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Science:Math Exam Resources/Courses/MATH105/April 2011/Question 07

2012-03-07T18:49:26Z

CameronChristou:

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Science:Math Exam Resources/Courses/MATH105/April 2011/Question 07/Solution 1

2012-03-07T18:48:53Z

CameronChristou:

The derivative of <math> f </math> with respect to <math> x </math> is
:<math> f_{x}(x,y) = (1/x) + 2bx + 3ay </math>.
Also, the derivative of <math> f </math> with respect to <math> y </math> is
:<math>f_{y}(x,y) = (2/y) + 3ax</math>/

For a critical point at <math> (1,3) </math>, we require that <math> f_{x}(1,3) = f_{y}(1,3) = 0 </math>. Using the above two equations, this gives us the following system to solve for <math> a </math> and <math> b </math>:
#<math> f_{x}(1,3) = (1/1) + 2b(1) + 3a(3) = 1 + 2b + 9a = 0</math>
#<math> f_{y}(1,3) = (2/3) + 3a(1) = 2/3 + 3a = 0</math>.

Equation 2 tells us that <math> a = -2/9 </math>.

Substituting this value for <math> a </math> into equation 1 tells us that <math>1 + 2b + 9(-2/9) = 1 + 2b - 2 = 2b - 1 = 0</math>. Therefore, we have that <math> b = 1/2 </math>.

This critical point is a saddle point. To show this, we solve for the second derivatives <math> f_{xx}, f_{xy} </math> and <math> f_{yy} </math>.

First,
:<math> f_{xx}(x,y) = -(1/x)^{2} + 2b </math>.
Therefore,
:<math>f_{xx}(1,3) = -(1) + 2(1/2) = 0</math>.

Second,
:<math> f_{yy}(x,y) = -2(1/y)^{2} </math>.
Therefore,
:<math> f_{yy}(1,3) = -2/9 </math>

Third,
:<math> f_{xy}(x,y) = 3a </math>.
Therefore,
:<math> f_{xy}(1,3) = -2/3</math>.

So the Hessian matrix of <math> f</math> is
:<math> \left[\begin{array}{cc} 0 & -2/3 \\ -2/3 & -2/9 \end{array}\right]</math>.

The eigenvalues of this matrix are given by the solving the formula <math> (-\lambda)(-2/9 - \lambda) -4/9 = 0</math> for <math> \lambda </math>.

That is,
:<math> \lambda^2 + (2/9)\lambda - 4/9 = 0</math>.
Solving gives us the answer
:<math> \lambda = (1/9)(\pm \sqrt{37} - 1) </math>.

Because one eigenvalue is positive and one eigenvalue is negative, we conclude that the Hessian matrix is indefinite, and thus the point <math> (1,3) </math> is a saddle point.

Science:Math Exam Resources/Courses/MATH105/April 2011/Question 07/Solution 1

2012-03-07T18:48:05Z

CameronChristou:

The derivative of <math> f </math> with respect to <math> x </math> is
:<math> f_{x}(x,y) = (1/x) + 2bx + 3ay </math>.
Also, the derivative of <math> f </math> with respect to <math> y </math> is
:<math>f_{y}(x,y) = (2/y) + 3ax</math>/

For a critical point at <math> (1,3) </math>, we require that <math> f_{x}(1,3) = f_{y}(1,3) = 0 </math>. Using the above two equations, this gives us the following system to solve for <math> a </math> and <math> b </math>:
#<math> f_{x}(1,3) = (1/1) + 2b(1) + 3a(3) = 1 + 2b + 9a = 0</math>
#<math> f_{y}(1,3) = (2/3) + 3a(1) = 2/3 + 3a = 0</math>.

Equation 2 tells us that <math> a = -2/9 </math>.

Substituting this value for <math> a </math> into equation 1 tells us that <math>1 + 2b + 9(-2/9) = 1 + 2b - 2 = 2b - 1 = 0</math>. Therefore, we have that <math> b = 1/2 </math>.

This critical point is a saddle point. To show this, we solve for the second derivatives <math> f_{xx}, f_{xy} </math> and <math> f_{yy} </math>.

First,
:<math> f_{xx}(x,y) = -(1/x)^{2} + 2b </math>.
Therefore,
:<math>f_{xx}(1,3) = -(1) + 2(1/2) = 0</math>.

Second,
:<math> f_{yy}(x,y) = -2(1/y)^{2} </math>.
Therefore,
:<math> f_{yy}(1,3) = -2/9 </math>

Third,
:<math> f_{xy}(x,y) = 3a </math>.
Therefore,
:<math> f_{xy}(1,3) = -2/3</math>.

So the Hessian matrix of <math> f</math> is
:<math> \left[\begin{array}{cc} 0 & -2/3 \\ -2/3 & -2/9 \end{array}\right]</math>.

The eigenvalues of this matrix are given by the solving the formula <math> (-\lambda)(-2/9 - \lambda) -4/9 = 0</math> for </math> \lambda </math>.

That is,
:<math> \lambda^2 + (2/9)\lambda - 4/9 = 0</math>.
Solving gives us the answer
:<math> \lambda = (1/9)(\pm \sqrt{37} - 1) </math>.

Because one eigenvalue is positive and one eigenvalue is negative, we conclude that the Hessian matrix is indefinite, and thus the point <math> (1,3) </math> is a saddle point.

Science:Math Exam Resources/Courses/MATH105/April 2011/Question 07/Solution 1

2012-03-07T18:26:08Z

CameronChristou: Created page with "The derivative of <math> f </math> with respect to <math> x </math> is :<math> f_{x}(x,y) = (1/x) + 2bx + 3ay </math>. Also, the derivative of <math> f </math> with respect t..."

Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 27

2012-03-07T00:33:17Z

CameronChristou:

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Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 27/Statement

2012-03-07T00:32:57Z

CameronChristou: Created page with "Which of the following statements are true for all invertible ''n''x''n'' invertible matrices <math> A</math>: * <math> A </math> has rank ''n'' * The reduced row echelon f..."

Which of the following statements are true for all invertible ''n''x''n'' invertible matrices <math> A</math>:

* <math> A </math> has rank ''n''
* The reduced row echelon form of <math> A </math> is the identity matrix
* The homogeneous problem <math> Ax = 0</math> has only the trivial solution <math> x = 0</math>
* <math> A </math> does not have <math> 0 </math> as an eigenvalue
* <math> \det(A) \neq 0 </math>

Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 21

2012-03-07T00:27:06Z

CameronChristou:

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Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 21/Solution 1

2012-03-07T00:26:41Z

CameronChristou: Created page with "One would use the backslash operator “\”, and store the answer into a vector called <math> \textbf{x} </math>. The full command in your MATLAB console would look like :..."

One would use the backslash operator “\”, and store the answer into a vector called <math> \textbf{x} </math>. The full command in your MATLAB console would look like

:x = A\b;

Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 21

2012-03-07T00:22:21Z

CameronChristou:

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Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 21/Statement

2012-03-07T00:21:52Z

CameronChristou: Created page with "Suppose that <math> \textbf{A} </math> is a 12 x12 invertible matrix and <math> \textbf{b} </math> is a column vector with 12 entries. Both <math> \textbf{A} </math> and <mat..."

Suppose that <math> \textbf{A} </math> is a 12 x12 invertible matrix and <math> \textbf{b} </math> is a column vector with 12 entries. Both <math> \textbf{A} </math> and <math> \textbf{b} </math> have been entered into MATLAB.

What command(s) would you use to compute the vector <math> \textbf{x} </math> that solves the linear system
:<math> \textbf{Ax} = \textbf{b} </math>?

Science:Math Exam Resources/Courses/MATH101/April 2005/Question 02 (d)

2012-03-07T00:12:26Z

CameronChristou:

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Science:Math Exam Resources/Courses/MATH101/April 2005/Question 02 (d)/Statement

2012-03-07T00:12:05Z

CameronChristou: Created page with "Let <math> \displaystyle R</math> be the finite region bounded by the curve :<math> \displaystyle y = 4 - x^{2} </math> and bounded below by :<math> \displaystyle y = 2-x </ma..."

Let <math> \displaystyle R</math> be the finite region bounded by the curve
:<math> \displaystyle y = 4 - x^{2} </math>
and bounded below by
:<math> \displaystyle y = 2-x </math>.
Express the length of the upper curve that bounds <math> \displaystyle R </math> as a definite integral, and using an appropriate substitution express your answer as an integral involving trigonometric functions. You do not need to evaluate this trigonometric integral.

Science:Math Exam Resources/Courses/MATH101/April 2005/Question 02 (c)

2012-03-07T00:09:39Z

CameronChristou:

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Science:Math Exam Resources/Courses/MATH101/April 2005/Question 02 (c)/Statement

2012-03-07T00:08:50Z

Let <math> \displaystyle R</math> be the finite region bounded by the curve
:<math> \displaystyle y = 4 - x^{2} </math>
and bounded below by
:<math> \displaystyle y = 2-x </math>.
Express the volume of the solid obtained by rotating <math> \displaystyle R</math> about the vertical line <math> \displaystyle x = 2</math> as a definite integral. You do not need to simplify or evaluate this integral.

Science:Math Exam Resources/Courses/MATH101/April 2005/Question 02 (b)

2012-03-07T00:07:06Z

CameronChristou:

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Science:Math Exam Resources/Courses/MATH101/April 2005/Question 02 (b)/Statement

2012-03-07T00:06:41Z

Let <math> \displaystyle R</math> be the finite region bounded by the curve
:<math> \displaystyle y = 4 - x^{2} </math>
and bounded below by
:<math> \displaystyle y = 2-x </math>.
Express the volume of the solid obtained by rotating <math>\displaystyle R</math> about the <math> x </math>-axis as a definite integral. You do not need to simplify or evaluate this integral.

Science:Math Exam Resources/Courses/MATH101/April 2005/Question 02 (a)

2012-03-07T00:04:11Z

CameronChristou:

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Science:Math Exam Resources/Courses/MATH101/April 2005/Question 02 (a)/Statement

2012-03-07T00:03:53Z

Science:Math Exam Resources/Courses/MATH101/April 2005/Question 01 (e)

2012-03-06T23:57:41Z

CameronChristou:

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Science talk:Math Exam Resources/Courses/MATH101/April 2005/Question 1 (e)

2012-03-06T23:57:26Z

CameronChristou: Talk page autocreated when first thread was posted

Thread:Science talk:Math Exam Resources/Courses/MATH101/April 2005/Question 1 (e)/question interpretation

2012-03-06T23:57:25Z

CameronChristou: New thread: question interpretation

The exam has a flaw in the question and is missing significant pieces. I chose to put in a + for the missing operator in the equation. Using a - instead only changes a sign in the solution, but otherwise is the exact same solution.

Science:Math Exam Resources/Courses/MATH101/April 2005/Question 01 (e)/Statement

2012-03-06T23:55:05Z

CameronChristou: Created page with "Find the general solution <math>\displaystyle y = y(x) </math> of the differential equation :<math>y^{\prime\prime} + 2y^{\prime} + y = x </math>."

Find the general solution <math>\displaystyle y = y(x) </math> of the differential equation
:<math>y^{\prime\prime} + 2y^{\prime} + y = x </math>.

Science:Math Exam Resources/Courses/MATH101/April 2005/Question 01 (g)

2012-03-06T23:49:48Z

CameronChristou:

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Science:Math Exam Resources/Courses/MATH101/April 2005/Question 01 (g)/Statement

2012-03-06T23:49:31Z

CameronChristou: Created page with "A continuous random variable <math> \displaystyle X </math> is exponentially distributed with a mean of <math> 4 </math>. Find the probability that <math> X \geq 8</math>."

A continuous random variable <math> \displaystyle X </math> is exponentially distributed with a mean of <math> 4 </math>.

Find the probability that <math> X \geq 8</math>.

Science:Math Exam Resources/Courses/MATH101/April 2008/Question 01 (b)

2012-03-06T20:35:48Z

CameronChristou:

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Science:Math Exam Resources/Courses/MATH101/April 2008/Question 01 (b)

2012-03-06T20:35:35Z

CameronChristou:

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Thread:Science talk:Math Exam Resources/Courses/MATH220/April 2011/Question 5/Be more explicit please/reply

2012-03-06T19:09:17Z

CameronChristou: Reply to Be more explicit please

Also, solutions should be separated into distinct answer boxes.

Thread:Science talk:Math Exam Resources/Courses/MATH103/April 2011/Question 3 (a)/Solution review/reply (3)

2012-03-06T19:07:16Z

CameronChristou: Reply to Solution review

The official exam solution for this question gave points for identifying the roots of the equation, the local max and min (they didn't have to compute derivatives and solve for critical points, but had to mark them on the graph), and for showing that the functions increased / decreased without bound appropriately as |x| gets bigger. At that point, as long as they connected the dots they were allotted full marks.

This was meant just as a visual aid for part b of this question, so that they could pick out the correct bounds for integration.

Science:Math Exam Resources/Courses/MATH220/April 2011/Question 10 (b)

2012-03-06T18:53:24Z

CameronChristou: Created page with "  <!-- second letter is for ..."

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Science:Math Exam Resources/Courses/MATH220/April 2011/Question 10 (a)

2012-03-06T18:52:20Z

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Science:Math Exam Resources/Courses/MATH220/April 2011/Question 09 (c)

2012-03-06T18:52:04Z

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Science:Math Exam Resources/Courses/MATH220/April 2011/Question 09 (b)

2012-03-06T18:51:51Z

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Science:Math Exam Resources/Courses/MATH220/April 2011/Question 09 (a)

2012-03-06T18:51:35Z

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Science:Math Exam Resources/Courses/MATH220/April 2011/Question 08

2012-03-06T18:50:20Z

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Science:Math Exam Resources/Courses/MATH220/April 2011/Question 07 (c)

2012-03-06T18:48:02Z

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Science:Math Exam Resources/Courses/MATH220/April 2011/Question 07 (b)

2012-03-06T18:47:26Z

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Science:Math Exam Resources/Courses/MATH220/April 2011/Question 07 (a)

2012-03-06T18:47:08Z

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Science:Math Exam Resources/Courses/MATH103/April 2011/Question 07 (c)

2012-03-05T05:48:14Z

CameronChristou:

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Science:Math Exam Resources/Courses/MATH103/April 2011/Question 07 (c)

2012-03-05T05:47:51Z

CameronChristou:

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Science:Math Exam Resources/Courses/MATH103/April 2011/Question 07 (c)/Solution 2

2012-03-05T05:27:20Z

CameronChristou: Created page with "The first solution assumes that the Taylor series for <math> \cos(x) </math> is not known in advance. The Taylor series for <math> \sin(x) </math> is known since it was given..."

The first solution assumes that the Taylor series for <math> \cos(x) </math> is not known in advance. The Taylor series for <math> \sin(x) </math> is known since it was given in the statement of a previous question.

If the Taylor series for <math> \cos(x) </math> is known, then we can start the solution from the starting point

<math> \cos(x) = \sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k)!} </math>.

At this point, we multiply the sum by <math> x </math> to obtain

<math> x\cos(x) = x \sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k)!} = \sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k+1}}{(2k)!} </math>.

Science:Math Exam Resources/Courses/MATH220/April 2011/Question 06

2012-03-05T05:17:19Z

CameronChristou:

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Science:Math Exam Resources/Courses/MATH220/April 2011/Question 06/Statement

2012-03-05T05:17:03Z

CameronChristou: Created page with "Let <math> n \in \mathbb{Z}</math>. Prove that <math> n \equiv 3 ({\rm mod} 5)</math> if and only if <math> 3n + 1 </math> is divisible by <math> 5</math>."

Let <math> n \in \mathbb{Z}</math>. Prove that <math> n \equiv 3 ({\rm mod} 5)</math> if and only if <math> 3n + 1 </math> is divisible by <math> 5</math>.

Science:Math Exam Resources/Courses/MATH220/April 2011/Question 06

2012-03-05T05:15:24Z

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Science:Math Exam Resources/Courses/MATH220/April 2011/Question 05

2012-03-05T05:14:35Z

CameronChristou:

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Science:Math Exam Resources/Courses/MATH220/April 2011/Question 05/Statement

2012-03-05T05:14:15Z

CameronChristou: Created page with "Prove that the function <math> f : \mathbb{R} - \{1\} \to \mathbb{R} - \{2\} </math> given by <math> f(x) = \frac{2x}{x-1} </math> is bijective."

Prove that the function <math> f : \mathbb{R} - \{1\} \to \mathbb{R} - \{2\} </math> given by <math> f(x) = \frac{2x}{x-1} </math> is bijective.

Science:Math Exam Resources/Courses/MATH220/April 2011/Question 05

2012-03-05T05:12:33Z

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Science:Math Exam Resources/Courses/MATH220/April 2011/Question 04

2012-03-05T05:12:01Z

CameronChristou:

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Science:Math Exam Resources/Courses/MATH220/April 2011/Question 04/Statement

2012-03-05T05:11:44Z

CameronChristou: Created page with "Prove that <math> P \Rightarrow ( Q \vee R ) \equiv ( P \Rightarrow Q ) \vee ( P \Rightarrow R) </math>."

Prove that

<math> P \Rightarrow ( Q \vee R ) \equiv ( P \Rightarrow Q ) \vee ( P \Rightarrow R) </math>.

Science:Math Exam Resources/Courses/MATH220/April 2011/Question 04

2012-03-05T05:08:48Z

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Science:Math Exam Resources/Courses/MATH220/April 2011/Question 01 (j)

2012-03-04T00:37:32Z

CameronChristou:

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