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Thread:Science talk:Math Exam Resources/Courses/MATH152/April 2016/Question A 26/Answer updated- please check the signs

2018-03-26T22:01:52Z

RobertFraser: New thread: Answer updated- please check the signs

Hi everyone,

I am pretty sure I got the signs right in this answer. The point is that a current pointing in the direction of i_1 will reduce the voltage V_1, because the current takes the positive charge away from the capacitor. A current pointing in the direction of i_2 will increase the voltage V_2 for the same reason.

Finally, Lenz's law states that a potential difference in an inductor induces a current pointing in the opposite direction, so dI/dt should point in the opposite direction of the potential difference.

-Rob

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 26

2018-03-26T21:57:42Z

RobertFraser:

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Thread:Science talk:Math Exam Resources/Courses/MATH152/April 2016/Question A 26/Hint and Solution Unclear/reply (2)

2018-03-26T21:57:08Z

RobertFraser: Reply to Hint and Solution Unclear

I don't like these notes because I thought that a "voltage drop" was the negative of the difference in voltage, and they seem to be using the opposite convention.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 26/Hint 1

2018-03-26T21:55:39Z

RobertFraser:

Recall that capacitor equation states that

<math> \frac{dV}{dt} = - \frac{i}{C} </math>

where the voltage across the capacitor is <math>V</math>, the capacitance is <math>C</math>, and that the constructive equation for inductors states

\frac{dI}{dt} = -\frac{V}{L} </math>

where <math>L</math> is the inductance of the inductor.

Take careful notice of the direction of the currents <math>i_1</math> and <math>i_2</math>. One of these currents is pointing in the direction of the voltage across the capacitor and the other is pointing in the opposite direction.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 26/Hint 1

2018-03-26T21:41:46Z

RobertFraser:

Recall that capacitor equation states that

<math> \frac{dV}{dt} = \frac{i}{C} </math>

where the voltage drop across the capacitor is <math>V</math>, the capacitance is <math>C</math>, and that the constructive equation for inductors states

\frac{dI}{dt} = -\frac{V}{L} </math>

where <math>L</math> is the inductance of the inductor.

Take careful notice of the direction of the currents <math>i_1</math> and <math>i_2</math>. One of these currents is pointing in the direction of the voltage drop and the other is pointing in the opposite direction.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 26/Hint 1

2018-03-26T21:38:27Z

RobertFraser:

Recall that capacitor equation states that

<math> \frac{dV}{dt} = -\frac{i}{C} </math>

where the voltage across the capacitor is <math>V</math> the capacitance is <math>C</math>, and that the constructive equation for inductors states

\frac{dI}{dt} = -\frac{V}{L} </math>

where <math>L</math> is the inductance of the inductor.

Take careful notice of the direction of the currents <math>i_1</math> and <math>i_2</math>. One of these currents is pointing in the direction of the voltage drop and the other is pointing in the opposite direction.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 26/Solution 1

2018-03-26T21:34:30Z

RobertFraser:

Following the hint, we have

<math> \begin{align} \frac{dV_1}{dt} & = & -\frac{i_1}{C_1} & = & I/3-V_1/6 + V_2/6 \\
\frac{dV_2}{dt} & =& \frac{i_2}{C_2} & = & I/6+V_1/6-V_2/6 \\
\frac{d I}{dt} & =& -\frac{V}{L} & =& - 110I/3 -20V_1/3 -10 V_2/3 \end{align}.</math>

'''Answer:'''

<math> \color{blue} \begin{bmatrix}
\frac{d V_1}{dt} \\ \frac{dV_2}{dt} \\ \frac{dI}{dt}
\end{bmatrix}
=
\begin{bmatrix}
1/3 & -1/6 & 1/6 \\
1/6 & 1/6 & -1/6 \\
-110/3 & -20/3 & -10/3
\end{bmatrix}
\begin{bmatrix}
I \\ V_1 \\ V_2
\end{bmatrix}

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 26/Solution 1

2018-03-26T21:23:58Z

RobertFraser:

Following the hint, we have

<math> \begin{align} \frac{dV_1}{dt} & = & -\frac{i_1}{C_1} & = & I/3-V_1/6 + V_2/6 \\
\frac{dV_2}{dt} & =& -\frac{i_2}{C_2} & = & I/6+V_1/6-V_2/6 \\
\frac{d I}{dt} & =& -\frac{V}{L} & =& - 110I/3 -20V_1/3 -10 V_2/3 \end{align}.</math>

'''Answer:'''

<math> \color{blue} \begin{bmatrix}
\frac{d V_1}{dt} \\ \frac{dV_2}{dt} \\ \frac{dI}{dt}
\end{bmatrix}
=
\begin{bmatrix}
1/3 & -1/6 & 1/6 \\
1/6 & 1/6 & -1/6 \\
-110/3 & -20/3 & -10/3
\end{bmatrix}
\begin{bmatrix}
I \\ V_1 \\ V_2
\end{bmatrix}

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 26/Solution 1

2018-03-26T21:22:47Z

RobertFraser:

Following the hint, we have

<math> \begin{align} \frac{dV_1}{dt} & = & -\frac{i_1}{C_1} & = & I/3+V_1/6-V_2/6 \\
\frac{dV_2}{dt} & =& -\frac{i_2}{C_2} & = & I/6+V_1/6-V_2/6 \\
\frac{d I}{dt} & =& -\frac{V}{L} & =& - 110I/3 -20V_1/3 -10 V_2/3 \end{align}.</math>

'''Answer:'''

<math> \color{blue} \begin{bmatrix}
\frac{d V_1}{dt} \\ \frac{dV_2}{dt} \\ \frac{dI}{dt}
\end{bmatrix}
=
\begin{bmatrix}
1/3 & -1/6 & 1/6 \\
1/6 & 1/6 & -1/6 \\
-110/3 & -20/3 & -10/3
\end{bmatrix}
\begin{bmatrix}
I \\ V_1 \\ V_2
\end{bmatrix}

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 26/Solution 1

2018-03-26T21:21:40Z

RobertFraser:

Following the hint, we have

<math> \begin{align} \frac{dV_1}{dt} & =-\frac{i_1}{C_1} & = I/3+V_1/6-V_2/6 \\
\frac{dV_2}{dt} & =-\frac{i_2}{C_2} & = I/6+V_1/6-V_2/6 \\
\frac{d I}{dt} & =-\frac{V}{L} & = - 110I/3 -20V_1/3 -10 V_2/3 \end{align}.</math>

'''Answer:'''

<math> \color{blue} \begin{bmatrix}
\frac{d V_1}{dt} \\ \frac{dV_2}{dt} \\ \frac{dI}{dt}
\end{bmatrix}
=
\begin{bmatrix}
1/3 & -1/6 & 1/6 \\
1/6 & 1/6 & -1/6 \\
-110/3 & -20/3 & -10/3
\end{bmatrix}
\begin{bmatrix}
I \\ V_1 \\ V_2
\end{bmatrix}

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 26/Solution 1

2018-03-26T21:21:04Z

RobertFraser:

Following the hint, we have

<math> \begin{align} \frac{dV_1}{dt} & =-\frac{i_1}{C_1} = I/3+V_1/6-V_2/6 \\
\frac{dV_2}{dt}=-\frac{i_2}{C_2} & = I/6+V_1/6-V_2/6 \\
\frac{d I}{dt}=-\frac{V}{L}=-\frac{V}{0.1} & = - 110I/3 -20V_1/3 -10 V_2/3 \end{align}.</math>

'''Answer:'''

<math> \color{blue} \begin{bmatrix}
\frac{d V_1}{dt} \\ \frac{dV_2}{dt} \\ \frac{dI}{dt}
\end{bmatrix}
=
\begin{bmatrix}
1/3 & -1/6 & 1/6 \\
1/6 & 1/6 & -1/6 \\
-110/3 & -20/3 & -10/3
\end{bmatrix}
\begin{bmatrix}
I \\ V_1 \\ V_2
\end{bmatrix}

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 26/Solution 1

2018-03-26T21:20:40Z

RobertFraser:

Following the hint, we have

<math> \begin{align} \frac{dV_1}{dt} & =-\frac{i_1}{C_1} = I/3+V_1/6-V_2/6
\frac{dV_2}{dt}=-\frac{i_2}{C_2} & = I/6+V_1/6-V_2/6;
\frac{d I}{dt}=-\frac{V}{L}=-\frac{V}{0.1} & = - 110I/3 -20V_1/3 -10 V_2/3 \end{align}.</math>

'''Answer:'''

<math> \color{blue} \begin{bmatrix}
\frac{d V_1}{dt} \\ \frac{dV_2}{dt} \\ \frac{dI}{dt}
\end{bmatrix}
=
\begin{bmatrix}
1/3 & -1/6 & 1/6 \\
1/6 & 1/6 & -1/6 \\
-110/3 & -20/3 & -10/3
\end{bmatrix}
\begin{bmatrix}
I \\ V_1 \\ V_2
\end{bmatrix}

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 26/Solution 1

2018-03-26T21:19:41Z

RobertFraser:

Following the hint, we have

<math> \begin{align} \frac{dV_1}{dt}=-\frac{i_1}{C_1} = I/3+V_1/6-V_2/6
\frac{dV_2}{dt}=-\frac{i_2}{C_2}= I/6+V_1/6-V_2/6;</math>
\frac{d I}{dt}=-\frac{V}{L}=-\frac{V}{0.1}= - 110I/3 -20V_1/3 -10 V_2/3 \end{align}.</math>

'''Answer:'''

<math> \begin{bmatrix}
\frac{d V_1}{dt} \\ \frac{dV_2}{dt} \\ \frac{dI}{dt}
\end{bmatrix}
=
\begin{bmatrix}
1/3 & -1/6 & 1/6 \\
1/6 & 1/6 & -1/6 \\
-110/3 & -20/3 & -10/3
\end{bmatrix}
\begin{bmatrix}
I \\ V_1 \\ V_2
\end{bmatrix}

Thread:Science talk:Math Exam Resources/Courses/MATH152/April 2016/Question A 25/Incorrect equation for I

2018-03-26T20:07:30Z

RobertFraser: New thread: Incorrect equation for I

Hi Aili and Brian,

There was an error in Aili's solution: the current entering the top node is I + i_1 and the current leaving the top node is i_2, so we get I = i_2 - i_1. A circuit very similar to this one also appears in the following problem:

http://wiki.ubc.ca/Science:Math_Exam_Resources/Courses/MATH152/April_2017/Question_B_03_(a)

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 25

2018-03-26T20:03:10Z

RobertFraser:

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{{MER Question page}}

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 25/Solution 1

2018-03-26T20:02:47Z

RobertFraser:

We follow the hint and apply Kirchoff's current law to the top node.

The current coming into the node from the left is is <math>i_1</math>, the current entering the node from the bottom is <math>I</math>, and the current leaving the node from the right is <math>i_2</math>. We therefore have <math>i_1 + I = i_2</math>. Solving for <math>I</math> gives <math>I=i_2 - i_1</math>.

'''Answer:''' <math>\color{blue} I = i_2 - i_1 </math>

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 25/Solution 1

2018-03-26T20:02:28Z

RobertFraser:

We follow the hint and apply Kirchoff's current law to the top node.

The current coming into the node from the left is is <math>i_1</math>, the current entering the node from the bottom is <math>I</math>, and the current leaving the node from the right is <math>i_2</math>. We therefore have <math>i_1 + I = i_2</math>. Solving for <math>I</math> gives <math>\color{blue} I=i_2 - i_1</math>.

'''Answer:''' <math>\color{blue} I = i_2 - i_1 </math>

Thread:Science talk:Math Exam Resources/Courses/MATH152/April 2016/Question A 11/Re-written solution

2018-03-26T19:43:48Z

RobertFraser: New thread: Re-written solution

I have modified the solution in order to include the calculations showing how to solve the linear system. I would like someone to review this for readability. I am sure the answer is correct because Li and I got the same values.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 11

2018-03-26T19:41:55Z

RobertFraser:

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[[Category:MER Tag Ohm's Law]]

{{MER Question page}}

Thread:Science talk:Math Exam Resources/Courses/MATH152/April 2016/Question A 11/Math Exam Resources/Courses/MATH152/April 2016/Question A 11/reply

2018-03-26T19:41:34Z

RobertFraser: Reply to Math Exam Resources/Courses/MATH152/April 2016/Question A 11

Hi Aili,

Current actually can be negative depending on the direction the loop is pointing. I have verified that Li has correctly computed the values of <math>i_1</math> and <math>i_2</math>.

However, I have made some major changes to the structure of the solution, so I am putting it up for a second review.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 11/Solution 1

2018-03-26T19:38:48Z

RobertFraser: Major overhaul in solution structure

By Kirchoff's voltage law, the sum of the voltage drops across the left and right loop must be zero. This leads to the equations

<math> \begin{align}
2i_1 + 1(i_1 - i_2) & = 9 \\
1(i_1 - i_2) - 3i_2 & = 14 \end{align}</math>

We solve this system using row reduction. This system is represented by the augmented matrix

<math> \left[ \begin{array}{cc|c}
3 & -1 & 9 \\
-1 & 4 & -14
\end{array} \right] </math>.

We add <math>1/3</math> of the first row to the second row:

<math> \left[ \begin{array}{cc|c}
3 & -1 & 9 \\
0 & 11/3 & -11
\end{array} \right] </math>.

Now, we multiply the second row by <math>3/11</math>:

<math> \left[ \begin{array}{cc|c}
3 & -1 & 9 \\
0 & 1 & -3
\end{array} \right] </math>.

We add the second row to the first row:

<math> \left[ \begin{array}{cc|c}
3 & 0 & 6 \\
0 & 1 & -3
\end{array} \right] </math>.

Finally, we divide the first row by 3:

<math> \left[ \begin{array}{cc|c}
1 & 0 & 2 \\
0 & 1 & -3
\end{array} \right] </math>.

Therrefore, we get that <math> i_1 = 2 \text{ Amperes}</math> and that <math> i_2 = -3 \text{ Amperes}</math>.

We now verify that this answer makes sense. The direction of <math> i_2 </math> is opposite the direction of the current, because the current is running from high voltage to low voltage. Therefore, we expect that <math> i_2 </math> should be negative and <math> i_1</math> should be positive. This is confirmed by our calculations.

'''Answer:''' <math>\color{blue} i_1 = 2 \text{A}, i_2 = -3 \text{A}</math>

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 11/Hint 1

2018-03-26T19:26:01Z

RobertFraser: Updated the hint

Consider Kirchoff's laws. Keep in mind that both <math>i_1</math> and <math>i_2</math> contribute to the current through the 1 <math>\Omega</math> resistor.

Thread:User talk:RobertFraser/Math 110 (2016) Q8b/reply (2)

2018-03-25T22:35:47Z

RobertFraser: Reply to Math 110 (2016) Q8b

Lol thanks for fixing the inequality in my solution

Thread:User talk:RobertFraser/Math 110 (2016) Q8b/reply

2018-03-25T21:11:59Z

RobertFraser: Reply to Math 110 (2016) Q8b

Hi Nina,

Thank you for pointing this out. I tend to write my solutions on my laptop, save them in plaintext files, and then copy them to the wiki. I copied the wrong one this time.

-Rob

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (b)

2018-03-25T21:10:54Z

RobertFraser:

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







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (b)/Solution 1

2018-03-25T21:10:20Z

RobertFraser:

A linear apprxoimation will be an underestimate if the function <math>f(x) = \frac{1}{x} </math> is concave up between 4 and 4.16, and will be an overestimate if <math>f(x)</math> is concave down on this interval. We computed in part (a) that <math>f'(x) = - \frac{1}{x^2}</math>. We can use this to compute <math>f''(x)</math>:

<math>f''(x) = \frac{2}{x^3}.</math>

This is positive between the values <math> x = 4 </math> and <math> x = 4.16 </math>. Therefore, the function is concave up and the estimate is an underestimate.

'''Answer:''' <math>\color{blue} \frac{1}{4.16} < 0.24.</math>

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (b)/Solution 1

2018-03-25T21:09:33Z

RobertFraser:

A linear apprxoimation will be an underestimate if the function <math>f(x) = \frac{1}{x} </math> is concave up between 4 and 4.16, and will be an overestimate if <math>f(x)</math> is concave down on this interval. We computed in part (a) that <math>f'(x) = - \frac{1}{x^2}</math>. We can use this to compute <math>f''(x)</math:

<math>f''(x) = \frac{2}{x^3}.</math>

This is positive between the values <math> x = 4 </math> and <math> x = 4.16 </math>. Therefore, the function is concave up and the estimate is an underestimate.

'''Answer:''' <math>\color{blue} \frac{1}{4.16} < 0.24.</math>

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (d)

2018-03-20T01:20:02Z

RobertFraser:

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (d)/Solution 1

2018-03-20T01:19:33Z

RobertFraser: Created page with "The second degree Taylor polynomial of <math>\frac{1}{x}</math> at 4 is given by <math>f(4) + f'(4) (x - 4) + \frac{f''(4)}{2} (x - 4)^2 </math> We already computed <math>f(..."

The second degree Taylor polynomial of <math>\frac{1}{x}</math> at 4 is given by

<math>f(4) + f'(4) (x - 4) + \frac{f''(4)}{2} (x - 4)^2 </math>

We already computed <math>f(4) = \frac{1}{4}</math> and <math> f'(4) = \frac{-1}{16}</math>. We have that <math>f''(x) = \frac{2}{x^3}</math>, so <math>f''(4) = \frac{2}{64} = \frac{1}{32} </math>.

So the Taylor polynomial is

<math>\frac{1}{4} - \frac{1}{16} (x-4) + \frac{1}{64} (x-4)^2 </math>.

'''Answer:''' <math>\color{blue} \frac{1}{4} - \frac{1}{16} (x-4) + \frac{1}{64} (x - 4)^2 </math>

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (d)/Hint 1

2018-03-20T01:19:00Z

RobertFraser: Created page with "Recall the formula <math> f(a) + f'(a) (x-a) + \frac{f''(a)}{2} (x-a)^2 </math> for the second-degree Taylor polynomial for <math>f(x)</math> at <math>x = a</math>."

Recall the formula

<math> f(a) + f'(a) (x-a) + \frac{f''(a)}{2} (x-a)^2 </math>

for the second-degree Taylor polynomial for <math>f(x)</math> at <math>x = a</math>.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (c)

2018-03-20T01:13:16Z

RobertFraser:

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (c)/Solution 1

2018-03-20T01:12:51Z

RobertFraser:

You should start by drawing a graph of <math>\frac{1}{x}</math>. We take the tangent line at the point <math>(4, \frac{1}{4})</math>. We compare the point <math>(4.16, \frac{1}{4.16})</math> on the graph of <math> y = \frac{1}{x}</math> and the point <math>(4.16, 0.24)</math> on the graph of the tangent line.

[[File:2016 Apr Math 110 8c.png]]

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (c)/Solution 1

2018-03-20T01:12:30Z

RobertFraser: Created page with "You should start by drawing a graph of <math>\frac{1}{x}</math>. We take the tangent line at the point <math>(4, \frac{1}{4}</math>. We compare the point <math>(4.16, \frac{1}..."

You should start by drawing a graph of <math>\frac{1}{x}</math>. We take the tangent line at the point <math>(4, \frac{1}{4}</math>. We compare the point <math>(4.16, \frac{1}{4.16})</math> on the graph of <math> y = \frac{1}{x}</math> and the point <math>(4.16, 0.24)</math> on the graph of the tangent line.

[[File:2016 Apr Math 110 8c.png]]

File:2016 Apr Math 110 8c.png

2018-03-20T01:11:29Z

RobertFraser: User created page with UploadWizard

=={{int:filedesc}}==
{{Information
|description={{en|1=Graph for Problem 8c of the 2016 Apr Math 110 Exam}}
|date=2018-03-19 18:10:57
|source={{own}}
|author=[[User:RobertFraser|Robert Fraser]]
|permission=
|other_versions=
}}

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

[[Category:Uploaded with UploadWizard]]

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (c)/Hint 1

2018-03-20T00:58:57Z

RobertFraser: Created page with "You should know how to draw the graph of <math>\frac{1}{x} </math> for <math> x > 0 </math>. If you forgot, just make sure you know whether the function is increasing/decreasi..."

You should know how to draw the graph of <math>\frac{1}{x} </math> for <math> x > 0 </math>. If you forgot, just make sure you know whether the function is increasing/decreasing and make sure you know the concavity.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (b)

2018-03-20T00:56:48Z

RobertFraser:

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (b)/Solution 1

2018-03-20T00:56:26Z

RobertFraser:

We want to approximate the value

<math> \frac{1}{4.16}.</math>

To this end, we define the function <math>f(x) = \frac{1}{x} </math>.

We will use a linear approximation:

<math> f(x) \approx f(a) + f'(a) (x-a) </math>

We want to use <math>x = 4.16</math> and <math>a = 4</math>. It is necessary to compute <math>f'(4)</math>. The derivative of <math>f(x)</math> is

<math> f'(x) = - \frac{1}{x^2} </math>

so

<math> f'(4) = - \frac{1}{4^2} = - \frac{1}{16} </math>

so

<math> f(4.16) \approx \frac{1}{4} - \frac{1}{16} (4.16 - 4) </math>

and therefore

<math> f(4.16) \approx 0.25 - \frac{0.16}{16} </math>

and we get the approximation

<math> f(4.16) \approx 0.24.</math>

This is believable because we expect to get something just a little less than <math>0.25</math>, since we are dividing by a number that is just a little more than 4.

'''Answer:''' <math> \color{blue} \frac{1}{4.16} \approx 0.24 </math>

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (b)/Solution 1

2018-03-20T00:56:01Z

RobertFraser: Created page with "We want to approximate the value <math> \frac{1}{4.16}.</math> To this end, we define the function <math>f(x) = \frac{1}{x} </math>. We will use a linear approximation: <..."

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (b)/Hint 1

2018-03-20T00:55:36Z

RobertFraser: Created page with "The answer to this question is dictated by the second derivative of <math>f(x) = \frac{1}{x}</math>. Try drawing a picture if you can't remember which sign corresponds to an o..."

The answer to this question is dictated by the second derivative of <math>f(x) = \frac{1}{x}</math>. Try drawing a picture if you can't remember which sign corresponds to an overestimate and which sign corresponds to an underestimate.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (a)

2018-03-20T00:47:57Z

RobertFraser:

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (a)/Hint 1

2018-03-20T00:47:31Z

RobertFraser: Created page with "Try using a linear approximation. You should set <math>f(x) = \frac{1}{x}</math> and center your approximation around <math>a = 4</math>."

Try using a linear approximation. You should set <math>f(x) = \frac{1}{x}</math> and center your approximation around <math>a = 4</math>.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (a)/Solution 1

2018-03-20T00:46:55Z

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 06 (b)

2018-03-19T23:51:24Z

RobertFraser:

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 06 (b)/Solution 1

2018-03-19T23:51:03Z

RobertFraser: Created page with "File:Math 110 Apr 2016 6.png We would like to relate the angle <math>\theta</math> to the height <math>h</math>. In order to do so, we use trigonometry. By basic trigonom..."

[[File:Math 110 Apr 2016 6.png]]

We would like to relate the angle <math>\theta</math> to the height <math>h</math>. In order to do so, we use trigonometry. By basic trigonometry, we have the equation

<math> \tan \theta = \frac{h}{1000}.</math>

We will differentiate both sides:

<math> \sec^2 \theta \frac{d \theta}{d t} = \frac{1}{1000} \frac{dh}{dt}.</math>

Now, we solve for <math>\frac{d \theta}{dt} </math>:

<math> \frac{d \theta}{dt} = \frac{1}{1000 \sec^2(\theta)} \frac{dh}{dt}. </math>

We use the fact that <math>\sec \theta = \frac{1}{\cos \theta} </math>.

<math> \frac{d \theta}{dt} = \frac{\cos^2 \theta}{1000} \frac{dh}{dt} </math>

Now, we plug in <math>\frac{dh}{dt} = 500</math>:

<math> \begin{align}
\frac{d \theta}{dt} & = \frac{500\cos^2 \theta}{1000} \\
\frac{d \theta}{dt} & = \frac{\cos^2 \theta}{2}
\end{align}</math>

Now, we observe from our picture that <math>\cos(\theta) = \frac{h}{D}</math>.

<math> \frac{d \theta}{dt} = \frac{h^2}{2 D^2} </math>

We plug in <math>h = 2000</math> and <math>D^2 = 5000000</math> (from part a):

<math> \begin{align}
\frac{d \theta}{dt} & = \frac{4000000}{10000000} \\
& = \frac{2}{5}
\end{align}</math>

So the rate of change is <math>\frac{2}{5} \frac{\text{radians}}{\text{sec}}</math>.

<math>\color{blue} \frac{d \theta}{dt} = \frac{2}{5} \frac{\text{radians}}{\text{sec}}</math>

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 06 (b)/Hint 1

2018-03-19T23:23:13Z

RobertFraser: Created page with "File:2016 Apr Math 110 6.png Try to find a trigonometric function that relates the angle <math>\theta</math> to the height <math>h</math> in the picture."

[[File:2016 Apr Math 110 6.png]]

Try to find a trigonometric function that relates the angle <math>\theta</math> to the height <math>h</math> in the picture.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 06 (a)

2018-03-19T23:17:54Z

RobertFraser:

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 06 (a)/Solution 1

2018-03-19T23:17:22Z

RobertFraser: Created page with "We refer to the following picture: File:2016 Apr Math 110 6.png We want to compute <math>\frac{dD}{dt}</math>. In order to compute this, we need to come up with an equat..."

We refer to the following picture:

[[File:2016 Apr Math 110 6.png]]

We want to compute <math>\frac{dD}{dt}</math>. In order to compute this, we need to come up with an equation relating the rate we know, <math>\frac{dh}{dt}</math>, to the rate we want to know, <math>\frac{dD}{dt}</math>.

By the Pythagorean theorem, we have

<math> D^2 = 1000^2 + h^2.</math>

We want the derivative with respect to <math>t</math>. Note that we should take the derivative before plugging in the value of <math>h</math> or <math>D</math>.

<math> 2D \frac{dD}{dt} = 2h \frac{dh}{dt} </math>

so

<math>\frac{dD}{dt} = \frac{h}{d} \frac{dh}{dt}.</math>

We would like to plug in <math>h = 2000 </math> and <math> \frac{dh}{dt} = 500</math>. However, we still need to figure out what value to plug in for <math>D</math>. Once again, we will use the Pythagorean theorem. We get

<math> \begin{align}
D^2 & = 1000^2 + 2000^2 \\
D^2 & = 5000000 \\
D & = \sqrt{5000000} \\
D & = 1000\sqrt{5}.
\end{align} </math>

So we plug in the values for <math>D</math>, <math>h</math>, and <math>\frac{dh}{dt}</math>, and get

<math> \begin{align}
\frac{dD}{dt} & = \frac{h}{D} \frac{dh}{dt} \\
\frac{dD}{dt} & = \frac{1000}{1000 \sqrt{5}} 500 \\
\frac{dD}{dt} & = \frac{500}{\sqrt 5} \\
\frac{dD}{dt} & = 100 \sqrt{5}
\end{align}</math>

So we have that the distance is changing at a rate of <math>100 \sqrt{5} \frac{\text{m}}{\text{sec}}</math>.

'''Answer:''' <math>\color{blue} 100 \sqrt{5} \frac{\text{m}}{\text{sec}} </math>

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 06 (a)/Hint 1

2018-03-19T23:01:04Z

RobertFraser: Created page with "Try drawing a picture to show the situation in the problem. It is usually helpful to label any changing quantity with a variable instead of a constant. Your picture should lo..."

Try drawing a picture to show the situation in the problem. It is usually helpful to label any changing quantity with a variable instead of a constant.

Your picture should look something like the following.

[[File:2016 Apr Math 110 6.png]]

File:2016 Apr Math 110 6.png

2018-03-19T23:00:25Z

RobertFraser: User created page with UploadWizard

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|date=2018-03-19 15:59:50
|source={{own}}
|author=[[User:RobertFraser|Robert Fraser]]
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{{self|cc-by-sa-3.0}}

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Science:Math Exam Resources/Courses/MATH110/April 2016/Question 07

2018-03-19T22:00:35Z

RobertFraser: