UBC Wiki - User contributions [en]

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 04 (b)

2015-04-21T23:38:13Z

MikeL:

[[Category:MER QGQ flag]][[Category:MER QGH flag]][[Category:MER QGS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 04 (b)/Solution 1

2015-04-21T23:37:42Z

MikeL:

We just need to check the local maximum, the local minimum and the points on the boundary. From question <math>4a</math> above, we know that it has only one local minimum <math>(3,6)</math> and no local maximum. Since <math>(3,6)</math> is not on <math>R</math>, we only need to find the min and max on the boundary.

[[File:4b-solution.jpg|thumb|Plot of region R]]

On <math>C_1</math>:

<math>C_1=\{x=0,y\in[-3,3]\},</math>

<math>\begin{align}
T(x,y)=\frac{1}{9}y^3-6y = f(y) \quad \quad f'(y) =\frac{1}{3}y^2-6 \end{align}</math>

The critical points from above occur when <math>\frac{1}{3}y^2 - 6 = 0</math> which is at <math>y = \pm \sqrt{18} = \pm 3 \sqrt{2}</math>, neither of which are in <math>[-3,3]</math>. Thus, we just test the endpoints: <math>f(-3) = 15</math> and <math>f(3) = -15</math>. These are two values we will consider later.

On <math>C_2</math>:

<math>C_2=\{x=\frac{1}{3}y^2-3,y\in[-3,3]\},</math>

<math>T(x,y)=\frac{1}{9}y^3+\left(\frac{1}{3}y^2-3\right)^2-2\left(\frac{1}{3}y^2-3\right)y+6\left(\frac{1}{3}y^2-3\right)-6y=\frac{y^4}{9}-\frac{5}{9}y^3-9 = g(y)</math>

<math>g'(y) =\frac{4}{9}y^3-\frac{15}{9}y^2=\frac{y^2}{9}(4y-15)=0</math> at <math>y=0</math> and <math>y = 15/4</math>. As <math>15/4</math> is not in <math>[-3,3]</math> we ignore it. We now compute <math>g(0) = -9</math> and the value at the endpoints with <math>g(-3) = 15</math> and <math>g(3) = -15</math>.

From all of this, we find <math>\max=15</math> and <math>\min=-15</math>.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 05 (a)

2015-04-21T23:21:50Z

MikeL:

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 05 (a)/Solution 1

2015-04-21T23:21:00Z

MikeL:

We first solve for the differential equation:

<math>\begin{align}
\frac{dB}{dt}&=aB-m\\
dB&=(aB-m)dt\\
\frac{dB}{aB-m}&=dt\quad (\text{separating the variables})\\
\int \frac{dB}{aB-m}&=\int dt\quad \\
\frac{1}{a}\ln |aB-m| &=t + C\\
\ln |aB - m| &= at + C \quad (\text{or aC, but C is arbitrary so we still call it C}) \\
|aB - m| &= e^{at+C} = A e^{at} \quad (\text{where } A = e^C) \\
aB - m &= \pm A e^{at} \quad (\text{removing the absolute values gives a } \pm) \\
B &= \frac{1}{a} (m - A e^{at}) \quad (\text{A here is } \pm A \text{ is arbitrary and still unknown}) \end{align} </math>

Now we can use the initial conditions with <math>B(0) = 30000</math> and <math>a=0.02=1/50</math> to find:

<math> B(0) = 30 000 = 50 (m - A) \implies A = m-600</math> and thus

<math>B(t) = 50(m - (m-600)e^{0. 02t}) = 50((600-m) e^{0.02t} + m)</math>.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 05 (a)/Solution 1

2015-04-21T23:20:13Z

MikeL:

We first solve for the differential equation:

<math>\begin{align}
\frac{dB}{dt}&=aB-m\\
dB&=(aB-m)dt\\
\frac{dB}{aB-m}&=dt\quad (\text{separating the variables})\\
\int \frac{dB}{aB-m}&=\int dt\quad \\
\frac{1}{a}\ln |aB-m| &=t + C\\
\ln |aB - m| &= at + C \quad (\text{or aC, but C is arbitrary so we still call it C}) \\
|aB - m| &= e^{at+C} = A e^{at} \quad (\text{where } A = e^C) \\
aB - m &= \pm A e^{at} \quad (\text{removing the absolute values gives a } \pm) \\
B &= \frac{1}{a} (m - A e^{at}) \quad (\text{A here is } \pm A \text{ is arbitrary and still unknown}) \end{align} </math>

Now we can use the initial conditions with <math>B(0) = 30000</math>, <math>a=0.02</math> and <math>m=1/50</math> to find:

<math> B(0) = 30 000 = 50 (m - A) \implies A = m-600</math> and thus

<math>B(t) = 50(m - (m-600)e^{0. 02t}) = 50((600-m) e^{0.02t} + m)</math>.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 05 (a)

2015-04-21T23:05:17Z

MikeL:

[[Category:MER QGQ flag]][[Category:MER QGH flag]][[Category:MER QBS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 04 (b)

2015-04-21T23:03:48Z

MikeL:

[[Category:MER QGQ flag]][[Category:MER QGH flag]][[Category:MER QBS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 06 (b)/Solution 1

2015-04-20T23:38:39Z

MikeL:

Since <math>\displaystyle \sum_{n=1}^{\infty}\frac{na_n-2n+1}{n+1}</math> converges, we have <math>\displaystyle \lim_{n\rightarrow\infty}\frac{na_n-2n+1}{n+1}=0</math>. Hence:

<math>\begin{align}
0&=\lim_{n\rightarrow\infty}\frac{na_n-2n+1}{n+1}\\
&=\lim_{n\rightarrow\infty}\left((a_n-2)\cdot\frac{n}{n+1}+\frac{1}{n+1}\right)\\
&=\lim_{n\rightarrow\infty}(a_n-2)\cdot\frac{n}{n+1}+\lim_{n\rightarrow\infty}\frac{1}{n+1}\\
&=\lim_{n\rightarrow\infty}(a_n-2)\cdot\frac{n}{n+1}+0\\
&=\lim_{n\rightarrow\infty}(a_n-2)\cdot\lim_{n\rightarrow\infty}\frac{n}{n+1}\\
&=\lim_{n\rightarrow\infty}(a_n-2)\cdot1\\
&=\lim_{n\rightarrow\infty}(a_n-2)\\\end{align}</math>

Therefore <math>\displaystyle \lim_{n\rightarrow\infty}(a_n-2)=0</math>. Or rather, <math>\displaystyle \lim_{n\rightarrow\infty}a_n=2</math>.

On the other hand, <math>\ln\left(\frac{a_n}{a_{n+1}}\right)=\ln a_n-\ln a_{n+1}</math>. Hence,

<math>\begin{align}
-\ln a_1+\sum_{n=1}^{k}\ln \left(\frac{a_n}{a_{n+1}}\right)&=-\ln a_1+\sum_{n=1}^{k}(\ln a_n-\ln a_{n+1})\\
&=-\ln a_1+(\ln a_1-\ln a_{2})+(\ln a_2-\ln a_{3})+\dots+(\ln a_k-\ln a_{k+1})\\
&=- \ln a_{k+1}\\\end{align}</math>

Hence,

<math>\begin{align}
-\ln a_1+\sum_{n=1}^{\infty}\ln \left(\frac{a_n}{a_{n+1}}\right)&=\lim_{k\rightarrow\infty}\left(-\ln a_1+\sum_{n=1}^{k}\ln \left(\frac{a_n}{a_{n+1}}\right)\right)\\
&=\lim_{k\rightarrow\infty} - \ln a_{k+1}\\
&=- \ln 2.\end{align}</math>

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 01 (b)

2015-04-20T15:26:49Z

MikeL:

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 01 (b)/Solution 1

2015-04-20T15:26:27Z

MikeL:

Setting <math>V=\pi</math>,

<math>\begin{align}
\frac{\pi x^2y}{3}&=\pi\\
x^2y&=3\\
y&=\frac{3}{x^2}\\\end{align}</math>

We plot the equation <math>y = \frac{3}{x^2}</math> noting that <math>x\geq 0</math> is necessary for the curve to make physical sense.

[[File:Level Curve.png|thumb|Plot of level curve for Math 105 Q1b of 2014 April exam.]]

File:Level Curve.png

2015-04-20T15:24:56Z

MikeL: User created page with UploadWizard

=={{int:filedesc}}==
{{Information
|description={{en|1=Plot of level curve for Math 105 Q1b of 2014 April exam.}}
|date=2015-04-20 08:24:10
|source={{own}}
|author=[[User:MikeL|MikeL]]
|permission=
|other_versions=
}}

=={{int:license-header}}==
{{self|cc-by-sa-3.0}}

[[Category:Uploaded with UploadWizard]]

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 01 (l)

2015-04-19T18:27:50Z

MikeL:

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 01 (l)/Hint 1

2015-04-19T18:27:12Z

MikeL:

Remember that <math>\displaystyle\sum_{n=0}^\infty r^n=\frac{1}{1-r}</math> if <math> |r|<1</math>.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 01 (l)/Hint 2

2015-04-19T18:26:17Z

MikeL:

Remember that you can re-index series. If you want a series to start from <math>n=0</math> instead of <math>n=1</math>, you can do this by replacing the <math>n</math> in the series terms by <math>n+1</math>.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 01 (l)/Solution 1

2015-04-19T18:23:55Z

MikeL:

We have

:<math>\begin{align} \sum_{n=1}^\infty \left[\left(\frac{1}{3}\right)^n+\left(-\frac{2}{5}\right)^{n-1}\right] &= \sum_{n=0}^\infty \left[\left(\frac{1}{3}\right)^{n+1}+\left(-\frac{2}{5}\right)^{n}\right] \\ &= \frac{1}{3} \sum_{n=0}^\infty \left(\frac{1}{3}\right)^{n} + \sum_{n=0}^{\infty} \left(-\frac{2}{5}\right)^{n} \\ &= \frac{1}{3} \frac{1}{1-1/3} + \frac{1}{1-(-2/5)} \\ &= \frac{1}{2}+\frac{5}{7} \\ &= \frac{17}{14}. \end{align} </math>

With the first equality, we rewrote the geometries series starting from <math>n=0</math> in order to apply <math>\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}</math> for <math>|r|<1</math>. This was accomplished by re-indexing. In the second equality, we split the sum into two and factored out <math.1/3</math> from the first sum.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 01 (l)

2015-04-19T18:14:48Z

MikeL:

[[Category:MER QGQ flag]][[Category:MER RH flag]][[Category:MER QBS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 01 (i)

2015-04-19T18:12:47Z

MikeL:

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 01 (i)/Solution 1

2015-04-19T18:11:20Z

MikeL:

First we complete the square in the expression <math>3-2x-x^2</math> by considering
<math>3 - 2x - x^2 = 3 - (x^2 + 2x) = 3 - ((x+1)^2 - 1) = 4 - (x+1)^2</math>.

This changes the integral to
<math>
\int\frac{dx}{\sqrt{3-2x-x^2}} = \int\frac{dx}{\sqrt{4-(x+1)^2}}
</math>

We substitute <math> t = x+1 </math> with <math>dx = dt</math> and obtain
<math>\int\frac{dt}{\sqrt{4 - t^2}}</math>

We do another substitution with <math>t = 2\sin(u)</math> (so that <math>u = \sin^{-1} ( \frac{t}{2} ) </math> ) and <math>dt = 2\cos(u)du</math> giving

<math>\begin{align}
\int\frac{2\cos(u)du}{\sqrt{4 - 4\sin(u)^2}}&= \int\frac{2\cos(u)du}{2\sqrt{1-\sin(u)^2}}\\
&= \int\frac{\cos(u)du}{\sqrt{\cos(u)^2}}
\end{align}</math>

where we used that <math> 1 - \sin(u)^2 = \cos(u)^2</math>.

The remaining integral is
<math>\int \frac{\cos(u)}{\cos(u)} du = \int du = u + C </math>.

Now we re-substitute <math>u + C = \sin^{-1}\left(\frac{t}{2}\right) + C = \sin^{-1}\left(\frac{x+1}{2}\right) + C</math>.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 01 (b)

2015-04-19T17:59:33Z

MikeL:

[[Category:MER QGQ flag]][[Category:MER RH flag]][[Category:MER QBS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 04 (a)

2015-04-19T17:58:17Z

MikeL:

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 04 (a)/Solution 1

2015-04-19T17:57:41Z

MikeL:

Compute the derivatives:

<math>T_x(x,y)=2x-2y+6 \quad \quad T_y(x,y)=\frac{1}{3}y^2-2x-6;</math>

<math>T_{xx}(x,y)=2,\; \quad T_{xy}(x,y)=-2, \; \quad T_{yy}(x,y)=\frac{2}{3}y.</math>

Set <math>T_x(x,y)=0</math> and <math>T_y(x,y)=0</math> to find critical points:

<math>2x-2y+6=0\;</math>

<math>\frac{1}{3}y^2-2x-6=0.</math>

From the first equality we get <math>x=y-3</math>. Plugging this into the second equality, we get <math>\frac{1}{3}y^2-2y=0</math>. Solving it gives <math>y=0</math> and <math> y = 6</math>. This yields the critical points <math>(x,y)=(-3,0)</math>, and <math>(x,y)=(3,6)</math>.

For <math>(-3,0)</math>, <math>T_{xx}=2</math>, <math>T_{yy}=\frac{2}{3}y=0</math> and <math>T_{xy}=-2</math>, so we have a saddle point because <math>T_{xx} T_{yy} - T_{xy}^2 = -4 < 0</math>.

For <math>(3,6)</math>, <math>T_{xx}=2</math>, <math>T_{yy}=\frac{2}{3}y=4</math>, and <math>T_{xx}\cdot T_{yy}-T_{xy}^2=2\cdot\frac{2}{3}y-(-2)^2=2\cdot4-(-2)^2=4>0</math> (and thus it could be a local max or local min). Then, because <math>T_{xx} = 2 > 0</math>, we conclude it is a local minimum.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 01 (f)

2015-04-19T17:51:08Z

MikeL:

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 01 (f)/Solution 2

2015-04-19T17:48:58Z

MikeL:

We use integration by parts and set <math> u = f'(x)</math> and <math> dv = f''(x)dx</math>.
Then

<math>
\int_1^2f'(x)f''(x)dx = \left[f'(x)^2\right]_1^2 - \int_1^2f''(x)f'(x)dx
</math>

Note that the integral on the left and right are the same and thus we can rewrite this as

<math> 2\int_1^2f'(x)f''(x)dx = \left[f'(x)^2\right]_1^2 = 3^2 - 2^2 = 9-4 = 5</math>

Dividing by 2 leads to
<math>\int_1^2 f'(x)f''(x)dx = \frac52</math>

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 01 (f)/Solution 3

2015-04-19T17:47:14Z

MikeL: Created page with "We use integration by substitution. If <math> u = f'(x)</math> then <math>du = f''(x) dx</math>. Thus, <math>f'(x) f''(x) dx = u du</math>. We also can transform the bounds so..."

We use integration by substitution. If <math> u = f'(x)</math> then <math>du = f''(x) dx</math>. Thus, <math>f'(x) f''(x) dx = u du</math>. We also can transform the bounds so that <math>1 \mapsto f'(1) = 2</math> and <math>2 \mapsto f'(2) = 3</math>. In terms of u, we can integrate:

:<math>\int_2^3 u du = \frac{1}{2} u^2 |_2^3 = \frac{5}{2}</math>. This is our final answer.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 05 (b)

2015-04-18T17:53:22Z

MikeL:

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 06 (b)

2015-04-18T17:46:46Z

MikeL:

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 06 (b)/Hint 3

2015-04-18T17:45:02Z

MikeL: Created page with "Consider writing out the terms of <math>-\ln a_1 + \sum_{n=1}^{k} \ln(\frac{a_n}{a_{n+1}})</math> for a general k. You may find it useful to recall <math>\ln(x/y) = \ln x - \l..."

Consider writing out the terms of <math>-\ln a_1 + \sum_{n=1}^{k} \ln(\frac{a_n}{a_{n+1}})</math> for a general k. You may find it useful to recall <math>\ln(x/y) = \ln x - \ln y</math>. You should find many terms cancel.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 06 (b)/Hint 2

2015-04-18T17:41:51Z

MikeL: Created page with "By the Divergence Test, we must have that the terms of <math>\frac{na_n-2n+1}{n+1}</math> approach zero as <math>n \rightarrow \infty</math>. Thus, <math>\lim_{n \rightarrow \..."

By the Divergence Test, we must have that the terms of <math>\frac{na_n-2n+1}{n+1}</math> approach zero as <math>n \rightarrow \infty</math>. Thus, <math>\lim_{n \rightarrow \infty} \frac{na_n-2n+1}{n+1} = 0</math>. Can you use this to obtain more information about the terms of <math>a_n</math>?

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 06 (b)/Hint 1

2015-04-18T17:39:05Z

MikeL:

If <math>\sum b_n</math> is convergent, what do you know about <math>\lim_{n \rightarrow \infty} b_n</math>?

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 06 (a)

2015-04-18T17:33:30Z

MikeL:

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 04 (a)

2015-04-18T17:19:35Z

MikeL:

[[Category:MER QGQ flag]][[Category:MER QBH flag]][[Category:MER QBS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 03 (b)

2015-04-18T17:14:10Z

MikeL:

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 03 (b)/Solution 1

2015-04-18T17:12:41Z

MikeL:

We want to find where <math>\sqrt{(x-(-1))^2+(y-2)^2}</math> attains its minimum on the circle <math>x^2 + y^2 = 125</math>.

Equivalently, we need to find where <math>(x-(-1))^2+(y-2)^2=(x+1)^2+(y-2)^2</math> is the smallest on the circle <math>x^2 + y^2 = 125</math>. From (a), we know that the minimum is at <math>(-5,10)</math>.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 03 (b)/Hint 2

2015-04-18T17:09:50Z

MikeL: Created page with "Distances are positive quantities and thus a point that minimizes the square of a distance will also be the point that minimizes the distance."

Distances are positive quantities and thus a point that minimizes the square of a distance will also be the point that minimizes the distance.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 03 (b)/Hint 1

2015-04-18T17:08:00Z

MikeL:

The distance from a point (x,y) to (-1,2) is <math>\sqrt{(x+1)^2 + (y-2)^2}</math>.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 03 (a)

2015-04-18T17:02:27Z

MikeL:

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 03 (a)/Solution 1

2015-04-18T17:01:16Z

MikeL:

The object function is <math>f(x,y) = (x+1)^2+(y-2)^2</math> and the constraint is <math>x^2+y^2=125</math>, i. e. <math>g(x,y) = x^2+y^2-125=0</math>.

By the method Lagrange multipliers, set <math>\nabla f = \lambda \nabla g, g = 0</math> which tells us <math>\langle 2(x+1), 2(y-2) \rangle = \lambda \langle 2x, 2y \rangle, \quad x^2 + y^2 - 125 = 0 </math>.

Looking at the vector equation in components, we have that

:<math> \begin{align} 2(x+1) = 2 \lambda x & \implies x+1 = \lambda x \quad (1) \\
2(y-1) = 2 \lambda y & \implies y-2 = \lambda y. \quad(2) \end{align} </math>

Provided we don't divide by zero (so that <math>y \neq 2</math>), we can divide (1) by (2) to yield <math>\frac{x+1}{y-2} = \frac{\lambda x}{\lambda y} = \frac{x}{y} \implies xy + y = xy - 2x \implies y=-2x </math> where the first implication came by cross multiplying.

If <math>y=x</math> then from the constraint <math>x^2 + y^2 - 125 = 0</math> we must have <math>x^2 + (-2x)^2 - 125 = 5x^2 - 125 = 0 \implies x = \pm 5</math> and <math>y = -2x = \mp 10</math>.

Evaluating <math>f(5,-10) = (5+1)^2 + (-10+2)^2 = 180</math> and <math>f(-5,10) = (-5+1)^2 + (10-2)^2 = 80</math>.

We still need to consider the possibility that <math>y=2</math>. If <math>y=2</math> then (2) reads <math>0 = 2 \lambda \implies \lambda = 0</math>. If <math>\lambda = 0</math> then (1) tells us that <math>x+1 = 0</math> so that <math>x=-1</math>. However, <math>(-1,2)</math> does not satisfy <math>g(x,y) = 0</math> so this is not a valid solution to the Lagrange system.

Overall have found that the maximum value is 180 and the minimum value is 80.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 03 (a)/Solution 1

2015-04-17T15:12:22Z

MikeL:

The object function is <math>f(x,y) = (x+1)^2+(y-2)^2</math> and the constraint is <math>x^2+y^2=125</math>, i. e. <math>g(x,y) = x^2+y^2-125=0</math>.

By the method Lagrange multipliers, set <math>\nabla f = \lambda \nabla g, g = 0</math> which tells us <math>\langle 2(x+1), 2(y-2) \rangle = \lambda \langle 2x, 2y \rangle, \quad x^2 + y^2 - 125 = 0 </math>.

Looking at the vector equation in components, we have that

:<math> \begin{align} 2(x+1) = 2 \lambda x & \implies x+1 = \lambda x \quad (1) \\
2(y-1) = 2 \lambda y & \implies y-2 = \lambda y. \quad(2) \end{align} </math>

Provided we don't divide by zero (so that <math>y \neq 2</math>), we can divide (1) by (2) to yield <math>\frac{x+1}{y-2} = \frac{\lambda x}{\lambda y} = \frac{x}{y} \implies xy + y = xy - 2x \implies y=-2x </math> where the first implication came by cross multiplying.

If <math>y=x</math> then from the constraint <math>x^2 + y^2 - 125 = 0</math> we must have <math>x^2 + (-2x)^2 - 125 = 5x^2 - 125 = 0 \implies x = \pm 5</math> and <math>y = -2x = \mp 10</math>.

Evaluating <math>f(5,-10) = (5+1)^2 + (-10+2)^2 = 180</math> and <math>f(-5,10) = (-5+1)^2 + (10-2)^2 = 80</math>.

We still need to consider the possibility that <math>y=2</math>. If <math>y=2</math> then (2) reads <math>0 = 2 \lambda \implies \lambda = 0</math>. If <math>\lambda = 0</math> then (1) tells us that <math>x+1 = 0</math> so that <math>x=-1</math>. However, <math>(2,-1)</math> does not satisfy <math>g(x,y) = 0</math> so this is not a valid solution to the Lagrange system.

Overall have found that the maximum value is 180 and the minimum value is 80.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 03 (a)/Hint 1

2015-04-17T14:57:00Z

MikeL:

The method of Lagrange multipliers states that the extreme values of the objective function <math>f(x,y)</math> subject to the constraint <math>g(x,y) = 0</math> will occur at the solutions to the system: <math>\nabla f = \langle f_x, f_y \rangle = \lambda \langle g_x, g_y \rangle = \lambda \nabla g</math>.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 03 (a)

2015-04-17T14:54:13Z

MikeL:

[[Category:MER QGQ flag]][[Category:MER QBH flag]][[Category:MER QBS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 02 (c)

2015-04-17T14:53:07Z

MikeL:

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 02 (b)

2015-04-17T14:51:01Z

MikeL:

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 02 (b)

2015-04-16T15:19:28Z

MikeL:

[[Category:MER QGQ flag]][[Category:MER QGH flag]][[Category:MER RS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 02 (b)/Solution 1

2015-04-16T15:18:32Z

MikeL:

Let <math>a_k = \frac{x^k}{10^{k+1} (k+1)!}</math> be the terms in the series. From the ratio test, we are guaranteed absolute convergence when <math>\lim_{k\rightarrow\infty} |\frac{a_{k+1}}{a_k}| < 1</math>.

With <math>a_k = \frac{x^k}{10^{k+1} (k+1)!}</math>, we have <math>a_{k+1} = \frac{x^{k+1}}{10^{k+2} (k+2)!}</math> and thus

:<math> \begin{align} |\frac{a_{k+1}}{a_k}| &= |\frac{ \frac{x^{k+1}}{10^{k+2} (k+2)!} }{\frac{x^k}{10^{k+1} (k+1)!}}| \\ &= |\frac{x^{k+1} 10^{k+1} (k+1)!}{x^k 10^{k+2} (k+2)!}| \\ &= |\frac{x}{10} \frac{(k+1)!}{(k+2)!}|. \end{align} </math>

We recall that <math>k! = k(k-1)(k-2)...(3)(2)(1)</math> and thus in general <math>k! = k(k-1)!</math>. This also means that <math>(k+2)! = (k+2)(k+1)!</math>. Thus,

:<math> \begin{align} |\frac{a_{k+1}}{a_k}| &= |\frac{x (k+1)!}{10 (k+2) (k+1)!}| \\ &= |\frac{x}{10 (k+2)}| \end{align} </math>.

Computing <math>\lim_{k \rightarrow \infty} |\frac{x}{10 (k+2)}| = 0 < 1</math> for all x. Hence, the series converges absolutely for all x-values and the radius of convergence is <math>\infty</math>.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 02 (b)/Hint 1

2015-04-16T15:05:30Z

MikeL:

Recall that the radius of convergence for a series can be found by determining the range of x-values where the series converges absolutely. The ratio test will give you information about absolute convergence.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 02 (a)

2015-04-16T15:04:07Z

MikeL:

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 02 (a)/Solution 1

2015-04-16T15:03:40Z

MikeL:

Since <math>\displaystyle\frac{\sqrt[3]{k^4+1}}{\sqrt{k^5+9}}\approx \frac{\sqrt[3]{k^4}}{\sqrt{k^5}}=\frac{k^{4/3}}{k^{5/2}}=\frac{1}{k^{5/2-4/3}}=\frac{1}{k^{7/6}}</math>, we want to compare <math>\displaystyle\frac{\sqrt[3]{k^4+1}}{\sqrt{k^5+9}}</math> with <math>\displaystyle\frac{1}{k^{7/6}}</math>.

Let <math>\displaystyle a_k=\frac{1}{k^{7/6}}>0</math> and <math>\displaystyle b_k=\frac{\sqrt[3]{k^4+1}}{\sqrt{k^5+9}}>0</math>. Recall <math>\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^\alpha}</math> converges if and only if <math>\alpha>1</math>. Hence <math>\displaystyle \sum_{k=1}^{\infty} a_k</math> converges.

On the other hand, by the Limit Comparison Test, if <math>\displaystyle \lim_{k\rightarrow\infty} \frac{b_k}{a_k}=L</math> for some nonzero finite value of L and both series are positive then either both <math>\displaystyle\sum_{k=1}^{\infty} a_k</math> and <math>\displaystyle\sum_{k=1}^{\infty} b_k</math> converge or both diverge. Hence, if we can prove <math>\displaystyle\lim_{k\rightarrow\infty} \frac{b_k}{a_k}=1</math>, then we know that <math>\displaystyle\sum_{k=1}^{\infty} b_k</math> converges. They key fact here is that we know the convergence properties of one of the two series we are comparing, namely the series with <math>\frac{1}{k^{7/6}}</math>.

<math>\begin{align}
\displaystyle
\lim_{k\rightarrow\infty} \frac{b_k}{a_k}&=\lim_{k\rightarrow\infty}k^{7/6}\frac{\sqrt[3]{k^4+1}}{\sqrt{k^5+9}}
=\lim_{k\rightarrow\infty}k^{7/6}\frac{\sqrt[3]{(1+k^{-4})\cdot k^4}}{\sqrt{(1+9k^{-5})\cdot k^{5}}}\\
&=\lim_{k\rightarrow\infty}k^{7/6}\frac{\sqrt[3]{(1+k^{-4})}}{\sqrt{(1+9k^{-5})}}\frac{k^{4/3}}{k^{5/2}}
=\lim_{k\rightarrow\infty}k^{7/6}\frac{\sqrt[3]{(1+k^{-4})}}{\sqrt{(1+9k^{-5})}}k^{4/3-5/2}\\
&=\lim_{k\rightarrow\infty}k^{7/6}\frac{\sqrt[3]{(1+k^{-4})}}{\sqrt{(1+9k^{-5})}}k^{-7/6}
=\lim_{k\rightarrow\infty}\frac{\sqrt[3]{(1+k^{-4})}}{\sqrt{(1+9k^{-5})}}\\
&=\lim_{k\rightarrow\infty}\frac{\sqrt[3]{(1+k^{-4})}}{\sqrt{(1+9k^{-5})}}=\frac{1}{1}=1.\\\end{align}</math>

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 02 (a)/Hint 2

2015-04-16T15:00:02Z

MikeL:

Compare the series with <math>\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^{7/6}}</math>.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 02 (a)/Hint 1

2015-04-16T14:59:43Z

MikeL:

When k is very large, the largest contribution to the numerator is effectively <math>k^{4/3}</math> and the largest contribution to the denominator is effectively <math>k^{5/2}</math>.

Science:Math Exam Resources/Courses/MATH105/April 2014/Question 02 (a)/Hint 2

2015-04-16T14:59:21Z

MikeL: Created page with "When k is very large, the largest contribution to the numerator is effectively <math>k^{4/3}</math> and the largest contribution to the denominator is effectively <math>k^{5/2..."

When k is very large, the largest contribution to the numerator is effectively <math>k^{4/3}</math> and the largest contribution to the denominator is effectively <math>k^{5/2}</math>.