Josh's copy of Find stationary point of a curve and determine its nature

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Josh Lim 6 years, 3 months ago

Created this as a copy of Find stationary point of a curve and determine its nature.

Name	Status	Author	Last Modified
Find stationary point of a curve and determine its nature	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:50
Josh's copy of Find stationary point of a curve and determine its nature	draft	Josh Lim	01/07/2019 12:20

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Question statement

For the following, find the stationary point and determine its nature. For your answers, where appropriate, write your solutions as fractions, NOT decimals, and cancel down where possible.

Give any introductory information the student needs.

			Name	Type	Generated Value

a1	integer	3
sx1	rational	1
m	integer	2
sy1	rational	-12
q	integer	-5
sy2	rational	81/8
sx2	rational	3/4
a2	integer	-2
b1	integer	-6
b2	integer	3
c2	integer	9
c1	integer	-9

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Description

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

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Generated value: `integer`

→ Used by:

Advice
"Part a)" - prompt
"Part a)" → "Gap 0." - Correct answer
"Part a)" → "Gap 3." - Correct answer

Gap-fill

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

Let $y = \simplify{{a1}x^2 + {b1}x + {c1}}$ .

$\dfrac{\mathrm{d}y}{\mathrm{d}x} =$ Gap 0.

Enter the coordinates of the stationary point of $y$ : $\big($ Gap 1. $,$ Gap 2. $\big)$

$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} =$ Gap 3.

What is the nature of the stationary point of $y$ ?

Gap 4.

Advice

a)

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \simplify{{2a1}x+{b1}}$ .

To find the $x$ -coordinate of the stationary point, solve $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ for $x$ :

$\begin{align} \frac{\mathrm{d}y}{\mathrm{d}x} = \simplify{{2a1}x + {b1}} &= 0 \\ x &= \simplify[all,fractionNumbers]{{sx1}} \end{align}$

Find the $y$ -coordinate by substituting this value of $x$ into the definition of $y(x)$ :

$\begin{align} \simplify[fractionnumbers]{y({sx1})} &= \simplify[basic,fractionnumbers]{{a1}{sx1}^2+{b1}{sx1}+{c1}} \\ &= \simplify[fractionnumbers]{{sy1}} \end{align}$

Finally, to determine the nature of the stationary point, look at $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ at $x = \simplify[fractionnumbers]{{sx1}}$ .

$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \simplify{{2*a1}}$

This is positive, so the stationary point is a minimum.

b)

$\dfrac{\mathrm{d}z}{\mathrm{d}x} = \simplify{{2a2}x+{b2}}$ .

To find the $x$ -coordinate of the stationary point, solve $\frac{\mathrm{d}z}{\mathrm{d}x} = 0$ for $x$ :

$\begin{align} \frac{\mathrm{d}z}{\mathrm{d}x} = \simplify{{2a2}x + {b2}} &= 0 \\ x &= \simplify[all,fractionNumbers]{{sx2}} \end{align}$

Find the $y$ -coordinate by substituting this value of $x$ into the definition of $z(x)$ :

$\begin{align} \simplify[fractionnumbers]{z({sx1})} &= \simplify[basic,fractionnumbers]{{a2}{sx2}^2+{b2}{sx2}+{c2}} \\ &= \simplify[fractionnumbers]{{sy2}} \end{align}$

Finally, to determine the nature of the stationary point, look at $\frac{\mathrm{d}^2z}{\mathrm{d}x^2}$ at $x = \simplify[fractionnumbers]{{sx2}}$ .

$\frac{\mathrm{d}^2z}{\mathrm{d}x^2} = \simplify{{2*a2}}$

This is negative, so the stationary point is a maximum.

c)

$\dfrac{\mathrm{d}t}{\mathrm{d}x} = \simplify{{3m}x^2}$ .

To find the $x$ -coordinate of the stationary point, solve $\frac{\mathrm{d}t}{\mathrm{d}x} = 0$ for $x$ :

$\begin{align} \frac{\mathrm{d}t}{\mathrm{d}x} = \simplify{{3m}x^2} &= 0 \\ x &= 0 \end{align}$

Find the $y$ -coordinate by substituting this value of $x$ into the definition of $t(x)$ :

$\begin{align} z(0) &= \simplify[basic,fractionnumbers]{{m}*0^3 + {q}} \\ &= \var{q} \end{align}$

Finally, to determine the nature of the stationary point, look at $\frac{\mathrm{d}^2t}{\mathrm{d}x^2}$ at $x = 0$ .

$\frac{\mathrm{d}^2t}{\mathrm{d}x^2} = \simplify{{6m}x}$

$\left.\frac{\mathrm{d}^2t}{\mathrm{d}x^2} \right\rvert_{x=0} = \var{6m} \times 0 = 0$

This is zero, so the stationary point is a point of inflection.

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This question is used in the following exam:

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Josh's copy of Find stationary point of a curve and determine its nature

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Josh Lim 6 years, 3 months ago

Error in variable testing condition

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Generated value: `integer`

Parts

Gap-fill

a)

b)

c)

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Josh's copy of Find stationary point of a curve and determine its nature

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Josh Lim 6 years, 3 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

a)

b)

c)

Generated value: `integer`