Differentiation: Stationary Points and their Nature 3

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Name	Status	Author	Last Modified
Differentiation: Stationary Points and their Nature 3	Ready to use	Ben McGovern	26/09/2022 11:56
Differentiation: Contextualized Stationary Points and their Nature	draft	Ed Southwood	31/01/2025 11:09

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Question statement

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	1
sol	rational	-108/3125
soldp	decimal	dec("-0.03")
b	integer	1
maxmin1	string	minimum
maxmin2	string	maximum
simp	number	1
y1	number	0
y2	number	-4

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Name

Data type

Value

xxxxxxxxxx
 
random(-4..4 except 0)

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Generated value: `integer`

→ Used by:

maxmin1
maxmin2
simp
sol
y1
y2

Advice
"Part a)" - prompt
"Part a)" → "Gap 0" - Correct answer
"Part b)" - prompt
"Part b)" → "Gap 0." - Minimum accepted value
"Part b)" → "Gap 0." - Maximum accepted value
"Part b)" → "Gap 2." - Minimum accepted value
"Part b)" → "Gap 2." - Maximum accepted value

Parts

Part a) Gap-fill
1. Gap Gap 0 Mathematical expression
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Part b) Gap-fill
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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4. Gap Gap 3. Number entry
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Ask the student a question, and give any hints about how they should answer this part.

Find the derivative of \[ \simplify{y=(x-{a})^2/({b}x)}.\]

$\frac{dy}{dx}=$Gap 0

Advice

Part (a):

Since the function is of the form $y=\tfrac{u(x)}{v(x)}$, we want to use the quotient rule to calculate the derivative $\tfrac{dy}{dx}$.

As \[ \simplify{y=(x-{a})^2/({b}x)}, \]

let \[ u(x) = \simplify{(x-{a})^2} \quad \text{and} \quad v(x)=\simplify{{b}x}.\]

Next, we need to find the derivatives, $\tfrac{du}{dx}$ and $\tfrac{dv}{dx}$:

\[ \dfrac{du}{dx} = \simplify{2(x-{a})}\quad \text{and} \quad\dfrac{dv}{dx}=\simplify{{b}}.\]

Substituting these results into the quotient rule formula we can obtain an expression for $\tfrac{dy}{dx}$:

\[ \begin{split} \dfrac{dy}{dx} &\,= \dfrac{v(x) \times \frac{du}{dx} - u(x) \times\frac{dv}{dx}}{[v(x)]^2} \\ \\&\,=\dfrac{(\simplify{{b}x}) \times\simplify{2(x-{a})} - \simplify{(x-{a})^2} \times \simplify{{b}}}{\simplify{({b}x)^2}}. \end{split}\]

Simplifying,

\[ \begin{split} \dfrac{dy}{dx} &\,=\dfrac{\simplify{2{b}x(x-{a})-{b}(x-{a})^2}}{\simplify{{b}^2x^2}} \\ &\,= \dfrac{\simplify{2x(x-{a})-(x-{a})^2}}{\simplify{{b}x^2}}\\ &\,=\dfrac{\simplify[!cancelTerms]{2x^2-{2a}x-x^2+{2a}x-{a^2}}}{\simplify{{b}x^2}}\\ &\,=\dfrac{\simplify{x^2-{a^2}}}{\simplify{{b}x^2}} \end{split} \]

Part (b):

A graph can help. You can pan and zoom this image if you would like to.

{geogebra_applet('https://www.geogebra.org/m/t6vskwck',[a:{a}, b:{b}])}

To find the stationary points of the function, we must solve $\tfrac{dy}{dx}=0$ for $x$. Setting $\tfrac{dy}{dx}=0$:

\[ \begin{split} \dfrac{\simplify{x^2-{a^2}}}{\simplify{{b}x^2}}&\,= 0 \\ \implies \simplify{x^2-{a^2}}&\,=0 \end{split} \]

Therefore, the stationary points are when $x=\var{abs(a)}$ and $x=\var{-(abs(a))}$.

We find the corresponding $y$-values of the stationary points by plugging these $x$-values into the initial equation:

When $x=\var{abs(a)}$,

\[\begin{split} y &\,=\dfrac{\simplify[unitFactor,!cancelTerms,fractionNumbers]{({abs(a)}-{a})^2}}{\simplify[alwaysTimes,unitFactor]{{b}{abs(a)}}} \\ &\,= \simplify[fractionNumbers]{{(abs(a)-a)^2/(b*abs(a))}} .\end{split}\]

When $x=\var{-abs(a)}$,

\[\begin{split} y &\,=\dfrac{\simplify[unitFactor,!cancelTerms,fractionNumbers]{({-abs(a)}-{a})^2}}{\simplify[alwaysTimes,unitFactor]{{b}({-abs(a)})}} \\ &\,= \simplify[fractionNumbers]{{(-abs(a)-a)^2/(b*-abs(a))}} .\end{split}\]

Therefore, the stationary points are \[ \left(\var{abs(a)},\simplify[fractionNumbers]{{(abs(a)-a)^2/(b*abs(a))}}\right)\quad \text{and} \quad \left(\var{-abs(a)},\simplify[fractionNumbers]{{(-abs(a)-a)^2/(b*(-abs(a)))}}\right).\]

Finally, we need to determine the nature of the stationary points. To do this we want to calculate the second derivative of the initial function and then evaluate it for each $x$-value of the stationary points.

Recall:

If $\tfrac{d^2y}{dx^2}<0$ the stationary point is a maximum;
If $\tfrac{d^2y}{dx^2}>0$ the stationary point is a minimum;
If $\tfrac{d^2y}{dx^2}=0$ the stationary point is a point of inflection.

To calculate $\tfrac{d^2y}{dx^2}$, we want to differentiate $\tfrac{dy}{dx}$ again with respect to $x$:

\[ \begin{split} &\frac{dy}{dx} &=\dfrac{\simplify{x^2-{a^2}}}{\simplify{{b}x^2}} \\ \\\implies &\frac{d^2y}{dx^2} &\,= \dfrac{\simplify{({b}x^2)(2x)-{2b}x(x^2-{a^2})}}{\simplify{({b}x^2)^2}} \\ &&\,=\dfrac{\simplify[!cancelTerms]{2x^3-2x^3+{2a^2}x}}{\simplify{{b}x^4}} \\ &&\,=\simplify[unitFactor]{{2a^2/simp}/({b/simp}x^3)}\end{split}\]

For $\left(\var{abs(a)},\simplify[fractionNumbers]{{(abs(a)-a)^2/(b*abs(a))}}\right)$, $\tfrac{d^2y}{dx^2} =\simplify[fractionNumbers]{{2a^2/(b*(abs(a))^3)}}$, so it is a {maxmin1};

For $\left(\var{-abs(a)},\simplify[fractionNumbers]{{(-abs(a)-a)^2/(b*-abs(a))}}\right)$, $\tfrac{d^2y}{dx^2} =\simplify[fractionNumbers]{{2a^2/(b*(-abs(a))^3)}}$, so it is a {maxmin2}.

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Differentiation: Stationary Points and their Nature 3

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Differentiation: Stationary Points and their Nature 3

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Taxonomy: mathcentre

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