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Thread:Science talk:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (c)/Hint 1/Need a better explanation/reply

2018-03-27T08:48:21Z

BRIANCHAN: Reply to Need a better explanation

Hi Han,

Thanks for the feedback, you might be right. However, the hint and solution for this question has already been approved by another reviewer (status for both is QG), so it may be a bit awkward for me to make further revisions to this question.

Brian

Thread:Science talk:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (f)/Answer in reference to previous question?/reply

2018-03-27T08:40:06Z

BRIANCHAN: Reply to Answer in reference to previous question?

Thanks for the feedback. However, for a differentiable function, a point in its domain is a critical point (local maximum or local minimum) iff the derivative at that point is zero, and the domain of this function excludes zero.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question B 01 (c)/Solution 2

2018-03-27T08:24:19Z

BRIANCHAN:

Translate this problem into a problem involving a system of linear equations. To do this, note that <math> L_1 </math> and <math> L_2 </math> intersect exactly when there are numbers <math> s </math> and <math> t </math> that satisfy

<math> [0,2,1] + s[-1,2,2] = [-1,0,3] + t[-2,1,1] </math>

rearranging this gives

<math> s[-1,2,2] - t[-2,1,1] = [-1,0,3] - [0,2,1] </math>

and

<math> s[-1,2,2] + t[2,-1,-1] = [-1 - 0,0 - 2,3 - 1] = [-1, -2, 2] </math>

Writing as a system of linear equations gives,

<math> [-s + 2t,2s - t, 2s - t] = [-1, -2, 2] </math>

In matrix form,

<math>
\begin{pmatrix}
-1 & 2 & | & -1 \\
2 & -1 & | & -2 \\
2 & -1 & | & 2
\end{pmatrix}
</math>

In other words, <math> L_1 </math> and <math> L_2 </math> intersect exactly when the above system of linear equations has at least one solution. Using Gaussian Elimination, we obtain

<math>
\begin{pmatrix}
-1 & 2 & | & -1 \\
2 & -1 & | & -2 \\
2 & -1 & | & 2
\end{pmatrix} \to
\begin{pmatrix}
-2 & 4 & | & -2 \\
2 & -1 & | & -2 \\
2 & -1 & | & 2
\end{pmatrix} \to
\begin{pmatrix}
-2 & 4 & | & -2 \\
0 & 3 & | & -4 \\
0 & 3 & | & 0
\end{pmatrix} \to
\begin{pmatrix}
-2 & 4 & | & -2 \\
0 & 0 & | & -4 \\
0 & 3 & | & 0
\end{pmatrix}
</math>

The second row of the last matrix reads

<math>
0s + 0t = -4
</math>

which is impossible. Hence, the system of linear equations has no solution and the lines <math> L_1 </math> and <math> L_2 </math> do not intersect.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 21/Hint 3

2018-03-27T08:20:51Z

BRIANCHAN:

Let <math> \pi_1 </math> denote the projection onto <math> L_1 </math> and let <math> \pi_2 </math> denote the projection onto <math> L_2 </math>. Use the geometric interpretation of projection onto a line to determine the range of <math> \pi_1 </math>. Then, use the same argument, and the fact that <math> L_1 </math> and <math> L_2 </math> are perpendicular, to determine the range of <math> \pi_2 \circ \pi_1 </math>.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 21/Solution 2

2018-03-27T08:20:20Z

BRIANCHAN:

As in the (third) hint, let <math> \pi_1 </math> denote the projection onto <math> L_1 </math> and let <math> \pi_2 </math> denote the projection onto <math> L_2 </math>. If <math> x </math> is a point on the plane, then (by the geometric interpretation of projection) <math> \pi_1(x) </math> is the point in <math> L_1 </math> such that the line containing <math> x </math> and <math> \pi_1(x) </math> is perpendicular to <math> L_1 </math>. It follows that the range of <math> \pi_1 </math> is <math> L_1 </math>.

Now, let <math> x </math> be a point on the plane like before. To determine <math> (\pi_2 \circ \pi_1)(x) </math> (recall that <math> (\pi_2 \circ \pi_1)(x) = \pi_2(\pi_1(x)) </math>), note the following. <math> x </math> is mapped to <math> \pi_1(x) </math>, which is in <math> L_1 </math>. Moreover, <math> \pi_1(x) </math> is mapped (by the geometric interpretation of projection) to the point <math> \pi_2(\pi_1(x)) </math> in <math> L_2 </math> such that the line containing <math> \pi_1(x) </math> and <math> \pi_2(\pi_1(x)) </math> is perpendicular to <math> L_2 </math>. But <math> L_1 </math> is perpendicular to <math> L_2 </math>, and <math> \pi_1(x) </math> is contained in <math> L_1 </math>, so <math> \pi_2(\pi_1(x)) </math> is contained in <math> L_1 </math> and in <math> L_2 </math>. As the only point contained in <math> L_1 </math> and in <math> L_2 </math> is the origin, the range of <math> \pi_2 \circ \pi_1 </math> is the origin.

And the only matrix that maps the entire plane to the origin is the following matrix, which is our answer

<math>
\color{blue}

\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}

</math>

Science:Math Exam Resources/Courses/MATH152/April 2016/Question B 01 (c)

2018-03-27T08:18:49Z

BRIANCHAN:

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







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH152/April 2016/Question B 01 (c)/Solution 2

2018-03-27T08:17:11Z

BRIANCHAN: Created page with "First translate this problem into a problem involving a system of linear equations. To do this, note that <math> L_1 </math> and <math> L_2 </math> intersect exactly when ther..."

First translate this problem into a problem involving a system of linear equations. To do this, note that <math> L_1 </math> and <math> L_2 </math> intersect exactly when there are numbers <math> s </math> and <math> t </math> that satisfy

<math> [0,2,1] + s[-1,2,2] = [-1,0,3] + t[-2,1,1] </math>

rearranging this gives

<math> s[-1,2,2] - t[-2,1,1] = [-1,0,3] - [0,2,1] </math>

and

<math> s[-1,2,2] + t[2,-1,-1] = [-1 - 0,0 - 2,3 - 1] = [-1, -2, 2] </math>

Writing as a system of linear equations gives,

<math> [-s + 2t,2s - t, 2s - t] = [-1, -2, 2] </math>

In matrix form,

<math>
\begin{pmatrix}
-1 & 2 & | & -1 \\
2 & -1 & | & -2 \\
2 & -1 & | & 2
\end{pmatrix}
</math>

In other words, <math> L_1 </math> and <math> L_2 </math> intersect exactly when the above system of linear equations has at least one solution. Using Gaussian Elimination, we obtain

<math>
\begin{pmatrix}
-1 & 2 & | & -1 \\
2 & -1 & | & -2 \\
2 & -1 & | & 2
\end{pmatrix} \to
\begin{pmatrix}
-2 & 4 & | & -2 \\
2 & -1 & | & -2 \\
2 & -1 & | & 2
\end{pmatrix} \to
\begin{pmatrix}
-2 & 4 & | & -2 \\
0 & 3 & | & -4 \\
0 & 3 & | & 0
\end{pmatrix} \to
\begin{pmatrix}
-2 & 4 & | & -2 \\
0 & 0 & | & -4 \\
0 & 3 & | & 0
\end{pmatrix}
</math>

The second row of the last matrix reads

<math>
0s + 0t = -4
</math>

which is impossible. Hence, the system of linear equations has no solution and the lines <math> L_1 </math> and <math> L_2 </math> do not intersect.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question B 01 (c)/Hint 2

2018-03-27T08:01:42Z

BRIANCHAN: Created page with "Translate this problem into a system of linear equations problem."

Translate this problem into a system of linear equations problem.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 24/Hint 1

2018-03-27T07:52:04Z

BRIANCHAN:

Definition of eigenvectors and eigenvalues.

In Matlab, command <math>[V,D] = eig(A)</math> returns diagonal matrix <math>D</math> of eigenvalues and matrix <math>V</math> whose columns are the corresponding right eigenvectors, so that <math>A*V = V*D</math>

Note also that there are no repeated diagonal entries in <math> D </math>. This can be used for the question.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 24/Hint 1

2018-03-27T07:51:19Z

BRIANCHAN:

Definition of eigenvectors and eigenvalues.

In Matlab, command <math>[V,D] = eig(A)</math> returns diagonal matrix <math>D</math> of eigenvalues and matrix <math>V</math> whose columns are the corresponding right eigenvectors, so that <math>A*V = V*D</math>

Note also that there are no repeated non-zero entries in <math> D </math>. This can be used for the question.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 24/Hint 1

2018-03-27T07:50:54Z

BRIANCHAN:

Definition of eigenvectors and eigenvalues.

In Matlab, command <math>[V,D] = eig(A)</math> returns diagonal matrix <math>D</math> of eigenvalues and matrix <math>V</math> whose columns are the corresponding right eigenvectors, so that <math>A*V = V*D</math>

Note also that there are no repeated non-zero entries in <math> D </math>. This can be used for this question.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 21

2018-03-27T07:26:42Z

BRIANCHAN:

[[Category:MER QGQ flag]][[Category:MER RH flag]][[Category:MER RS flag]][[Category:MER QGT flag]]





[[Category:MER Tag Orthogonal projection]]

{{MER Question page}}

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 21/Solution 2

2018-03-27T07:24:47Z

BRIANCHAN: Created page with "As in the (third) hint, let <math> \pi_1 </math> denote the projection onto <math> L_1 </math> and let <math> \pi_2 </math> denote the projection onto <math> L_2 </math>. I..."

As in the (third) hint, let <math> \pi_1 </math> denote the projection onto <math> L_1 </math> and let <math> \pi_2 </math> denote the projection onto <math> L_2 </math>. If <math> x </math> is a point on the plane, then (by the geometric interpretation of projection) <math> \pi_1(x) </math> is the point in <math> L_1 </math> such that the line containing <math> x </math> and <math> \pi_1(x) </math> is orthogonal to <math> L_1 </math>. It follows that the range of <math> \pi_1 </math> is <math> L_1 </math>.

Now, let <math> x </math> be a point on the plane like before. To determine <math> (\pi_2 \circ \pi_1)(x) </math> (recall that <math> (\pi_2 \circ \pi_1)(x) = \pi_2(\pi_1(x)) </math>), note the following. <math> x </math> is mapped to <math> \pi_1(x) </math>, which is in <math> L_1 </math>. Moreover, <math> \pi_1(x) </math> is mapped (by the geometric interpretation of projection) to the point <math> \pi_2(\pi_1(x)) </math> in <math> L_2 </math> such that the line containing <math> \pi_1(x) </math> and <math> \pi_2(\pi_1(x)) </math> is orthogonal to <math> L_2 </math>. But <math> L_1 </math> is orthogonal to <math> L_2 </math>, and <math> \pi_1(x) </math> is contained in <math> L_1 </math>, so <math> \pi_2(\pi_1(x)) </math> is contained in <math> L_1 </math> and in <math> L_2 </math>. As the only point contained in <math> L_1 </math> and in <math> L_2 </math> is the origin, the range of <math> \pi_2 \circ \pi_1 </math> is the origin.

And the only matrix that maps the entire plane to the origin is the following matrix, which is our answer

<math>
\color{blue}

\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}

</math>

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 21/Hint 3

2018-03-27T07:07:31Z

BRIANCHAN: Created page with "Let <math> \pi_1 </math> denote the projection onto <math> L_1 </math> and let <math> \pi_2 </math> denote the projection onto <math> L_2 </math>. Use the geometric interpret..."

Let <math> \pi_1 </math> denote the projection onto <math> L_1 </math> and let <math> \pi_2 </math> denote the projection onto <math> L_2 </math>. Use the geometric interpretation of projection onto a line to determine the range of <math> \pi_1 </math>. Then, use the same argument, and the fact that <math> L_1 </math> and <math> L_2 </math> are orthogonal, to determine the range of <math> \pi_2 \circ \pi_1 </math>.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 02 (d)

2018-03-21T08:33:39Z

BRIANCHAN:

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 02 (d)/Solution 1

2018-03-21T08:33:10Z

BRIANCHAN:

Recall that the derivative of <math> \cos(x) </math> is <math> -\sin(x) </math>. So the derivative of <math> -\cos(x) </math> is <math> \sin(x) </math>. Hence,<math> -\cos(x) + c </math> is the antiderivative of <math> \sin(x) </math>. Moreover, the Power Rule states that the derivative of <math> x^a </math> is <math> ax^{a-1} </math> so in particular, the derivative of <math> x^{1/2} </math> is <math> \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} </math>. So as <math> x^{1/2} </math> is the antiderivative of <math> \frac{1}{2}x^{-1/2} </math>, it follows that <math> 2/\sqrt{x} = 2x^{-1/2} = 4 \times \frac{1}{2} x^{-1/2} </math> and that the antiderivative of <math> 2/\sqrt{x} </math> is <math> 4 x^{1/2} + c = 4 \sqrt{x} + c. </math> Thus, the antiderivative of <math> \sin(x) + 2/\sqrt{x} </math> is <math> \color{blue} -\cos(x) + 4 \sqrt{x} + c. </math>

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 02 (d)/Solution 1

2018-03-21T08:32:43Z

BRIANCHAN:

Recall that the derivative of <math> \cos(x) </math> is <math> -\sin(x) </math>. So the derivative of <math> -\cos(x) </math> is <math> \sin(x) </math>. Hence,<math> -\cos(x) + c </math> is the antiderivative of <math> \sin(x) </math>. Moreover, the Power Rule states that the derivative of <math> x^a </math> is <math> ax^{a-1} </math> so in particular, the derivative of <math> x^{1/2} </math> is <math> \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} </math>. So as <math> x^{1/2} </math> is the antiderivative of <math> \frac{1}{2}x^{-1/2} </math>, it follows that <math> 2/\sqrt{x} = 2x^{-1/2} = 4 \times \frac{1}{2} x^{-1/2} </math> and that the antiderivative of <math> 2/\sqrt{x} </math> is <math> 4 x^{1/2} + c = 4 \sqrt{x} + c. </math> Thus, the antiderivative of <math> \sin(x) + 2/\sqrt{x} </math> is

<math> \color{blue} -\cos(x) + 4 \sqrt{x} + c </math>

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 02 (d)/Solution 1

2018-03-21T08:31:39Z

BRIANCHAN:

Recall that the derivative of <math> \cos(x) </math> is <math> -\sin(x) </math>. So the derivative of <math> -\cos(x) </math> is <math> \sin(x) </math>. Hence,<math> -\cos(x) </math> is the antiderivative of <math> \sin(x) </math>. Moreover, the Power Rule states that the derivative of <math> x^a </math> is <math> ax^{a-1} </math> so in particular, the derivative of <math> x^{1/2} </math> is <math> \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} </math>. So as <math> x^{1/2} </math> is the antiderivative of <math> \frac{1}{2}x^{-1/2} </math>, it follows that <math> 2/\sqrt{x} = 2x^{-1/2} = 4 \times \frac{1}{2} x^{-1/2} </math> and that the antiderivative of <math> 2/\sqrt{x} </math> is <math> 4 x^{1/2} = 4 \sqrt{x}. </math> Thus, the antiderivative of <math> \sin(x) + 2/\sqrt{x} </math> is

<math> \color{blue} -\cos(x) + 4 \sqrt{x} + c </math>

where <math> c </math> is a constant.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 02 (d)/Solution 1

2018-03-21T08:31:15Z

BRIANCHAN: Created page with "Recall that the derivative of <math> \cos(x) </math> is <math> -\sin(x) </math>. So the derivative of <math> -\cos(x) </math> is <math> \sin(x) </math>. Hence,<math> -\cos(x)..."

Recall that the derivative of <math> \cos(x) </math> is <math> -\sin(x) </math>. So the derivative of <math> -\cos(x) </math> is <math> \sin(x) </math>. Hence,<math> -\cos(x) </math> is the antiderivative of <math> \sin(x) </math>. Moreover, the Power Rule states that the derivative of <math> x^a </math> is <math> ax^{a-1} </math> so in particular, the derivative of <math> x^{1/2} </math> is <math> \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} </math>. So as <math> x^{1/2} </math> is the antiderivative of <math> \frac{1}{2}x^{-1/2} </math>, it follows that <math> 2/\sqrt{x} = 2x^{-1/2} = 4 \times \frac{1}{2} x^{-1/2} </math> and that the antiderivative of <math> 2/\sqrt{x} </math> is <math> 4 x^{1/2} = 4 \sqrt{x}. </math> Thus, the antiderivative of <math> \sin(x) + 2/\sqrt{x} </math> is

<math> -\cos(x) + 4 \sqrt{x} + c </math>

where <math> c </math> is a constant.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 02 (d)/Hint 1

2018-03-21T08:24:50Z

BRIANCHAN: Created page with "Recall the derivatives of the trigonometric functions and the Power Rule."

Recall the derivatives of the trigonometric functions and the Power Rule.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (g)/Solution 1

2018-03-21T08:22:03Z

BRIANCHAN:

We could use the quotient rule to calculate the derivative, but instead we will break the function into a sum of parts and differentiate them individually. From part (d), we have <math> f'(x) = \frac{2 - x}{x^3} = \frac{2}{x^3} - \frac{x}{x^3} = \frac{-1}{x^2} + \frac{2}{x^3} = -x^{-2} + 2x^{-3} </math>. Then <math> f''(x) = \frac{d}{dx}(-x^{-2}) + \frac{d}{dx}(2x^{-3}). </math> And using the Power Rule of differentiation gives <math> \frac{d}{dx}(-x^{-2}) + \frac{d}{dx}(2x^{-3}) = 2x^{-3} - 6x^{-4} = \frac{2}{x^3} - \frac{6}{x^4}. </math>

Hence,

<math> \color{blue} f''(x) = \frac{2}{x^3} - \frac{6}{x^4}. </math>

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (g)/Solution 1

2018-03-21T08:21:08Z

BRIANCHAN:

We could use the quotient rule to calculate the derivative, but instead we will break the function into a sum of parts and differentiate them individually. From part (d), we have <math> f'(x) = \frac{2 - x}{x^3} = \frac{-1}{x^2} + \frac{2}{x^3} = x^{-2} + 2x^{-3} </math>. Then <math> f''(x) = \frac{d}{dx}(-x^{-2}) + \frac{d}{dx}(2x^{-3}). </math> And using the Power Rule of differentiation gives <math> \frac{d}{dx}(-x^{-2}) + \frac{d}{dx}(2x^{-3}) = 2x^{-3} - 6x^{-4} = \frac{2}{x^3} - \frac{6}{x^4}. </math>

Hence,

<math> \color{blue} f''(x) = \frac{2}{x^3} - \frac{6}{x^4}. </math>

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (g)/Solution 1

2018-03-21T08:20:44Z

BRIANCHAN:

We could use the quotient rule to calculate the derivative, but instead we will break the function into a sum of parts and differentiate them individually. From part (d), we have <math> f'(x) = \frac{2 - x}{x^3} = \frac{-1}{x^2} + \frac{2}{x^3} = x^{-2} + 2x^{-3} </math>. Then <math> f''(x) = \frac{d}{dx}(-x^{-2}) + \frac{d}{dx}(2x^{-3}). </math> And using the Power Rule of differentiation gives <math> \frac{d}{dx}(-x^{-2}) + \frac{d}{dx}(2x^{-3}) = -2x^{-3} - 6x^{-4} = \frac{2}{x^3} - \frac{6}{x^4}. </math>

Hence,

<math> \color{blue} f''(x) = \frac{2}{x^3} - \frac{6}{x^4}. </math>

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (g)/Solution 1

2018-03-21T08:19:47Z

BRIANCHAN:

We could use the quotient rule to calculate the derivative, but instead we will break the function into a sum of parts and differentiate them individually. From part (d), we have <math> f'(x) = \frac{2 - x}{x^3} = \frac{-1}{x^2} + \frac{2}{x^3} = x^{-2} + 2x^{-3} </math>. Then <math> f''(x) = \frac{d}{dx}(x^{-2}) + \frac{d}{dx}(2x^{-3}). </math> And using the Power Rule of differentiation gives <math> \frac{d}{dx}(x^{-2}) + \frac{d}{dx}(2x^{-3}) = -2x^{-3} - 6x^{-4} = \frac{-2}{x^3} - \frac{6}{x^4}. </math>

Hence,

<math> \color{blue} f''(x) = \frac{-2}{x^3} - \frac{6}{x^4}. </math>

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (g)

2018-03-21T08:19:03Z

BRIANCHAN:

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (g)/Solution 1

2018-03-21T08:18:29Z

BRIANCHAN: Created page with "We could use the quotient rule to calculate the derivative, but instead we will break the function into a sum of parts and differentiate them individually. From part (d), we h..."

We could use the quotient rule to calculate the derivative, but instead we will break the function into a sum of parts and differentiate them individually. From part (d), we have

<math> f'(x) = \frac{2 - x}{x^3} = \frac{-1}{x^2} + \frac{2}{x^3} = x^{-2} + 2x^{-3} </math>

Then

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

And using the Power Rule of differentiation gives

<math> \frac{d}{dx}(x^{-2}) + \frac{d}{dx}(2x^{-3}) = -2x^{-3} - 6x^{-4} = \frac{-2}{x^3} - \frac{6}{x^4}. </math>

Hence,

<math> \color{blue} f''(x) = \frac{-2}{x^3} - \frac{6}{x^4} </math>.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (f)/Solution 1

2018-03-21T08:13:31Z

BRIANCHAN:

From part (d), <math> f'(x) = \frac{2 - x}{x^3} </math>. So to solve <math> f'(x) = 0 </math>, we calculate <math> \frac{2 - x}{x^3} = 0 </math>, <math> 2 - x = 0 </math>, <math> x = 2 </math>. So the only local maxima or minima is <math> x = 2 </math>. At <math> x = 2 </math>, <math> f(2) = \frac{2-1}{2^2} = 1/4 </math>. Moreover, <math> f(1000) = \frac{999}{1000000} < 0.001 </math>, so <math> x = 2 </math> is a local minima.

Answer: <math> \color{blue} x = 2 \text{ and it is a local minima.} </math>

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (g)/Hint 1

2018-03-21T08:11:33Z

BRIANCHAN: Created page with "Use the answer to part (d). Then use the quotient rule or break the function into a sum of functions and differentiate each of them individually."

Use the answer to part (d). Then use the quotient rule or break the function into a sum of functions and differentiate each of them individually.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (f)

2018-03-21T08:06:41Z

BRIANCHAN:

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (f)/Solution 1

2018-03-21T08:05:50Z

BRIANCHAN: Created page with "From part (d), <math> f'(x) = \frac{-1}{x^2} + \frac{2}{x^3} </math>. So to solve <math> f'(x) = 0 </math>, we calculate <math> \frac{-1}{x^2} + \frac{2}{x^3} = 0 </math>, <m..."

From part (d), <math> f'(x) = \frac{-1}{x^2} + \frac{2}{x^3} </math>. So to solve <math> f'(x) = 0 </math>, we calculate <math> \frac{-1}{x^2} + \frac{2}{x^3} = 0 </math>, <math> \frac{2}{x^3} = \frac{-1}{x^2} </math>, <math> 2 = x </math>. So the only local maxima or minima is <math> x = 2 </math>. At <math> x = 2 </math>, <math> f(2) = \frac{2-1}{2^2} = 1/4 </math>. Moreover, <math> f(1000) = \frac{999}{1000000} < 0.001 </math>, so <math> x = 2 </math> is a local minima.

Answer: <math> \color{blue} x = 2 \text{ and it is a local minima.} </math>

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (f)/Hint 1

2018-03-21T08:00:20Z

BRIANCHAN: Created page with "Recall that the local maxima and minima are exactly the points at which <math> f'(x) = 0 </math>. To see whether <math> x = s </math> local maxima or local minima, calculate <..."

Recall that the local maxima and minima are exactly the points at which <math> f'(x) = 0 </math>. To see whether <math> x = s </math> local maxima or local minima, calculate <math> f(s) </math> and calculate other values of <math> f(x) </math> for different <math> x </math>.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (c)

2018-03-21T07:51:58Z

BRIANCHAN:

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (c)/Hint 1

2018-03-21T07:51:23Z

BRIANCHAN:

Determine the values of <math> x </math> for which the numerator <math> x - 1 </math> is zero, then determine the values of <math> x </math> for which the denominator <math> x^2 </math> is zero. Moreover, determine the behavior of <math> f(x) </math> when the absolute value of <math> x </math> is large.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (c)/Solution 1

2018-03-21T07:49:41Z

BRIANCHAN:

The numerator <math> x - 1 </math> is zero exactly when <math> x = 1 </math> and the denominator <math> x^2 </math> is zero exactly when <math> x = 0 </math>. So, as the numerator <math> x - 1 </math> is not zero at <math> x = 0 </math>, <math> f(x) </math> has exactly one vertical asymptote, which is <math> x = 0 </math>. Moreover, we also see that

<math> \lim_{x \to - \infty} f(x) = \lim_{x \to - \infty} \frac{x - 1}{x^2} = \lim_{x \to -\infty} \frac{1 - 1/x}{x} = 0 </math>

and

<math> \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x - 1}{x^2} = \lim_{x \to \infty} \frac{1 - 1/x}{x} = 0 </math>

and it follows that it has <math> y = 0 </math> as its horizontal asymptote. Answer, the asymptotes are: <math> \color{blue} x = 0, y = 0 </math>.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (c)/Solution 1

2018-03-21T07:48:59Z

BRIANCHAN: Created page with "The numerator <math> x - 1 </math> is zero exactly when <math> x = 1 </math> and the denominator <math> x^2 </math> is zero exactly when <math> x = 0 </math>. So, as the numer..."

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (c)/Hint 1

2018-03-21T07:44:30Z

BRIANCHAN: Created page with "Determine the values of <math> x </math> for which the numerator <math> x - 1 </math> is zero, then determine the values of <math> x </math> for which the denominator <math> x..."

Determine the values of <math> x </math> for which the numerator <math> x - 1 </math> is zero, then determine the values of <math> x </math> for which the denominator <math> x^2 </math> is zero.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 09 (b)

2018-03-21T07:35:54Z

BRIANCHAN:

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 09 (a)

2018-03-21T07:35:35Z

BRIANCHAN:

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 09 (b)/Solution 1

2018-03-21T07:34:35Z

BRIANCHAN:

No such function exists. To see this, use the Mean Value Theorem twice. Let <math> a = -1 </math> and <math> b = 2 </math>. Because <math> f'' < 0 </math> everywhere, <math> f </math> and <math> f' </math> are differentiable everywhere and we can use the Mean Value Theorem for <math> f </math> and <math> f' </math> on the closed interval <math> [a, b] </math>. By the Mean Value Theorem applied to <math> f </math> on <math> [a, b] </math>, there is a number <math> s </math> such that <math> -1 = a < s and <math> f'(s) = \frac{f(b) - f(a)}{b - a} = \frac{f(2) - f(-1)}{2 - (-1)} = \frac{5 - (-1)}{2 - (-1)} = \frac{6}{3} = 2 </math>.

Now let <math> a = s </math> and <math> b = 2 </math>, where <math> s </math> is as defined above. By the Mean Value Theorem applied to <math> f' </math> on <math> [a, b] </math>, there is a number <math> t </math> such that <math> s = a < t and <math> f''(t) = \frac{f'(b) - f'(a)}{b - a} = \frac{f'(2) - f'(s)}{2 - s} = \frac{3 - 2}{2 - s} = \frac{1}{2 - s} > 0 </math>. But then, <math> f''(t) > 0 </math> contradicting the assumption that <math> f'' < 0 </math> everywhere. Hence, no such function <math> f </math> exists.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 09 (b)/Solution 1

2018-03-21T07:33:28Z

BRIANCHAN:

No such function exists. To see this, we will use the Mean Value Theorem twice. Let <math> a = -1 </math> and <math> b = 2 </math>. Because <math> f'' < 0 </math> everywhere, <math> f </math> and <math> f' </math> are differentiable everywhere and we can use the Mean Value Theorem for <math> f </math> and <math> f' </math> on the closed interval <math> [a, b] </math>. By the Mean Value Theorem applied to <math> f </math> on <math> [a, b] </math>, there is a number <math> s </math> such that <math> -1 = a < s and <math> f'(s) = \frac{f(b) - f(a)}{b - a} = \frac{f(2) - f(-1)}{2 - (-1)} = \frac{5 - (-1)}{2 - (-1)} = \frac{6}{3} = 2 </math>.

Now let <math> a = s </math> and <math> b = 2 </math>, where <math> s </math> is as defined above. By the Mean Value Theorem applied to <math> f' </math> on <math> [a, b] </math>, there is a number <math> t </math> such that <math> s = a < t and <math> f''(t) = \frac{f'(b) - f'(a)}{b - a} = \frac{f'(2) - f'(s)}{2 - s} = \frac{3 - 2}{2 - s} = \frac{1}{2 - s} > 0 </math>. But then, <math> f''(t) > 0 </math> contradicting the assumption that <math> f'' < 0 </math> everywhere. Hence, no such function <math> f </math> exists.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 09 (b)/Solution 1

2018-03-21T07:32:13Z

BRIANCHAN:

No such function exists. To see this, we will use the Mean Value Theorem twice. Let <math> a = -1 </math> and <math> b = 2 </math>. Because <math> f'' < 0 </math> everywhere, <math> f </math> and <math> f' </math> are differentiable everywhere and we can use the Mean Value Theorem for <math> f </math> and <math> f' </math> on the closed interval <math> [a, b] </math>. By the Mean Value Theorem applied to <math> f </math> on <math> [a, b] </math>, there is a number <math> s </math> such that <math> -1 = a < s and <math> f'(s) = \frac{f(b) - f(a)}{b - a} = \frac{f(2) - f(-1)}{2 - (-1)} = \frac{5 - (-1)}{2 - (-1)} = \frac{6}{3} = 2 </math>.

Now let <math> a = s </math> and <math> b = 2 </math>, where <math> s </math> is as defined above. By the Mean Value Theorem applied to <math> f' </math> on <math> [a, b] </math>, there is a number <math> t </math> such that <math> s = a < t and <math> f''(t) = \frac{f'(b) - f'(a)}{b - a} = \frac{f'(2) - f'(s)}{2 - s} = \frac{1}{2 - s} > 0 </math>. But then, <math> f''(t) > 0 </math> contradicting the assumption that <math> f'' < 0 </math> everywhere. Hence, no such function <math> f </math> exists.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 09 (b)/Hint 1

2018-03-21T07:26:24Z

BRIANCHAN:

Note that <math> \frac{f(2) - f(-1)}{2 - (-1)} = 2 </math>, and try to imagine a function that does satisfy <math> f'' < 0 </math> everywhere and <math> f'(2) = 3 </math>. To determine whether such a function exists, sketch an example of a differentiable function that satisfies <math> f(-1) = -1 </math>, <math> f(2) = 2 </math>, and <math> f'(2) = 3 </math>, then try to use the Mean Value Theorem (as many times as you need to) to explain why your drawings end up the way they do.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 09 (b)/Solution 1

2018-03-21T07:23:20Z

BRIANCHAN: Created page with "No such function exists. To see this, we will use the Mean Value Theorem twice. Let <math> a = -1 </math> and <math> b = 2 </math>. Because <math> f'' < 0 </math> everywhere,..."

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 09 (b)/Hint 1

2018-03-21T07:16:55Z

BRIANCHAN:

Note that <math> \frac{f(2) - f(-1)}{2 - (-1)} = 3 </math>, and try to imagine a function that does satisfy <math> f'' < 0 </math> everywhere and <math> f'(2) = 3 </math>. To determine whether such a function exists, sketch an example of a differentiable function that satisfies <math> f(-1) = -1 </math>, <math> f(2) = 2 </math>, and <math> f'(2) = 3 </math>, then try to use the Mean Value Theorem (as many times as you need to) to explain why your drawings end up the way they do.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 09 (b)/Hint 1

2018-03-21T07:13:04Z

BRIANCHAN:

Note that <math> \frac{f(2) - f(-1)}{2 - (-1)} = 3 </math>, and try to imagine a function that does satisfy <math> f'' < 0 </math> everywhere and <math> f'(2) = 3 </math>. To determine whether such a function exists, try to sketch an example of such a function, then try to use the Mean Value Theorem (as many times as you need to) to explain why your drawings end up the way they do.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 09 (b)/Hint 1

2018-03-21T07:12:29Z

BRIANCHAN: Created page with "Note that <math> \frac{f(2) - f(-1)}{2 - (-1)} = 3 </math>, and try to imagine a function that does satisfy <math> f'' < 0 </math> everywhere and <math> f'(2) = 3 </math>. To..."

Note that <math> \frac{f(2) - f(-1)}{2 - (-1)} = 3 </math>, and try to imagine a function that does satisfy <math> f'' < 0 </math> everywhere and <math> f'(2) = 3 </math>. To determine whether such a function exists, try to sketch an example of such a function, and try to use the Mean Value Theorem to explain why your drawings end up the way they do.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 09 (a)/Solution 1

2018-03-21T07:02:11Z

BRIANCHAN:

The tangent line at the point <math> (x, f(x)) </math> is parallel to the line with the equation <math> y = 2x </math> exactly when <math> f'(x) = 2 </math>. So it is enough to prove that there is a value of <math> x </math> such that <math> f'(x) = 2 </math>. We would like to use the Mean Value Theorem. Let <math> a = -1 </math> and <math> b = 2 </math>. Then <math> b - a = 3 </math>, <math> f(2) - f(-1) = 5 - (-1) = 6 </math>, and <math> \frac{f(b) - f(a)}{b - a} = 6/3 = 2 </math>. Because <math> f(x) </math> is differentiable everywhere, it is continuous on the closed interval <math> [a, b] </math> and differentiable on the open interval <math> (a,b) </math>. Thus, by the Mean Value Theorem, there exists an <math> x </math> such that <math> a < x and <math> f'(x) = \frac{f(b) - f(a)}{b - a} </math>. And as <math> \frac{f(b) - f(a)}{b - a} = 2 </math>, that means that there exists an <math> x </math> such that <math> a < x and <math> f'(x) = 2 </math>. This finishes the proof.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 09 (a)/Solution 1

2018-03-21T07:01:43Z

BRIANCHAN: Created page with "The tangent line at the point <math> (x, f(x)) </math> is parallel to the line with the equation <math> y = 2x </math> exactly when <math> f'(x) = 2 </math>. So it is enough t..."

The tangent line at the point <math> (x, f(x)) </math> is parallel to the line with the equation <math> y = 2x </math> exactly when <math> f'(x) = 2 </math>. So it is enough to prove that there is a value of <math> x </math> such that <math> f'(x) = 2 </math>.

We would like to use the Mean Value Theorem. Let <math> a = -1 </math> and <math> b = 2 </math>. Then <math> b - a = 3 </math>, <math> f(2) - f(-1) = 5 - (-1) = 6 </math>, and <math> \frac{f(b) - f(a)}{b - a} = 6/3 = 2 </math>. Because <math> f(x) </math> is differentiable everywhere, it is continuous on the closed interval <math> [a, b] </math> and differentiable on the open interval <math> (a,b) </math>. Thus, by the Mean Value Theorem, there exists an <math> x </math> such that <math> a < x and <math> f'(x) = \frac{f(b) - f(a)}{b - a} </math>. And as <math> \frac{f(b) - f(a)}{b - a} = 2 </math>, that means that there exists an <math> x </math> such that <math> a < x and <math> f'(x) = 2 </math>. This finishes the proof.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 10/Solution 1

2018-03-21T06:51:51Z

BRIANCHAN:

Let <math> a </math> and <math> b </math> be positive. The line <math> y = -ax + b </math> is tangent to <math> y = 3/x </math> at some point exactly when <math> 3/x = -ax + b </math>. That is, <math> 3 = -ax^2 + bx </math> which is the same as <math> -ax^2 + bx - 3 </math>. Since this is <math> 0 = Ax^2 + Bx + C </math> with <math> A = -a </math> and <math> c = -3 </math>, the discriminant is <math>B^2 - 4AC = b^2 + 12a </math>. So the discriminant is zero exactly when <math> b^2 = -12a </math>. Substituting this into <math> y = -ax + b </math> gives <math> y = \frac{-b^2}{12}x + b </math>, and such lines characterize the tangent lines of <math> y = 3/x </math>. Setting <math> x = 0 </math> gives <math> y = \frac{-b^2}{12}0 + b = b </math>, and setting <math> y = 0 </math> gives <math> 0 = \frac{-b^2}{12}x + b </math>, <math> -b = \frac{-b^2}{12}x </math>, and <math> x = 12/b </math>. Hence, the endpoints are <math> (0, b) </math> and <math> (12/b, 0) </math>, implying that the length of the line segment is <math> \sqrt{b^2 + (12/b)^2} </math>.

To minimize this function, it is enough to minimize the following function <math> b^2 + (12/b)^2 </math>. This function is differentiable with respect to <math> b </math>, tends to positive infinity as <math> b \to 0 </math>, and tends to positive infinity as <math> b \to \infty </math>. Hence, if <math> b </math> that minimizes the above length, then it satisfies <math> \frac{d}{db}(b^2 + (12/b)^2) = 0 </math> (<math> \frac{d}{db} </math> means differentiate with respect to <math> b </math>).

Using the power and chain rules, gives <math> \frac{d}{db}(b^2 + (12/b)^2) = 2b + (2(12/b) \times -12/b^2) = 2(b - 12^2/b^3) = 0 </math>. So <math>b - 12^2/b^3 = 0</math>, <math> b^4 - 12^2 = 0 </math>, <math> b^4 = 12^2 </math>, and <math> b^2 = 12 </math>. Substituting this value of <math> b </math> into <math> \sqrt{b^2 + (12/b)^2} </math> gives <math> \sqrt{b^2 + (12/b)^2} = \sqrt{b^2 + 12^2/b^2} = \sqrt{12 + 12^2/12} = \sqrt{12 + 12} = \sqrt{24} </math>.

Hence, the minimum length is <math>
\color{blue}

\sqrt{24}

</math>.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 10/Solution 1

2018-03-21T06:49:36Z

BRIANCHAN:

Let <math> a </math> and <math> b </math> be positive. The line <math> y = -ax + b </math> is tangent to <math> y = 3/x </math> at some point exactly when <math> 3/x = -ax + b </math>. That is, <math> 3 = -ax^2 + bx </math> which is the same as <math> -ax^2 + bx - 3 </math>. Since this is <math> 0 = Ax^2 + Bx + C </math> with <math> A = -a </math> and <math> c = -3 </math>, the discriminant is <math>B^2 - 4AC = b^2 + 12a </math>. So the discriminant is zero exactly when <math> b^2 = -12a </math>. Substituting this into <math> y = -ax + b </math> gives <math> y = \frac{-b^2}{12}x + b </math>, and such lines characterize the tangent lines of <math> y = 3/x </math>. Setting <math> x = 0 </math> gives <math> y = \frac{-b^2}{12}0 + b = b </math>, and setting <math> y = 0 </math> gives <math> 0 = \frac{-b^2}{12}x + b </math>, <math> -b = \frac{-b^2}{12}x </math>, and <math> x = 12/b </math>. Hence, the endpoints are <math> (0, b) </math> and <math> (12/b, 0) </math>, implying that the length of the line segment is <math> \sqrt{b^2 + (12/b)^2} </math>.

To minimize this function, it is enough to minimize the following function <math> b^2 + (12/b)^2 </math>. This function is differentiable with respect to <math> b </math>, tends to positive infinity as <math> b \to 0 </math>, and tends to positive infinity as <math> b \to \infty </math>. Hence, if <math> b </math> that minimizes the above length, then it satisfies <math> \frac{d}{db}(b^2 + (12/b)^2) = 0 </math> (<math> \frac{d}{db} </math> means differentiate with respect to <math> b </math>).

Using the power and chain rules, gives <math> \frac{d}{db}(b^2 + (12/b)^2) = 2b + (2(12/b) \times -12/b^2) = 2(b - 12^2/b^3) = 0 </math>. So <math>b - 12^2/b^3 = 0</math>, <math> b^4 - 12^2 = 0 </math>, <math> b^4 = 12^2 </math>, and <math> b^2 = 12 </math>. Substituting this value of <math> b </math> into <math> \sqrt{b^2 + (12/b)^2} </math> gives <math> \sqrt{b^2 + (12/b)^2} = \sqrt{(\sqrt{12})^2 + (12/\sqrt{12})^2 } = \sqrt{(\sqrt{12})^2 + (\sqrt{12})^2 } = \sqrt{12 + 12} = \sqrt{24} </math>.

Hence, the minimum length is <math>
\color{blue}

\sqrt{24}

</math>.