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Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 03/Solution 1

2025-04-23T20:33:56Z

MatthewColes:

If we imagine that the integrand is the side of a right-angled triangle, then the other sides of the triangle have length <math> x </math> and 3, and 3 is the length of the hypotenuse. We use this picture to guess the following substitution:

<math display="block>\begin{align}
\frac{x}{3} &= \sin(\theta)\\
\mathrm{d}x &= 3\cos(\theta) \mathrm{d}\theta
\end{align}</math>

We thus have

<math display="block">
\int \sqrt{9-x^2} \mathrm{d}x = \int \sqrt{9 - 9\sin^2(\theta)} \cdot 3 \cos(\theta)\mathrm{d}\theta = \int 9 \cos^2(\theta).\mathrm{d}\theta
</math>

We now use the trigonometric identity <math> \cos^2(\theta) = \frac{1+\cos(2\theta)}{2}</math> to evaluate

<math display="block">
\int 9\cos^2(\theta) \mathrm{d}\theta = \frac{9}{2} \int \left(1 + \cos(2\theta) \right) \mathrm{d}\theta = \frac{9}{2} \left( \theta + \frac{\sin(2\theta)}{2} \right) +C.
</math>

We have <math> \theta = \arcsin\left(\frac{x}{3}\right)</math>, so, we can rewrite the answer in terms of <math>x</math>:

<math display="block">
\int \sqrt{9-x^2} \mathrm{d}x = \frac{9}{2}\left( \arcsin\left(\frac{x}{3}\right) + \frac{\sin\left(2\arcsin\left(\frac{x}{3}\right)\right)}{2}\right) +C = \frac{9}{2}\arcsin\left(\frac{x}{3}\right) + \frac{9}{4} \left(2\sin\left(\arcsin\left(\frac{x}{3}\right)\right)\cos\left(\arcsin\left(\frac{x}{3}\right)\right)\right) +C,
</math>

where, in the last step, we used the trigonometric formula <math> \sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha)</math> with <math> \alpha = \beta = \arcsin(x/3)</math>. By definition of the arcsine, <math>\sin\left(\arcsin\left(\frac{x}{3}\right)\right) = \frac{x}{3}</math>. But this is also to be expected, given our interpretation of <math>x</math> as the length of the side opposite to <math>\theta</math> in a right-angled triangle with hypotenuse of length 3. This picture also shows that

<math display="block"> \cos\left(\arcsin\left(\frac{x}{3}\right)\right) = \frac{\sqrt{9-x^2}}{3}.</math>

Putting it all together, we find

<math display="block">
\int\sqrt{9-x^2} \mathrm{d}x = \frac{9}{2}\arcsin\left(\frac{x}{3}\right) + \frac{9}{2}\left(\frac{x}{3} \cdot \frac{\sqrt{9-x^2}}{3}\right) + C,
</math>

which is an equation that makes sense for <math> -3 \leq x \leq 3</math>.

Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 03/Solution 1

2025-04-23T20:32:29Z

MatthewColes:

If we imagine that the integrand is the side of a right-angled triangle, then the other sides of the triangle have length <math> x </math> and 3, and 3 is the length of the hypotenuse. We use this picture to guess the following substitution:

<math display="block>\begin{align}
\frac{x}{3} &= \sin(\theta)\\
\mathrm{d}x &= 3\cos(\theta) \mathrm{d}\theta
\end{align}</math>

We thus have

<math display="block">
\int \sqrt{9-x^2} \mathrm{d}x = \int \sqrt{9 - 9\sin^2(\theta)} \cdot 3 \cos(\theta)\mathrm{d}\theta = \int 9 \cos^2(\theta).\mathrm{d}\theta
</math>

We now use the trigonometric identity <math> \cos^2(\theta) = \frac{1+\cos(2\theta)}{2}</math> to evaluate

<math display="block">
\int 9\cos^2(\theta) \mathrm{d}\theta = \frac{9}{2} \int \left(1 + \cos(2\theta) \right) \mathrm{d}\theta = \frac{9}{2} \left( \theta + \frac{\sin(2\theta)}{2} \right).
</math>

We have <math> \theta = \arcsin\left(\frac{x}{3}\right)</math>, so, we can rewrite the answer in terms of <math>x</math>:

<math display="block">
\int \sqrt{9-x^2} \mathrm{d}x = \frac{9}{2}\left( \arcsin\left(\frac{x}{3}\right) + \frac{\sin\left(2\arcsin\left(\frac{x}{3}\right)\right)}{2}\right) = \frac{9}{2}\arcsin\left(\frac{x}{3}\right) + \frac{9}{4} \left(2\sin\left(\arcsin\left(\frac{x}{3}\right)\right)\cos\left(\arcsin\left(\frac{x}{3}\right)\right)\right),
</math>

where, in the last step, we used the trigonometric formula <math> \sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha)</math> with <math> \alpha = \beta = \arcsin(x/3)</math>. By definition of the arcsine, <math>\sin\left(\arcsin\left(\frac{x}{3}\right)\right) = \frac{x}{3}</math>. But this is also to be expected, given our interpretation of <math>x</math> as the length of the side opposite to <math>\theta</math> in a right-angled triangle with hypotenuse of length 3. This picture also shows that

<math display="block"> \cos\left(\arcsin\left(\frac{x}{3}\right)\right) = \frac{\sqrt{9-x^2}}{3}.</math>

Putting it all together, we find

<math display="block">
\int\sqrt{9-x^2} \mathrm{d}x = \frac{9}{2}\arcsin\left(\frac{x}{3}\right) + \frac{9}{2}\left(\frac{x}{3} \cdot \frac{\sqrt{9-x^2}}{3}\right) + C,
</math>

which is an equation that makes sense for <math> -3 \leq x \leq 3</math>.

Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 02

2024-06-12T23:52:11Z

MatthewColes:

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Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 02/Solution 1

2024-06-12T23:50:46Z

MatthewColes:

The two parts of <math display="inline">f(x)</math> are continuous, so we only need to make sure that <math display="inline">f(x)</math> is continuous at <math display="inline">x=3</math>. That is <math display="inline">\lim_{x \to 3^{-}}f(x) = \lim_{x \to 3^{+}}f(x) = f(3)</math>. Again, since both parts are continuous this equality will be achieved if the two parts are equal. We set <math display="inline">c\cdot3+3 = c\cdot3^2 - 3</math>, and solve for <math display="inline">c</math>:

<math display="block"> \begin{align}
c\cdot3 + 3 &= c(3)^2 - 3 \\
3c + 3 &= 9c - 3 \\
6c &= 6 \\
c &= 1
\end{align} </math>

Therefore, <math display="inline">c=1</math> makes <math display="inline">f(x)</math> continuous.

Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 02/Solution 1

2024-06-12T23:50:08Z

MatthewColes:

The two parts of <math display="inline">f(x)</math> are continuous, so we only need to make sure that <math display="inline">f(x)</math> is continuous at <math display="inline">x=3</math>. That is <math display="inline">\lim_{x \to 3^{-}}f(x) = \lim_{x \to 3^{+}f(x) = f(3)</math>. Again, since both parts are continuous this equality will be achieved if the two parts are equal. We set <math display="inline">c\cdot3+3 = c\cdot3^2 - 3</math>, and solve for <math display="inline">c</math>:

<math display="block"> \begin{align}
c\cdot3 + 3 &= c(3)^2 - 3 \\
3c + 3 &= 9c - 3 \\
6c &= 6 \\
c &= 1
\end{align} </math>

Therefore, <math display="inline">c=1</math> makes <math display="inline">f(x)</math> continuous.

Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 01

2024-06-12T23:39:45Z

MatthewColes:

Science:Math Exam Resources/Courses/MATH102/December 2013/Question B 01

2014-11-15T01:18:31Z

MatthewColes:

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Science:Math Exam Resources/Courses/MATH102/December 2013/Question B 01/Solution 1

2014-11-15T01:17:12Z

MatthewColes: Created page with "Differentiate both sides of <math> \displaystyle \tan y =x </math> with respect to <math> \displaystyle x </math>. Observe <math> \displaystyle \begin{align} \fra..."

Differentiate both sides of <math> \displaystyle \tan y =x </math> with respect to <math> \displaystyle x </math>. Observe
 
 
<math> \displaystyle
\begin{align}
\frac{d}{dx} \tan y &= \frac{d}{dx} x \\
\sec^2 y \frac{dy}{dx} &=1 \\
\frac{dy}{dx} &= \frac{1}{\sec^2x}.
\end{align}
</math>
 
 
Now make use of the trig identity <math> \displaystyle \sec^2 y =1 + \tan^2 y </math> to see
 
 
<math> \displaystyle
\begin{align}
\frac{dy}{dx} = \frac{1}{1+\tan^2y} = \frac{1}{1+x^2}
\end{align}
</math>
after noting that <math> \displaystyle \tan y =x </math>.

Science:Math Exam Resources/Courses/MATH104/Syllabus

2014-10-06T18:39:10Z

MatthewColes: Created page with "Introduction: Review of Exponentials, Logarithms, and Inverse Functions A standard business problem from managerial economics An Introduction to Limits Continuo..."

Introduction: Review of Exponentials, Logarithms, and Inverse Functions 
A standard business problem from managerial economics 
An Introduction to Limits 
Continuous Functions 
the intermediate Value Thm 
The Derivative 
Rules of Differentiation 
Derivative as rate of change 
The Chain Rule 
Implicit Differentiation 
Derivatives of Logarithms and Exponentials 
Derivatives of Logarithms and Exponentials Continued 
Applications: Elasticity of Demand 
Exponential Growth and Compound Interest 
Related Rates 
Maxima and Minima 
Asymptotes 
Graphing functions 
Optimization problems I 
Optimization Problems Continued 
Linear Approximation 
Approximating Functions with polynomials 
Approximating Functions with polynomials Continued 
Inverse Trigonometric Functions

Science:Math Exam Resources/Courses/MATH102/Syllabus

2014-10-06T18:32:29Z

MatthewColes: Created page with "Cell size: volume, area. Power functions Power functions (cont). Sketching simple polynomials (y=x^3-ax) Sketching simple polynomials (cont). Rational functions, Mic..."

Cell size: volume, area. Power functions 
Power functions (cont). Sketching simple polynomials (y=x^3-ax) 
Sketching simple polynomials (cont). Rational functions, Michaelis-Menten and Hill functions, “limits” at infinity 
Average rate of change and secant lines. Definition of the derivative. Instantaneous rate of change 
Limits and continuity, examples. One example of computing derivative of y=ct2 from the definition 
Derivatives: analytic, and geometric (zoom in on a point). Sketching f′(x) given f(x) (intro) 
Derivatives (cont): computational (spreadsheet example in class). More examples of sketching f′(x) given f(x) (intro) 
Rules of differentiation: Product and quotient rules. Antiderivatives of power functions. Application to falling ball, motion of Listeria. Sketching f(x) given f′(x) (intro using polynomial) 
Tangent lines and linear approximation 
Introduction to Newton’s method. Sketching the graph of a function using calculus tools: increasing, decreasing, critical points 
Sketching the graph of a function using calculus tools (cont): concavity and inflection points 
Putting it all together - sketching 
Finish sketching functions. Introduce simple optimization problem(s) 
More optimization examples including those with a constraint and those on bounded intervals. Distinction between absolute (global) and local minima and maxima 
Kepler's Wedding - A wine optimization problem 
Optimal Foraging 
Least Squares - finding the mean of a data set 
Least Squares - finding the best fitting line y=ax through a set of data points 
Chain Rule: examples, applications to optimization problems 
More Chain Rule: Related Rates and Implicit differentiation 
Exponential functions: intro and motivation, derivative of exponential functions 
Inverse functions and logarithm, applications of logs 
Exponential growth and decay, intro to differential equations, population growth and/or other examples 
Solving differential equations of the type dy/dt=a−by 
Newton's Law of Cooling (Murder Mystery example) 
Complete and/or review above topics 
Solving differential equations approximately using Euler's Method 
Introduction to nonlinear ODEs, qualitative analysis 
Slope fields with logistic equation as example 
State-space diagrams and examples (logistic) 
Disease dynamics 
Review of differential equations and/or complete above topics 
Introduction to Trigonometric Functions 
Trigonometric Functions and cyclic processes, phase, amplitude, etc. (fitting a sin or cos to a cyclic process), Inverse trig functions 
Derivatives of trig functions, related rates examples 
The Escape Response and inverse trig functions 
Second order ODEs. Complete and/or review trig

Science:Math Exam Resources/Courses/MATH100/Syllabus

2014-10-06T18:22:03Z

MatthewColes:

The Tangent and Velocity Problems 
The Limit of a Function 
Calculating Limits Using the Limit Laws 
Continuity 
Limits at Infinity; Horizontal Asymptotes 
Derivatives and Rates of Change 
The Derivative as a Function 
Derivatives of Polynomials and Exponential Functions 
The Product and Quotient Rules 
Derivatives of Trigonometric Functions 
The Chain Rule 
Inverse Functions and Logarithms 
Implicit Differentiation 
Derivatives of Logarithmic Functions 
Rates of Change in the Natural and Social Sciences 
Exponential Growth and Decay 
Related Rates 
Linear Approximations and Differentials 
Taylor Polynomials 
Taylor’s Formula with Remainder 
Maximum and Minimum Values 
The Mean Value Theorem 
How Derivatives Affect the Shape of a Graph (First and Second Derivative Tests) 
Summary of Curve Sketching 
Optimization Problems 
Newton’s Method 
Antiderivatives

Science:Math Exam Resources/Courses/MATH100/Syllabus

2014-10-06T18:16:50Z

MatthewColes: Created page with "Syllabus"

Syllabus

Science:Math Exam Resources/Courses/MATH110/Syllabus

2014-10-06T18:11:08Z

MatthewColes:

Lines in the Plane, Functions 
Tangent Lines, Velocities, Growth 
Evaluating limits 
Continuity 
Tangent lines and the idea of derivative 
The definition of derivative 
The Power, Product and Quotient Rules 
Trigonometry 
Exponentials 
The Chain Rule 
Implicit differentiation 
Extrema 
The Mean Value Theorem 
The First and Second Derivative Tests 
Concavity 
Asymptotes 
Curve sketching 
Optimization I 
Optimization II 
Related rates I 
Related rates II 
Approximations I 
Approximations II

Science:Math Exam Resources/Courses/MATH110/Syllabus

2014-10-06T18:08:03Z

MatthewColes:

Lines in the Plane, Functions

Tangent Lines, Velocities, Growth
Evaluating limits
Continuity
Tangent lines and the idea of derivative
The definition of derivative
The Power, Product and Quotient Rules
Trigonometry
Exponentials
The Chain Rule
Implicit differentiation
Extrema
The Mean Value Theorem
The First and Second Derivative Tests
Concavity
Asymptotes
Curve sketching
Optimization I
Optimization II
Related rates I
Related rates II
Approximations I
Approximations II

Science:Math Exam Resources/Courses/MATH110/Syllabus

2014-10-06T18:06:49Z

MatthewColes:

Lines in the Plane, Functions
Tangent Lines, Velocities, Growth
Evaluating limits
Continuity
Tangent lines and the idea of derivative
The definition of derivative
The Power, Product and Quotient Rules
Trigonometry
Exponentials
The Chain Rule
Implicit differentiation
Extrema
The Mean Value Theorem
The First and Second Derivative Tests
Concavity
Asymptotes
Curve sketching
Optimization I
Optimization II
Related rates I
Related rates II
Approximations I
Approximations II

Science:Math Exam Resources/Courses/MATH110/Syllabus

2014-10-06T18:03:43Z

MatthewColes:

Lines in the Plane, Functions

Tangent Lines, Velocities, Growth

Evaluating limits

Continuity

Tangent lines and the idea of derivative

The definition of derivative

The Power, Product and Quotient Rules

Trigonometry

Exponentials

The Chain Rule

Implicit differentiation

Extrema

The Mean Value Theorem

The First and Second Derivative Tests

Concavity

Asymptotes

Curve sketching

Optimization I

Optimization II

Related rates I

Related rates II

Approximations I

Approximations II

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (g)

2014-09-30T05:09:10Z

MatthewColes:

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Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (f)

2014-09-30T05:08:40Z

MatthewColes:

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Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (e)

2014-09-30T05:08:11Z

MatthewColes:

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Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (d)

2014-09-30T05:07:30Z

MatthewColes:

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Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (c)

2014-09-30T05:07:04Z

MatthewColes:

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Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (b) iii

2014-09-30T05:06:40Z

MatthewColes:

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Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (b) ii

2014-09-30T05:06:12Z

MatthewColes:

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Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (b) i

2014-09-30T05:05:42Z

MatthewColes:

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Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (a) ii

2014-09-30T05:04:56Z

MatthewColes:

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Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (a) i

2014-09-30T05:04:19Z

MatthewColes:

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Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (g)/Statement

2014-09-30T04:54:18Z

MatthewColes: Created page with "Find the total mass of the rectangular box <math> \displaystyle [0, 1] \times [0, 2] \times [0, 3] </math> (that is, the box defined by the inequalities <math> \displaystyle 0..."

Find the total mass of the rectangular box <math> \displaystyle [0, 1] \times [0, 2] \times [0, 3] </math> (that is, the box
defined by the inequalities <math> \displaystyle 0 \leq x \leq 1, 0 \leq y \leq 2, 0 \leq z \leq 3 </math>), with density function
<math> \displaystyle h(x, y, z) = x </math>.

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (f)/Statement

2014-09-30T04:51:28Z

MatthewColes: Created page with "Find all the points on the surface <math> \displaystyle x^2 + 9y^2 +4z^2=17 </math> where the tangent plane is parallel to the plane <math> \displaystyle x-8z=0 </math>."

Find all the points on the surface <math> \displaystyle x^2 + 9y^2 +4z^2=17 </math> where the tangent plane is
parallel to the plane <math> \displaystyle x-8z=0 </math>.

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (e)/Statement

2014-09-30T04:47:23Z

MatthewColes:

Suppose it is known that the direction of the fastest increase of the function <math> \displaystyle f(x,y) </math> at the origin is given by the vector <math> \displaystyle \langle 1,2 \rangle </math>. Find a unit vector <math> \displaystyle \textbf{u} </math> that is tangent to the level curve of <math> \displaystyle f(x,y) </math> that passes through the origin.

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (e)/Statement

2014-09-30T04:46:31Z

MatthewColes:

Suppose it is known that the direction of fastest increase of the function <math> \displaystyle f(x,y) </math> at the origin is given by the vector <math> \displaystyle \langle 1,2 \rangle </math>. Find a unit vector <math> \displaystyle \textbf{u} </math> that is tangent to the level curve of <math> \displaystyle f(x,y) </math> that passes through the origin.

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (e)/Statement

2014-09-30T04:46:03Z

MatthewColes: Created page with "Suppose it is know that the direction of fastest increase of the function <math> \displaystyle f(x,y) </math> at the origin is given by the vector <math> \displaystyle \langle..."

Suppose it is know that the direction of fastest increase of the function <math> \displaystyle f(x,y) </math> at the origin is given by the vector <math> \displaystyle \langle 1,2 \rangle </math>. Find a unit vector <math> \displaystyle \textbf{u} </math> that is tangent to the level curve of <math> \displaystyle f(x,y) </math> that passes through the origin.

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (d)/Statement

2014-09-30T04:41:52Z

MatthewColes: Created page with "Let <math> \displaystyle f(x) </math> be a differentiable function, and suppose it is given that <math> \displaystyle f'(0)=10 </math>. Let <math> \displaystyle g(s,t)= f(as-b..."

Let <math> \displaystyle f(x) </math> be a differentiable function, and suppose it is given that <math> \displaystyle f'(0)=10 </math>. Let <math> \displaystyle g(s,t)= f(as-bt) </math>, where <math> \displaystyle a </math> and <math> \displaystyle b </math> are constants. Evaluate <math> \displaystyle \frac{\partial g}{\partial s} </math> at the point <math> \displaystyle (s,t)=(b,a) </math>, that is, find <math> \displaystyle \frac{\partial g}{\partial s} \bigg|_{(b,a)} </math>.

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (c)/Statement

2014-09-30T04:36:16Z

MatthewColes:

Suppose that <math> \displaystyle u =x^2 +y z, \ x = \rho r \cos(\theta), \ y= \rho r \sin(\theta) </math> and <math> \displaystyle z= \rho r </math>. Find <math> \displaystyle \frac{\partial u}{\partial r} </math> at the point <math> \displaystyle (\rho_0, r_0,\theta_0)=(2,3,\pi/2) </math>.

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (c)/Statement

2014-09-30T04:32:57Z

MatthewColes: Created page with "Suppose that <math> \displaystyle u =x^2 +y z, x = \rho r \cos(\theta) </math>,"

Suppose that <math> \displaystyle u =x^2 +y z, x = \rho r \cos(\theta) </math>,

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (b) iii/Statement

2014-09-30T04:28:46Z

MatthewColes: Created page with "Let <math> \displaystyle z </math> be the same implicit function as in part (ii), defined by the equation <math> \displaystyle x^2 z^3 + y \sin(\pi x) =-y^2 </math>. Let <mat..."

Let <math> \displaystyle z </math> be the same implicit function as in part (ii), defined by the equation <math> \displaystyle x^2 z^3 + y \sin(\pi x) =-y^2 </math>. Let <math> \displaystyle x=0.97 </math>, and <math> \displaystyle y=1 </math>. Find the approximate value of <math> \displaystyle z </math>.

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (b) ii/Statement

2014-09-30T04:21:25Z

MatthewColes: Created page with "Let <math> \displaystyle z </math> be defined implicitly by <math> \displaystyle x^2 z^3 + y \sin(\pi x) =-y^2 </math>. Find <math> \displaystyle \frac{\partial z}{\partial x..."

Let <math> \displaystyle z </math> be defined implicitly by <math> \displaystyle x^2 z^3 + y \sin(\pi x) =-y^2 </math>. Find <math> \displaystyle \frac{\partial z}{\partial x} </math> at the point <math> \displaystyle P=(1,1,-1) </math>.

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (b) i/Statement

2014-09-30T04:18:32Z

MatthewColes: Created page with "Find the equation of the tangent plane to the surface <math> \displaystyle x^2 z^3 +y \sin(\pi x) = -y^2 </math> at the point <math> \displaystyle P=(1,1,-1) </math>."

Find the equation of the tangent plane to the surface <math> \displaystyle x^2 z^3 +y \sin(\pi x) = -y^2 </math> at the point <math> \displaystyle P=(1,1,-1) </math>.

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (a) ii/Statement

2014-09-30T04:06:40Z

MatthewColes:

The line <math> \displaystyle L </math> has vector parametric equation <math> \displaystyle \textbf{r}(t) = (2+3t) \textbf{i} + 4t \textbf{j} - \textbf{k} </math>

 ii Let <math> \displaystyle \alpha </math> be the angle between the line <math> \displaystyle L </math> and the plane given by the equation <math> \displaystyle x - y + 2z=0 </math>. Find <math> \displaystyle \alpha </math>.

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (a) i/Statement

2014-09-30T04:01:18Z

MatthewColes:

The line <math> \displaystyle L </math> has vector parametric equation <math> \displaystyle \textbf{r}(t) = (2+3t) \textbf{i} + 4t \textbf{j} - \textbf{k} </math>

 i. Write the symmetric equations for <math> \displaystyle L </math>.

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (a) ii/Statement

2014-09-30T04:00:37Z

MatthewColes: Created page with "The line <math> \displaystyle L </math> has vector parametric equation <math> \displaystyle \textbf{r}(t) = (2+3t) \textbf{i} + 4t \textbf{j} - \textbf{k} </math> ii "

The line <math> \displaystyle L </math> has vector parametric equation <math> \displaystyle \textbf{r}(t) = (2+3t) \textbf{i} + 4t \textbf{j} - \textbf{k} </math>

 ii

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (a) i/Statement

2014-09-30T03:59:26Z

MatthewColes:

The line <math> \displaystyle L </math> has vector parametric equation <math> \displaystyle \textbf{r}(t) = (2+3t) \textbf{i} + 4t \textbf{j} - \textbf{k} </math>

 (a) Write the symmetric equations for <math> \displaystyle L </math>.

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (a) i/Statement

2014-09-30T03:55:41Z

MatthewColes:

The line ''L'' has vector parametric equation <math> \displaystyle \textbf{r}(t) = (2+3t) \textbf{i} + 4t \textbf{j} - \textbf{k} </math>

Write the symmetric equations for ''L''.

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (a) i/Statement

2014-09-30T03:52:40Z

MatthewColes:

The line ''L'' has vector parametric equation <math> \displaystyle r(t) = (2+3t) i + 4t j - k </math>

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (a) i/Statement

2014-09-30T03:50:31Z

MatthewColes:

The line ''L'' has vector parametric equation <math> \displaystyle r(t) = (2+3t) {\bf i} + 4t {\bf j} - {\bf k}

Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (a) i/Statement

2014-09-30T03:45:35Z

MatthewColes: Created page with "The line ''L''"

The line ''L''

Science:Math Exam Resources/Courses/MATH110/Syllabus

2014-09-29T23:53:05Z

MatthewColes:

Science:Math Exam Resources/Courses/MATH110/Syllabus

2014-09-29T23:52:39Z

MatthewColes:

Science:Math Exam Resources/Courses/MATH110/Syllabus

2014-09-29T23:45:55Z

MatthewColes:

Science:Math Exam Resources/Courses/MATH110/Syllabus

2014-09-18T01:39:38Z

MatthewColes: Blanked the page

Science:Math Exam Resources/Courses/MATH110/Syllabus

2014-09-18T01:39:17Z

MatthewColes:

syllabus goes here?