$ .

On solving these we get

$x = \var{c},\;\;\;y=\var{d}$

So the stationary point is $(\var{c},\var{d}) \in D$ .

On substituting these values into $f(x,y)$ we get:

$f(\var{c},\var{d})=\simplify[std]{{a}+{b}*e^0={a+b}}$

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This question is used in the following exam:

Partial derivatives and stationary points by Simon Thomas in Calculus.

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Simon's copy of Functions of two variables: Stationary points 3

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Simon Thomas 6 years, 1 month ago

Error in variable testing condition

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Generated value: `integer`

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Simon's copy of Functions of two variables: Stationary points 3

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 6 years, 1 month ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`