// Numbas version: finer_feedback_settings {"name": "Standard time -- 2hrs 2024-25 WM175 Assessment 1", "metadata": {"description": "

Topics:
Algebra, Calculus, Trigonometeric Equations, Complex numbers & Partial Derivatives

Students must complete the exam within 120 mins (standard time).

Questions have variables to produce randomised questions.

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Some quadratics are to be solved by factorising

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Solve the quadratic equations by factorising. Enter answers into the answer matrix in ASCENDING ORDER.

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Factorise $\\simplify{{a[0]*c[0]}x^2+{a[0]*d[0]+b[0]*c[0]}x+ {b[0]*d[0]}}$. [[0]]

\n

Hence solve $\\simplify{{a[0]*c[0]}x^2+{a[0]*d[0]+b[0]*c[0]}x+ {b[0]*d[0]}}=0$.

\n

[[1]]

\n

Use fractions or integers for your answers.

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Factorise $\\simplify{{a[1]*c[1]}x^2+{a[1]*d[1]+b[1]*c[1]}x+ {b[1]*d[1]}}$. [[0]]

\n

Hence solve $\\simplify{{a[1]*c[1]}x^2+{a[1]*d[1]+b[1]*c[1]}x+ {b[1]*d[1]}}=0$.

\n

[[1]]

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This question tests the student's ability to solve simple linear equations by elimination. Part a) involves only having to manipulate one equation in order to solve, and part b) involves having to manipulate both equations in order to solve. 

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Value of $y$ in part b

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$x$ coefficient of the second equation in part a. An integer multiple of the $x$ coefficient of the second equation.

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Value of $x$ in part b

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Constant part of the LHS of the first equation in part a

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RHS of the first equation in part a

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$x$ coefficient in the second equation of part b. Never an integer multiple of the $x$ coefficient in the first equation.

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Value of $x$ in part a

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RHS of the second equation in part a

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$y$ coefficient of the second equation in part b. Never an integer multiple of the $y$ coefficient in the first equation.

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Coefficient of $y$ in the first equation of part b.

\n

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Constant part of the LHS of the second equation in part a

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$x$ coefficient of the first equation in part a

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Value of $x$ in part a

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\n

\\begin{align}
\\simplify{{h}x+{k}y} &= \\var{m} \\text{,} \\\\
\\simplify{{j}x+{l}y} &= \\var{n} \\text{.}
\\end{align}

\n

$x =$ [[0]]

\n

$y =$ [[1]]

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Differentiate $\\displaystyle (ax^m+b)^{n}$.

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Find the derivative of $y$ using the function of a function rule (also known as the chain rule).

\n

Note, you should check that the preview of your answer shown is correct before submitting.  

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\\[\\simplify[std]{y = ({a} * x^{m}+{b})^{n}}\\]

\n

\n

$\\displaystyle{u}=$  [[0]]

\n

$\\displaystyle \\frac{\\mathrm{d}u}{\\mathrm{d}x}=$  [[1]]

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$\\displaystyle \\frac{\\mathrm{d}y}{\\mathrm{d}u}=$  [[2]]

\n

Hence,

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$\\displaystyle \\frac{\\mathrm{d}y}{\\mathrm{d}x}=$  [[3]]

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For help, you may wish to click on Show steps - but be aware this will cost you 2 marks. 

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This example shows how to differentiate a similar function of a function - use it to help you answer the one above.

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\n

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Definite integation of basic functions.

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Make sure your calculator is in radians for this question.

\n

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(a) and (b):

\n

\\[\\int_\\var{d-2}^\\var{d}\\simplify[all]{{a}*x-sin({b}*x)}\\,dx=\\left[\\simplify{{a}*x^2/2+cos({b}*x)/{b}}\\right]_\\var{d-2}^\\var{d}=\\var{a1}\\]

\n

(c) and (d):

\n

\\[\\int_\\var{d}^\\var{d+3}\\simplify[all]{{c}/x^{a+1}+{b}*sqrt(x)}\\,dx=\\left[\\simplify{{-c/a}/x^{a}+{2*b/3}*x^(3/2)}\\right]_\\var{d}^\\var{d+3}=\\var{a2}\\]

\n

(e) and (f):

\n

\\[\\int_{-2}^\\var{a-2}\\simplify[all]{{c+1}*exp(x/{b})-{a}}\\,dx=\\left[\\simplify{{b*(c+1)}*exp(x/{b})-{a}*x}\\right]_{-2}^\\var{a-2}=\\var{a3}\\]

\n

(g) and (h):

\n

\\[\\int_\\var{b}^\\var{b+a}\\simplify[all]{{c}/({b}*x)+{b}/{a+1}*cos({c}*x)}\\,dx=\\left[\\simplify{{c}/{b}*ln(x)+{b}/{c*(a+1)}*sin({c}*x)}\\right]_\\var{b}^\\var{b+a}=\\var{a4}\\]

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(i) and (j):

\n

\\[\\int_\\var{a}^\\var{a+1}\\simplify[all]{x^{b-1}/{b+2}-{d}+{c}*exp({-a}*x)}\\,dx=\\left[\\simplify{x^{b}/{b*(b+2)}-{d}*x-{c/a}*exp({-a}*x)}\\right]_\\var{a}^\\var{a+1}=\\var{a5}\\]

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Integrate this expression with respect to x

\n

Don't forget the constant of integration + c

\n

\\[\\int\\simplify[all]{{a}*x-sin({b}*x)}\\,dx\\]

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Remember that

\n


the integral of    $x^n$   is    $x^{n+1} / {n+1}$ ,  and

\n

the integral of    sin$(ax)$   is    $-$cos$(ax) / a$

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You have forgotten the constant of integration (condoned on this occasion).

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Hence evaluate

\n

\\[\\int_\\var{d-2}^\\var{d}\\simplify[all]{{a}*x-sin({b}*x)}\\,dx\\]

\n

\n

Remember to use radians and give your answer correct to 3 significant figures.

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Example:

\n

The definite intgeral is calculated by substituting $x$ with the values of the limits.
e.g. 

\n

$ \\int_1^4\\ {{x}^4}\\,dx = 4^5 / 5 - 1^4/5=204.6=205$ (3 significant figures)

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Convert degrees to radians and radians to degrees.

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Find the value of

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An angle of $\\var{degrees}^{\\circ}$ corresponds in radians to [[0]] radians.

\n

(Note: you should leave your answer as a fraction in terms of $\\pi,$ which can be entered by typing \"pi\" without the quotation marks)

\n

An angle of $\\frac{\\pi}{\\var{radians}}$ radians corresponds to [[1]] degrees.

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Simple trig equations with radians

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Solve the following equation in radians, for $\\theta$ in the range $0\\leq\\theta\\leq2\\pi$.

\n

Give your answers correct to 2 decimal places in ascending order.

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\\[\\var{c+3}\\sin(\\theta)=-\\var{c}\\]

\n

$\\sin(\\theta)=$ [[0]]

\n

$\\theta=$ [[1]] or [[2]]

\n

Discard any solutions out of range.

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Trigonometric equations with degrees

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Solve the following equation in degrees, for $x$ in the range $0^\\circ\\leq x\\leq720^\\circ$.

\n

\\[\\var{a+1}\\cos(x-\\var{b})=\\var{a-2}\\]

\n

Give your final answers correct to the nearest degree in ascending order.

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$\\cos(x-\\var{b})=$ [[0]]

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$x-\\var{b}=$ [[0]]$^\\circ$, [[1]]$^\\circ$, [[2]]$^\\circ$ or [[3]]$^\\circ$

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$x=$ [[0]]$^\\circ$, [[1]]$^\\circ$, [[2]]$^\\circ$ or [[3]]$^\\circ$

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More difficult trigonometric equations with radians

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Solve the following equations in radians, for $\\theta$ in the range $0\\leq\\theta\\leq\\pi$:

\n

\\[\\var{b}\\cos(3\\theta+\\var{d})=\\var{c}\\]

\n

Give your answers for parts (b) (c) and (d) correct to 2 decimal places in ascending order.

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$\\cos(3\\theta+\\var{d})=$ [[0]]

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$3\\theta+\\var{d}=$ [[0]] , [[1]], [[2]] or [[3]]

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$3\\theta=$ [[0]] , [[1]], [[2]] or [[3]]

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$\\theta=$ [[0]] , [[1]] or [[2]]

\n

(Discard any solutions not in range.)

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Elementary examples of multiplication and addition of complex numbers. Four parts.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Express the answer following in the form $a+bi\\;$ where $a$ and $b$ are real.

\n

Your answer should only include fractions or integers. 

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real part of a+ bi

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imaginary part of a+bi

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$(\\simplify[std]{{a}})(\\simplify[std]{{b}})\\;=\\;$[[0]] + [[1]] i

\n

 

\n

 

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Do not include decimals in your answers, only fractions or integers. Also do not include brackets in your answers.

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Multiplication and addition of complex numbers. Four parts.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Calculate the answer in the form $a+bi\\;$ where $a$ and $b$ are real.

\n

Input numbers as fractions or integers only.

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$\\simplify[std,!otherNumbers]{{a}*({a3} + {b3} * i + {c3} * i ^ 2 + {d3} * i ^ 3)}\\;=\\;$[[0]] +[[1]] i

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Input all numbers as fractions or integers. Also do not include brackets in your answers.

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Inverse and division of complex numbers.  Four parts.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Express the following in the form $a+bi$.

\n

Input $a$ and $b$ as fractions or integers and not as decimals.

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$\\displaystyle \\simplify[std]{{c2}/{z2}}\\;=\\;$[[0]] + [[1]] i

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$\\displaystyle \\simplify[std]{{z3}/{z2}}\\;=\\;$[[0]] + [[1]] i

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Direct calculation of low positive and negative powers of complex numbers. Calculations involving a complex conjugate. Powers of $i$. Four parts.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Express the answers to parts (a) and (b) in the form $a+bi$.

\n

Input $a$ and $b$ as fractions or integers only.

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\\[z=(\\var{z2})^4\\]
$z=\\;\\;$[[0]] + [[1]] i

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\\[z=(\\var{z3})^{\\var{-d1}}\\]
$z=\\;\\;$[[0]] + [[1]] i

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Make sure that you input the real and imaginary parts as fractions and not as decimals

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Use integers or letters only for part (c) answers below.

\n

\n

when  \\[z=i^0\\] $z=\\;\\;$ [[0]]

\n

when  \\[z=i^1\\] $z=\\;\\;$ [[1]]

\n

when  \\[z=i^2\\] $z=\\;\\;$ [[2]]  

\n

when  \\[z=i^3\\]$z=\\;\\;$ [[3]]

\n

Hence, or otherwise calculate

\n

\\[z=i^{\\var{n}}\\]
$z=\\;\\;$[[4]]

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Finding the modulus and argument (in radians) of four complex numbers; the arguments between $-\\pi$ and $\\pi$ and careful with quadrants!

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Find the modulus and argument (in radians) of the following complex number, where the argument lies between $-\\pi$ and $\\pi$.

\n

When calculating the argument pay particular attention to the quadrant in which the complex number lies.

\n

Input final answers to 3 decimal places.

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$|\\var{z3}|=\\;\\;$[[0]], $\\arg(\\var{z3})=\\;\\;$[[1]] radians

\n

Input both answers to 3 decimal places.

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HINT

Remember that

\n

$|\\var{z3}|=\\;\\;sqrt (\\var{c2}^2+\\var{d2}^2)$   and   $\\arg(\\var{z})=\\;\\;tan (\\var{d2}/\\var{c2})$

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Using either algebraic division or the factor theorem, choose which one of these complex numbers is a root of the given equation $f(z)=0$ , and hence find all the roots.

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Given  $\\displaystyle f(z) = \\simplify[std]{z ^ 3 + {( -2) * a1 -c1}*z ^ 2 + {2 * a1 * c1 + a1 ^ 2 + b1 ^ 2} * z -{c1 * (a1 ^ 2 + b1 ^ 2)}}$, one of the following complex numbers is a root $z_1$ of the equation $f(z)=0$.

\n

Choose the correct value for $z_1$:[[0]]

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$\\simplify{{a1}+{b1}i}$

", "

$\\simplify{{x1a1}+{x1b1}i}$

", "

$\\simplify{{x2a1}+{x2b1}i}$

", "

$\\simplify{{x3a1}+{x3b1}i}$

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The remaining roots of $f(z)$ are:

\n

$z_2=\\;\\;$[[0]] (enter the complex root here)

\n

$z_3=\\;\\;$[[1]] (enter the real root here)

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Given two complex numbers, find by inspection the one that is a root of a given quartic real polynomial and hence find the other roots. 

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Consider both methods shown to get the correct answer to the question below.
After looking through both methods, choose which one you prefer and say why, giving a full explanation of your choice.

\n

Remember to compare efficiency, clarity of setting out, the reader's ability to follow the logic used easily in your answer.

Question

\n

Given  $\\displaystyle f(z) = \\simplify[std]{z ^ 4+ 4*z ^ 3+ 28 * z^2 +240z+800}$, one of the following complex numbers is a root $z_1$ of the equation $f(z)=0$.

\n

Choose the correct value for $z_1$:     (i)  $1-4i$      or     (ii)  $-4+2i$

\n


   

\n

\n

\n

", "advice": "

\n

 

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Please give your full reasoning for which of the methods shown above you prefer.

\n

You should aim to write about 50 words in your answer.

\n

Then send it to Shaheen Charlwood via Teams direct chat before the submission time given (usually the end time of the exam).

\n

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Find the stationary point $(p,q)$ of the function: $f(x,y)=ax^2+bxy+cy^2+dx+gy$. Calculate $f(p,q)$.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

In the following question find the $(x,y)$ coordinates of the single stationary point $(a,b)$ of the function

\n

\\[f(x,y)=\\simplify[std]{{a}*x^2+{b}*x*y+{c1}*y^2+{d}*x+{f}*y}\\]

", "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 $y$:

\n

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

\n

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

\n

\\[\\begin{eqnarray*} \\simplify[std]{{2*a}x+{b}y+{d}}&=&0\\\\ \\\\ \\simplify[std]{{b}x+{2*c1}y+{f}}&=&0 \\end{eqnarray*} \\]
On solving these we get \\[ x = \\simplify[std]{{2*c1*d-b*f}/{b^2-4*a*c1}},\\;\\;\\;y=\\simplify[std]{{2*a*f-b*d}/{b^2-4*a*c1}}\\]
On substituting these values into $f(x,y)$ we get:

\n

\\[f\\left(\\simplify[std]{{2*c1*d-b*f}/{b^2-4*a*c1}},\\simplify[std]{{2*a*f-b*d}/{b^2-4*a*c1}}\\right) = \\var{rawstatval}\\approx\\var{statval}\\]
to 2 decimal places.

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Input both cooordinates as fractions or integers and not decimals.

\n

$x$–coordinate, $a=$ [[0]].

\n

$y$–coordinate, $b=$ [[1]].

\n

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Input answer as a fraction or an integer, not a decimal

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Input answer as a fraction or an integer, not a decimal

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Find the stationary points of the function: $f(x,y)=a x ^ 3 + b x ^ 2 y + c y ^ 2 x + dy$ by choosing from a list of points.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "\n

Answer the following questions about the function:

\n

\\[f(x,y)=\\simplify[std]{ ({a} / 3) * x ^ 3 + ({b} / 2) * x ^ 2 * y + {c} * y ^ 2 * x + {d} * y}\\]

\n ", "advice": "

a)

\n

\\[\\begin{eqnarray*} {\\partial f \\over \\partial x} &=&\\simplify[std]{(({a} * (x ^ 2)) + ({b} * x * y) + ({c} * (y ^ 2)))}\\\\ \\\\ \\partial f \\over \\partial y &=&\\simplify[std]{((({b} / 2) * (x ^ 2)) + ({(2 * c)} * x * y) + {d})} \\end{eqnarray*}\\]

\n

b)

\n

$(a,b)$ is a stationary point for the function $f(x,y)$ if $f_x=0,\\;\\;f_y=0$,where the partial derivatives are evaluated at $x=a,\\;\\;y=b$.
So you have to make sure that both these partial derivatives are $0$ at the stationary point.

\n

For this example we have from the above equations that:
\\[\\begin{eqnarray*} \\simplify[std]{(({a} * (x ^ 2)) + ({b} * x * y) + ({c} * (y ^ 2)))}&=&0,\\qquad &\\mathbf{(1)}&\\\\ \\\\ \\simplify[std]{((({b} / 2) * (x ^ 2)) + ({(2 * c)} * x * y) + {d})}&=&0, \\qquad &\\mathbf{(2)}& \\end{eqnarray*}\\]

\n

The left hand side of equation (1) can be factorised as:

\n

\\[\\simplify[std]{({a1}x+{b1}y)*({c1}x+{d1}y)=0}\\]

\n

and so we have:
\\[y=\\simplify[std]{{-a1}/{b1}*x},\\mbox{ or } y= \\simplify[std]{{-c1}/{d1}*x}\\]

\n
First case: $y= \\simplify[std]{{-a1}/{b1}*x}$
\n

Substituting this into equation (2) gives:
\\[\\simplify[std]{{b}/2*x^2-{2c*a1}/{b1}*x^2+{d}}=0 \\Rightarrow \\simplify[std]{{-b*b1+4*c*a1}/{2*b1}*x^2={d}}\\]

\n

Hence $x=\\var{m}\\mbox{ or } x = \\var{-m}$ and the stationary points which are on the list and which you had to choose are:
\\[\\left(\\var{m},\\simplify[std]{-{a1*m}/{b1}}\\right)\\mbox{ and }\\left(\\var{-m},\\simplify[std]{{a1*m}/{b1}}\\right)\\]

\n
Second case: $y= \\simplify[std]{{-c1}/{d1}*x}$
\n

{check}

\n

Substituting this into equation (2) gives:
\\[\\simplify[std]{{b}/2*x^2-{2c*c1}/{d1}*x^2+{d}}=0 \\Rightarrow \\simplify[std]{{-b*d1+4*c*c1}/{2*d1}*x^2={d}}\\]

\n

{other}

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Enter the partial derivatives here. Note if you want to enter a product of unknowns, such as $xy$ then you input the expression in the form x*y.

\n

$\\displaystyle { \\partial f \\over \\partial x}=$ [[0]]

\n

$\\displaystyle {\\partial f \\over \\partial y}=$ [[1]]

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Finding Stationary Points.

\n

Tick the two choices which give stationary points for $f(x,y)$.

\n

Note that the easiest way to do this question is to substitute the values for $x$ and for $y$ into the expressions for $\\displaystyle {\\partial f \\over \\partial x}$ and $\\displaystyle{\\partial f \\over \\partial y}$ and see if you get $0$ for both.

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$x=\\var{m},\\;\\;y=\\simplify[std]{-{a1*m}/{b1}}$

", "

$x=\\var{-m},\\;\\;y=\\simplify[std]{{a1*m}/{b1}}$

", "

$x=\\var{m+1},\\;\\;y=\\simplify[std]{-{c1*(m+1)}/{d1}}$

", "

$x=\\var{-m-1},\\;\\;y=\\simplify[std]{{c1*(m+1)}/{d1}}$

", "

$x=\\var{m-1},\\;\\;y=\\simplify[std]{-{a1+2*b1}/{b1}}$

", "

$x=\\var{-m+1},\\;\\;y=\\simplify[std]{{a1+2*b1}/{b1}}$

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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 $y$

\n

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

\n

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

\n

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

\n

On solving these we get \\[ x = \\var{c},\\;\\;\\;y=\\var{d}\\]

\n

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

\n

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

\n

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

\n ", "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "parts": [{"stepsPenalty": 0, "prompt": "

$x$–coordinate, $a=$ [[0]]

\n

$y$–coordinate, $b=$ [[1]]

\n

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

\n

$f(a,b)=$ [[2]]

\n

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

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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 $y$

\n \n \n \n

\\[\\begin{eqnarray*}\n \n \\partial f \\over \\partial x &=&0\\\\\n \n \\\\\n \n \\partial f \\over \\partial y &=&0\n \n \\end{eqnarray*}\n \n \\]

\n \n \n \n

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

\n \n \n ", "marks": 0}], "type": "gapfill"}], "statement": "

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

\n

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

\n

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

\n

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

\n

Input both cooordinates as fractions or integers and not decimals.

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10/07/2012:

\n \t\t

Added tags.

\n \t\t

Question appears to be working correctly.

\n \t\t

 

\n \t\t", "description": "

Find the coordinates of the stationary point for $f: D \\rightarrow \\mathbb{R}$: $f(x,y) = a + be^{-(x-c)^2-(y-d)^2}$, $D$ is a disk centre $(c,d)$.

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This assessment has now finished. 

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