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Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 3 (c)/Solution 1

2019-04-12T22:10:17Z

HyunjuKwon:

We have

<math>

P_1: \ \ \ \bold{x} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + s_1 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + s_2 \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}

</math>

and

<math>
P_2: \ \ \ \bold{x} = t_1 \begin{pmatrix} -1 \\ 0 \\ -1 \end{pmatrix} + t_2 \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}
</math>

for some <math>s_1</math>, <math>s_2</math>, <math>t_1</math> and <math>t_2</math> from the question and part (b).

It is easy to see that the line <math>l_2</math> is just the intersection of plane <math>P_1, P_2</math>, i. e., any point <math>\bold{x} </math> on the line has to satisfy both equations. Thus we have following equation for line <math>l_2</math> .

<math> \bold{x}= \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + s_1 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + s_2 \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = t_1 \begin{pmatrix} -1 \\ 0 \\ -1 \end{pmatrix} + t_2 \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} </math>.

Now let's find out the relation among parameters <math>s_1,s_2,t_1,t_2</math> .
The first coordinate shows <math>1+s_2 = -t_1</math>, while the third shows <math>1-s_2 = -t_1+2t_2</math>. Summing these equations gives <math>2 = -2t_1+2t_2</math>, hence <math>t_2 = 1+t_1</math>. Substituting it back gives: for any point <math>\bold{x}</math> on line it satisfies

<math>

l_2: \bold{x}= t_1 \begin{pmatrix} -1 \\ 0 \\ -1 \end{pmatrix} + (1+t_1) \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}
= t_1\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}+\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix},
</math>

which has is equivalent equation form; for any point <math>\bold{x} = (x_1,x_2,x_3)^T</math> on the line <math>l_2</math> satisfies

<math>\color{blue}
\begin{align}
\begin{cases}
x_1+x_2 = 1\\
x_1+x_3 = 2
\end{cases}
\end{align}
</math>

Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 3 (c)/Solution 1

2019-04-12T22:08:09Z

HyunjuKwon:

We have

<math>

P_1: \ \ \ \bold{x} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + s_1 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + s_2 \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}

</math>

and

<math>
P_2: \ \ \ \bold{x} = t_1 \begin{pmatrix} -1 \\ 0 \\ -1 \end{pmatrix} + t_2 \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}
</math>

for some <math>s_1</math>, <math>s_2</math>, <math>t_1</math> and <math>t_2</math> from the question and part (b).

It is easy to see that the line <math>l_2</math> is just the intersection of plane <math>P_1, P_2</math>, i. e., any point <math>\bold{x} </math> on the line has to satisfy both equations. Thus we have following equation for line <math>l_2</math> .

<math> \bold{x}= \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + s_1 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + s_2 \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = t_1 \begin{pmatrix} -1 \\ 0 \\ -1 \end{pmatrix} + t_2 \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} </math>.

Now let's find out the relation among parameters <math>s_1,s_2,t_1,t_2</math> .
The first coordinate shows <math>1+s_2 = -t_1</math>, while the third shows <math>1-s_2 = -t_1+2t_2</math>. Summing these equations gives <math>2 = -2t_1+2t_2</math>, hence <math>t_2 = 1+t_1</math>. Substituting it back gives: for any point <math>\bold{x}</math> on line it satisfies

<math>

l_2: \bold{x}= t_1 \begin{pmatrix} -1 \\ 0 \\ -1 \end{pmatrix} + (1+t_1) \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}
= t_1\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}+\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix},
</math>

which has is equivalent equation form

<math>\color{blue}
\begin{align}
\begin{cases}
x_1+x_2 = 1\\
x_1+x_3 = 2
\end{cases}
\end{align}
</math>

Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 3 (c)/Solution 1

2019-04-12T22:07:50Z

HyunjuKwon:

We have

<math>

P_1: \ \ \ \bold{x} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + s_1 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + s_2 \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}

</math>

and

<math>
P_2: \ \ \ \bold{x} = t_1 \begin{pmatrix} -1 \\ 0 \\ -1 \end{pmatrix} + t_2 \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}
</math>

for some <math>s_1</math>, <math>s_2</math>, <math>t_1</math> and <math>t_2</math> from the question and part (b).

It is easy to see that the line <math>l_2</math> is just the intersection of plane <math>P_1, P_2</math>, i. e., any point <math>\bold{x} </math> on the line has to satisfy both equations. Thus we have following equation for line <math>l_2</math> .

<math> \bold{x}= \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + s_1 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + s_2 \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = t_1 \begin{pmatrix} -1 \\ 0 \\ -1 \end{pmatrix} + t_2 \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} </math>.

Now let's find out the relation among parameters <math>s_1,s_2,t_1,t_2</math> .
The first coordinate shows <math>1+s_2 = -t_1</math>, while the third shows <math>1-s_2 = -t_1+2t_2</math>. Summing these equations gives <math>2 = -2t_1+2t_2</math>, hence <math>t_2 = 1+t_1</math>. Substituting it back gives: for any point <math>\bold{x}</math> on line it satisfies

<math>

l_2: \bold{x}= t_1 \begin{pmatrix} -1 \\ 0 \\ -1 \end{pmatrix} + (1+t_1) \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}
= t_1\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}+\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix},
</math>

which has is equivalent equation form

<math>\color{blue}
\begin{align}
\begin{cases}
x_1+x_2 = -1\\
x_1+x_3 = 2
\end{cases}
\end{align}
</math>

Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 20/Solution 1

2019-04-12T03:38:47Z

HyunjuKwon:

We compute

:<math>
\begin{align}
\begin{vmatrix}
1 & 3 & 2 & 0& 1 \\
2 & 7 & 1 & 0& 2 \\
1 & 5 & 6 & -1 & 7 \\
0 & 0 & 3 & 0 & 0\\
1 & 1 & 4 & 0 &4 \\
\end{vmatrix} &= -3 \cdot \begin{vmatrix}1 & 3 & 0& 1 \\
2 & 7 & 0& 2 \\
1 & 5 & -1 & 7 \\
1 & 1 & 0 &4 \\
\end{vmatrix}
&& \text{expanding row 4} \\
&=
-3 \cdot -1 \cdot
\begin{vmatrix}
1 & 3 & 1 \\
2 & 7 & 2 \\
1 & 1 & 4
\end{vmatrix}
&& \text{expanding column 3} \\
&=
-3 \cdot -1 \cdot \left(
\begin{vmatrix}
7 & 2 \\
1 & 4
\end{vmatrix}
-3 \cdot
\begin{vmatrix}
2 & 2 \\
1 & 4
\end{vmatrix}
+
\begin{vmatrix}
2 & 7 \\
1 & 1
\end{vmatrix}
\right)
&& \text{expanding row 1} \\
&=
-3 \cdot -1 \cdot ((7 \cdot 4 - 2 \cdot 1) - 3(2 \cdot 4 - 2 \cdot 1) + (2 \cdot 1 - 7 \cdot 1)) \\
&= \color{blue} 9.
\end{align}
</math>

Science:Math Exam Resources/Courses/MATH152/April 2017/Question A 30/Solution 1

2019-04-12T03:31:36Z

HyunjuKwon:

The key idea of this problem is to realize that the plane <math>P</math> contains <math>3</math> linearly independent vectors; For example, <math>(1,0,0)^T, (0,1,0)^T, (0,0,1)^T</math>.
Therefore, if <math>T</math> is a linear transformation that maps <math>P</math> into <math>L</math>, <math>T</math> in fact sends all of <math>\mathbb{R}^3</math> to <math>L</math>. In particular, we have
<math display ="block">
T\begin{bmatrix}1\\0\\0 \end{bmatrix} = a\begin{bmatrix}1\\1\\1 \end{bmatrix}, \quad
T\begin{bmatrix}0\\1\\0 \end{bmatrix} = b\begin{bmatrix}1\\1\\1 \end{bmatrix}, \quad
T\begin{bmatrix}0\\0\\1 \end{bmatrix} = c\begin{bmatrix}1\\1\\1 \end{bmatrix}, \quad
</math>
for some constants <math>a</math>, <math>b</math>, and <math>c</math>.

On the other hand, we also need to ensure that at least one point of <math>P</math> is not mapped to the zero vector: otherwise the image of <math>P</math> under <math>T</math> would simply be the zero vector instead of all of <math>L</math>. In other words, at least one of <math>a</math>, <math>b</math>, and <math>c</math> is non-zero.

Therefore, the matrix of <math>T</math> must be of the form
<math display = "block">T=\left[\begin{array}{ccc}
a & b & c \\
a & b & c \\
a & b & c
\end{array}\right],</math>where at least one of <math>a</math>, <math>b</math>, and <math>c</math> is non-zero.

Science:Math Exam Resources/Courses/MATH152/April 2017/Question A 30/Solution 1

2019-04-12T03:16:22Z

HyunjuKwon:

The key to this problem is to realize that the plane <math>P</math> contains <math>3</math> linearly independent vectors. Therefore, if <math>T</math> is a linear transformation that maps <math>P</math> into <math>L</math>, <math>T</math> in fact sends all of <math>\mathbb{R}^3</math> to <math>L</math>. In particular, we have
<math display ="block">
T\begin{bmatrix}1\\0\\0 \end{bmatrix} = a\begin{bmatrix}1\\1\\1 \end{bmatrix}, \quad
T\begin{bmatrix}0\\1\\0 \end{bmatrix} = b\begin{bmatrix}1\\1\\1 \end{bmatrix}, \quad
T\begin{bmatrix}0\\0\\1 \end{bmatrix} = c\begin{bmatrix}1\\1\\1 \end{bmatrix}, \quad
</math>
for some non-zero constant <math>a</math>, <math>b</math>, and <math>c</math>.

On the other hand, we also need to ensure that at least one point of <math>P</math> is not mapped to the zero vector: otherwise the image of <math>P</math> under <math>T</math> would simply be the zero vector instead of all of <math>L</math>. In other words, at least one of <math>a</math>, <math>b</math>, and <math>c</math> is non-zero.

Therefore, the matrix of <math>T</math> must be of the form
<math display = "block">T=\left[\begin{array}{ccc}
a & b & c \\
a & b & c \\
a & b & c
\end{array}\right],</math></br>
where at least one of <math>a</math>, <math>b</math>, and <math>c</math> is non-zero.

Science:Math Exam Resources/Courses/MATH152/April 2017/Question A 30/Hint 1

2019-04-12T02:55:15Z

HyunjuKwon:

The plane <math>P</math> does not pass through the origin, it follows that <math>P</math> contains three points that, when viewed as vectors in <math>\mathbb{R}^3</math>, are linearly independent.

Science:Math Exam Resources/Courses/MATH152/April 2017/Question A 30/Statement

2019-04-12T02:43:19Z

HyunjuKwon:

Let <math>P</math> be a plane and <math>L</math> a line in <math>\mathbb{R}^3</math> given by equations

<math> \begin{align}
&P : x + y + z = 1\\
&L : u\begin{bmatrix}1\\ 1\\ 1\end{bmatrix} \quad u\in \R.
\end{align}
</math>

Find all linear transformations such that the image of <math>P</math> is <math>L</math> (that is the set of all
outputs is the line <math>L</math> when all points on the plane <math>P</math> are taken as inputs).

Science:Math Exam Resources/Courses/MATH101/April 2017/Question 03 (b)/Solution 1

2019-04-11T16:38:17Z

HyunjuKwon:

From the question we have <math display="block">\sum_{n=0}^\infty nx^{n+1}=x^2\sum_{n=0}^\infty nx^{n-1}.</math>

Therefore, it is enough to find an explicit expression of <math>\sum_{n=0}^\infty nx^{n-1}</math>.
For simplicity we denote <math display="inline">f(x)=\sum_{n=0}^\infty nx^{n-1}</math>.

Observe that <math>nx^{n-1} = \frac{d}{dx}x^n</math>. Using this, <math>f</math> can be written as
<math display = "block">
f(x) = \sum_{n=0}^\infty nx^{n-1} = \sum_{n=0}^\infty \frac{d}{dx}x^n.</math>

Since the interval of convergence of the power series is <math>(-1,1),</math> we can reverse the order of summation and derivative on this interval to get
<math display="block">
f(x) = \frac{d}{dx}\left( \sum_{n=0}^\infty x^n\right).</math>

By the hint (which can be easily obtained from the explicit expression of a geometric series), this implies that
<math display = "block">
f(x) = \frac{d}{dx}\left( \frac 1{1-x}\right).</math>

Therefore, computing the derivative based on the chain rule and power rule, the explicit formula for <math>f</math> is given by
<math display = "block">
f(x) = \frac{d}{dx}\left( \frac 1{1-x}\right) = \frac 1{(1-x)^2}.</math>

Combining with the first equation, we get <math display="inline">\sum_{n=0}^\infty nx^{n+1}=x^2f(x)=\frac{x^2}{(1-x)^2}.</math>

Answer: <math display="inline">\color{blue}\sum_{n=0}^\infty nx^{n+1}=\frac{x^2}{(1-x)^2}.</math>

Science:Math Exam Resources/Courses/MATH101/April 2017/Question 03 (b)/Hint 1

2019-04-11T16:23:47Z

HyunjuKwon:

Recall that <math>\sum_{n=0}^\infty x^n = \frac 1{1-x}</math> when <math>x\in (-1,1)</math>.

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

2019-04-10T19:17:37Z

HyunjuKwon:

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 part (a), we know that this function has its minimum on the circle 80 at the point <math>(-5,10).</math>

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

2019-04-10T19:17:01Z

HyunjuKwon:

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 part (a), we know that this function has its minimum 80 at the point <math>(-5,10)</math> on the circle.

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

2019-04-10T19:10:55Z

HyunjuKwon:

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

2019-04-10T19:10:41Z

HyunjuKwon:

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 this function has its minimum 80 at the point <math>(-5,10)</math>.

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

2019-04-10T17:30:38Z

HyunjuKwon:

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}</math>

Science:Math Exam Resources/Courses/MATH105/April 2016/Question 02 (a)/Solution 1

2019-04-10T17:22:44Z

HyunjuKwon:

Recall the fundamental theorem of calculus tells us that <math> \frac {d}{dx} \int_0^{x} f(t) dt = f(x)</math >

As we have variables at both the upper limit and lower limit of the integral , we first break the integral into two parts.

<math> \begin{align} \frac{d}{dx} \int_{x}^{x^2} \sin(t^2)dt = \frac{d}{dx}\int_{0}^{x^2} \sin(t^2) dt + \frac{d}{dx} \int_{x}^{0} \sin(t^2) dt
= {\color{blue} 2x\sin(x^4) - \sin(x^2)}

\end{align}

</math>

The second equality is directly followed from fundamental theorem of Calulus, because the variables are in the lower limit, we need to have a minus sign. We used chain rule to to evaluate the derivative of the first integral , let <math> u = x^2, \frac{du}{dx} = 2x </math>

<math>\frac{d}{dx}\int_{0}^{x^2} \sin(t^2) dt = 2x \frac{d}{du}\int_{0}^{u} \sin (t^2) dt = 2x\sin(u^2) = 2x\sin(x^4) </math>

'''''Answer:''''' <math>\color{blue} 2x\sin(x^4) - \sin(x^2)</math>

Science:Math Exam Resources/Courses/MATH105/April 2016/Question 02 (a)/Solution 1

2019-04-10T17:22:21Z

HyunjuKwon:

Recall the fundamental theorem of calculus tells us that <math> \frac {d}{dx} \int_0^{x} f(t) dt = f(x)</math >

As we have variables at both the upper limit and lower limit of the integral , we first break the integral into two parts.

<math> \begin{align} \frac{d}{dx} \int_{x}^{x^2} \sin(t^2)dt = \frac{d}{dx}\int_{0}^{x^2} \sin(t^2) dt + \frac{d}{dx} \int_{x}^{0} \sin(t^2) dt
= {\color{blue} 2x\sin(x^4) - \sin(x^2)}

\end{align}

</math>

The second equality is directly followed from fundamental theorem of Calulus, because the variables are in the lower limit, we need to have a minus sign. We used chain rule to to evaluate the derivative of the first integral , let <math> u = x^2, \frac{du}{dx} = 2x </math>

<math>\frac{d}{dx}\int_{0}^{x^2} \sin(t^2) dt = 2x \frac{d}{du}\int_{0}^{u} \sin (t^2) dt = 2x\sin(u^2) = 2x\sin(x^4) </math>

'''Answer:''''''Bold text''' <math>\color{blue} 2x\sin(x^4) - \sin(x^2)</math>

Science:Math Exam Resources/Courses/MATH101/April 2015/Question 06 (b)/Solution 1

2019-04-10T17:18:42Z

HyunjuKwon:

Since the series <math>\sum_{n=0}^\infty (1 - b_n)</math> converges, we have

<math>\lim_{n\to \infty} (1-b_n) = 0 \iff \lim_{n\to \infty} b_n = 1 </math>.

Applying the ratio test, the series <math>\sum_{n=0}^\infty b_nx^n</math> converges when

<math>
\lim_{n\to \infty}\left|\frac{b_{n+1} x^{n+1}}{b_n x^n} \right| = \lim_{n\to \infty}\left|\frac{b_{n+1} }{b_n } \right||x|
=\left|\frac{\lim_{n\to \infty}b_{n+1} }{\lim_{n\to \infty}b_n } \right||x| = |x|<1,
</math>

and diverges when <math>|x|>1</math>. The last equality follows from
<math>
\lim_{n\to \infty} b_n = 1 \implies \lim_{n\to \infty} b_{n+1} = 1.
</math>

Therefore, the radius of convergence of the power series <math>\sum_{n=0}^\infty b_nx^n</math> is equal to <math>\color{blue}1</math>.

Science:Math Exam Resources/Courses/MATH105/April 2018/Question 02 (c)/Solution 1

2019-04-10T17:16:04Z

HyunjuKwon:

Since the given equation is separable, we can rewrite is as
<math display = "block"> \frac{dy}{e^y} = e^{3x} dx.</math>

Taking integral on the both side of the equation, we have
<math display = "block"> \int e^{-y}dy = \int e^{3x} dx.</math>

Note that using substitution <math>u = ax</math> for any fixed number <math>a\neq 0</math>, we have
<math display = "block"> \int e^{ax} dx = \int e^{u} \frac{du}{a} = \frac 1a e^u+C = \frac 1a e^{ax}+C.</math>

Applying this for <math>a=-1</math> and <math>a=3</math>, we obtain
<math display="block"> \int e^{-y}dy = \frac 1{-1} e^{-y}+ C_1 = -e^{-y} + C_1,</math>and
<math display = "block"> \int e^{3x} dx=\frac 13 e^{3x} + C_2,</math>
where <math>C_1</math> and <math>C_2</math> are arbitrary constants.

Therefore, we get
<math display="block"> -e^{-y}+C_1=\int e^{-y}dy = \int e^{3x} dx= \frac 13 e^{3x} + C_2,</math>
which can be simplified as follows
<math display="block"> e^{-y}= -\frac 13 e^{3x} + C.</math>

We plug <math>x=0</math> to find the constant <math>C</math>,
<math display="block"> \frac 15=e^{-y(0)}= -\frac 13 + C \iff C = \frac {8}{15}.</math>

Finally, taking a logarithm to the simplified equation, we can find the solution
<math display="block"> y= -\ln\left(-\frac 13 e^{3x} + \frac 8{15}\right).</math>

'''Answer:''' <math>\color{blue} y=-\ln\left(-\frac 13 e^{3x} + \frac 8{15}\right).</math>

Science:Math Exam Resources/Courses/MATH105/April 2018/Question 02 (c)/Solution 1

2019-04-10T17:14:02Z

HyunjuKwon:

Since the given equation is separable, we can rewrite is as
<math display = "block"> \frac{dy}{e^y} = e^{3x} dx.</math>

Taking integral on the both side of the equation, we have
<math display = "block"> \int e^{-y}dy = \int e^{3x} dx.</math>

Note that using substitution <math>u = ax</math> for any fixed number <math>a\neq 0</math>, we have
<math display = "block"> \int e^{ax} dx = \int e^{u} \frac{du}{a} = \frac 1a e^u+C = \frac 1a e^{ax}+C.</math>

Applying this for <math>a=-1</math> and <math>a=3</math>, we obtain
<math display="block"> \int e^{-y}dy = \frac 1{-1} e^{-y}+ C_1 = -e^{-y} + C_1,</math>and
<math display = "block"> \int e^{3x} dx=\frac 13 e^{3x} + C_2,</math>
where <math>C_1</math> and <math>C_2</math> are arbitrary constants.

Therefore, we get
<math display="block"> -e^{-y}+C_1=\int e^{-y}dy = \int e^{3x} dx= \frac 13 e^{3x} + C_2,</math>
which can be simplified as follows
<math display="block"> e^{-y}= -\frac 13 e^{3x} + C.</math>

We plug <math>x=0</math> to find the constant <math>C</math>,
<math display="block"> \frac 15=e^{-y(0)}= -\frac 13 + C \iff = C = \frac {16}3.</math>

Finally, taking a logarithm to the simplified equation, we can find the solution
<math display = "block"> y= \ln\left(-\frac 13 e^{3x} + \frac {16}3\right).</math>

'''Answer:''' <math>\color{blue} y= \ln\left(-\frac 13 e^{3x} + \frac {16}3\right).</math>

Science:Math Exam Resources/Courses/MATH101/April 2018/Question 10 (iii)/Solution 1

2019-04-10T17:10:35Z

HyunjuKwon:

Since <math>c</math> is given by <math>2</math> in part (ii), it is enough to evaluate the integral
<math display="block">\int_{0}^{1} \frac{ \left[2x - \log(1+2x)\right]}{x^2} dx.</math>

We apply the integration by parts for <math>u(x) = 2x - \log(1+2x)</math> and <math>v'(x) = \frac 1{x^2}</math>. Then, <math>u'(x) = 2-\frac 2{1+2x} = \frac{4x}{1+2x}</math> , <math>v(x)= -\frac 1x</math>, and
<math display="block">
\begin{align}
\int_{0}^{1} \frac{ \left[2x - \log(1+2x)\right]}{x^2} dx
&=\lim_{t\to 0}\int_{t}^{1} \frac{ \left[2x - \log(1+2x)\right]}{x^2} dx\\
&=\lim_{t\to 0}\int_{t}^{1} u(x)v'(x) dx
=\lim_{t\to 0} u(x)v(x)\bigg|_{t}^{1} - \int_{0}^{1} u'(x) v(x) dx\\
&=\lim_{t\to 0}(2x - \log(1+2x))\cdot\left(-\frac 1x\right)\bigg|_{t}^{1}
-\int_{0}^{1} \frac{4x}{1+2x}\cdot \left(-\frac 1x\right)dx\\
&=-2+\log 3 + \lim_{t\to 0}\frac{(2t - \log(1+2t))}t
+\int_{0}^{1} \frac{4}{1+2x}dx.
\end{align}
</math>

By L'hospital rule, we can compute the limit as
<math display="block">
\lim_{t\to 0}\frac{(2t - \log(1+2t))}t
=\lim_{t\to 0}\frac{(2t - \log(1+2t))'}{(t)'}
=\lim_{t\to 0} \frac{2 - \frac 2{1+2t}}{1} = 0.
</math>

On the other hand, using the substitution <math>u=1+2x</math> (so <math>du=2dx</math>), the last term can be evaluated;
<math display="block">
\int_{0}^{1} \frac{4}{1+2x}dx
=\int_{1}^{3} \frac{2}{u}du = 2\log u\bigg|_{1}^{3} = 2\log 3.
</math>

Combining all the information, the integral <math>I</math> for <math>c=2</math> has the value
<math display="block">
I = -2 +\log 3+ \lim_{t\to 0}\frac{(2t - \log(1+2t))}t
+\int_{0}^{1} \frac{4}{1+2x}dx = -2 +\log 3 + 0 + 2\log 3 = \log(27)-2.</math>

'''Answer:''' <math>\color{blue}\log(27) -2</math>

Science:Math Exam Resources/Courses/MATH105/April 2018/Question 04 (b)/Solution 1

2019-04-10T17:02:26Z

HyunjuKwon:

Let <math>f(x) = x^2 e^{-3x^3}</math>. Obviously, <math>f(x)\geq 0</math> and <math>f(x)</math> is continuous for all <math>x\geq 1</math>. Since <math>f'(x) = (2x-9x^4)e^{-3x^3}=x(2-9x^3)e^{-3x^3}<0</math> when <math>x\geq 1</math>, <math>f</math> decreases for all <math>x\geq 1</math>. Therefore, we can apply the integral test for the series <math>\sum_{k=1}^\infty f(k)=\sum_{k=1}^\infty k^2 e^{-3k^3}</math>.

To see whether the integral
<math>\int_1^\infty f(x) dx</math> converges or not, we use the substitution <math>u = 3x^3</math>. Then, <math>du = 9 x^2 dx</math> and
<math display="block">
\begin{align}
\int_1^\infty f(x) dx
&=\lim_{b\to \infty}\int_1^b x^2 e^{-3x^3} dx\\
&=\frac 19\lim_{b\to \infty}\int_3^{3b^3} e^{-u} du
=\frac 19\lim_{b\to \infty} \left[-e^{-u}\right]_3^{3b^3}\\
&=\frac 19\lim_{b\to \infty} (-e^{-3b^3} + e^{-3})
=\frac 19 e^{-3}<+\infty.
\end{align}
</math>

Since the integral converges, the series <math>\sum_{k=1}^\infty k^2 e^{-3k^3}</math> also converges.

'''Answer:''' <math>\color{blue}\mbox{converges}</math>

Science:Math Exam Resources/Courses/MATH101/April 2018/Question 04 (ii)/Solution 1

2019-04-10T16:57:17Z

HyunjuKwon:

To find the positive points at which local minimums of <math>g</math> occur, let's find the derivative of <math>g</math> first.

Since <math>g</math> can be written as a composite of two functions, <math>g(x) = F(x^2) = F(h(x))</math>, where <math>h(x)=x^2</math>, we can apply the chain rule to find its derivative. In this process, we need the derivative of <math>F</math>.

By the Fundamental theorem of Calculus, we have
<math display="block">F'(x)=\frac d{dx}\left(\int_{0}^{x} (t^4 + 1) \sin{t} dt\right) = (x^4+1)\sin x.</math>

Then, the derivative of <math display="inline">g</math> is

<math display="block">g'(x)= (F(h(x))' = F'(h(x))h'(x) = F'(x^2) (x^2)' = 2x(x^8+1)\sin (x^2).</math>

Since <math>x^8+1\geq 1</math> for any <math>x</math> on the real line and we consider only positive <math>x</math>, the sign of <math>g'</math> is determined by <math>\sin (x^2)</math>.

Note that <math>\sin (x^2) =0</math> at <math>x^2=\pi, 2\pi, \cdots</math> (i.e., <math>x=\sqrt{\pi}, \sqrt{2\pi}, \cdots</math>). Also,

<math display="block">
\begin{align}
&\sin (x^2) >0, \quad \text{on } (0,\sqrt\pi), (\sqrt{2\pi}, \sqrt{3\pi}),\cdots\\
&\sin (x^2) <0, \quad \text{on } (\sqrt{\pi},\sqrt{2\pi}), (\sqrt{3\pi}, \sqrt{4\pi}),\cdots.
\end{align}
</math>

Since the sign of <math>\sin (x^2)</math> is changed from minus to plus at <math>x=\sqrt{2\pi}</math> for the first time among positive <math>x</math>, so is the sign of <math>g'</math>. Therefore, the smallest number <math>x</math> at which <math>g</math> has a local minimum is <math>x=\sqrt{2\pi}</math>.

'''Answer:''' <math>\color{blue} \sqrt{2\pi}</math>

Science:Math Exam Resources/Courses/MATH101/April 2009/Question 07 (a)/Solution 1

2019-04-10T04:24:18Z

HyunjuKwon:

''Hydrostatic Force was not covered in the 2012 offering of this course. Time might better be spent solving other problems given how soon the exam is coming.''

Consider the triangular side pictured to the right [[file:Math101Apr20097a.png|300px|thumbnail|right]]. It has height 2 and length 1. Also consider a little rectangular sliver in the picture of length L(y) and height dy. The reason we consider rectangular pieces with constant height is because the pressure depends on height and so we would like the effect from height to be the same on any given rectangular piece. We will use the labelling in the diagram with the bottom vertex being (0,0), the right vertex being (1/2,2) and the left vertex being (-1/2,2). With our origin at the bottom, we will label y from there. This means that the height (as measured from the top of the surface) is (2-y). We then have that the force on any given rectangle, dF, is,

:<math>
\displaystyle{}dF=\rho{g}(2-y)L(y)\text{d}y.
</math>

We could then get the total force by integrating dF but first we need an expression for the length in terms of the height, y. The length of any rectangle will start at the left edge of the triangle and end at the right edge. We can use the coordinates that we listed to find equations for these lines. The right edge is given by

:<math>
\displaystyle y=4x
</math>

which written in terms of y is,

:<math>
x_R=\frac{y}{4}
</math>

where we have added a subscript R to indicate it's the right edge. Notice we compute the line by finding the slope using the right vertex and bottom vertex and then notice that the line goes through our origin to its y-intercept is zero. We can similarly get that the left edge is

:<math>
x_L=-\frac{y}{4}.
</math>

This can also be argued with symmetry of the right edge. We then have that

:<math>
L(y)=x_R-x_L=\frac{y}{4}-\left(-\frac{y}{4}\right)=\frac{y}{2}.
</math>

Therefore we can now get F via integration where we see that y starts at zero and ends at y=2. Therefore,

:<math>
F=\int_0^2\rho{g}(2-y)L(y)\text{d}y=\int_0^2\rho{g}\frac{2y-y^2}{2}\text{d}y=\rho{g}\left.\left(\frac{y^2}{2}-\frac{y^3}{6}\right)\right|_0^2=\frac{2\rho{g}}{3}.
</math>

We are told in the problem that the density is 1000kg/m^3 and we take the gravitational constant as 9.8N/kg. Therefore,

:<math>
F=\frac{2}{3}\cdot 9800=\frac{19600}{3}N.
</math>

This is the hydrostatic force on each of the triangular sides.

Science:Math Exam Resources/Courses/MATH101/April 2017

2019-04-10T04:19:21Z

HyunjuKwon:

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Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)./Statement

2019-04-10T04:18:49Z

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Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Statement

2019-04-10T04:18:49Z

Evaluate <math>\int \frac{3x}{\sqrt{x^2-3}} dx.</math>

Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)./Solution 2

2019-04-10T04:18:49Z

HyunjuKwon: HyunjuKwon moved page Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)./Solution 2 to Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Solution 2 over redirect

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Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Solution 2

2019-04-10T04:18:49Z

By the chain rule, we have <math>(\sqrt{x^ 2-3})^{\prime} = \frac 1{2\sqrt{x^ 2-3}} \cdot (x^2-3)' =\frac{x}{\sqrt{x^ 2-3}}</math>.

Then, by the definition of anti-derivative, we get

<math> \int \frac{3x}{\sqrt{x^ 2-3}} = 3\int \frac{x}{\sqrt{x^ 2-3}} = 3\int(\sqrt{x^ 2-3})^{\prime} = 3\sqrt{x^ 2-3}+C</math>.

'''Answer:''' <math>\color{blue}{3\sqrt{x^ 2-3}}+C</math>

Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)./Solution 1

2019-04-10T04:18:49Z

HyunjuKwon: HyunjuKwon moved page Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)./Solution 1 to Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Solution 1 over redirect

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Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Solution 1

2019-04-10T04:18:49Z

Consider the substitution <math> u = x^2 - 3 </math>. Then <math> du = 2x \, dx </math> because the derivative of <math> x^2 - 3 </math> with respect to <math> x </math> is <math> 2x </math>. Hence,

<math> \int \dfrac{3x}{\sqrt{x^2 - 3}} \,dx = \dfrac{3}{2} \int \dfrac{2x \,dx}{\sqrt{x^2 - 3}} = \dfrac{3}{2} \int \dfrac{du}{\sqrt{u}} </math>

By the Power Rule for the integration, we get

<math> \int \dfrac{du}{\sqrt{u}} = \int u^{-\frac 12} du = \frac 1{-\frac 12 + 1} u^{-\frac 12 + 1} + C= 2 \sqrt{u} + C </math>

where <math>C</math> is an arbitrary constant (the constant of integration)

So,

<math> \int \dfrac{3x}{\sqrt{x^2 - 3}} \,dx = \dfrac{3}{2} \int \dfrac{du}{\sqrt{u}} = \dfrac{3}{2} (2 \sqrt{u} + C) </math>

and substituting <math> x^2 - 3 </math> for <math> u </math> into the above expression now gives

<math> \int \dfrac{3x}{\sqrt{x^2 - 3}} \,dx = 3 \sqrt{u} + C =\color{blue} 3 \sqrt{x^2 - 3} + C </math>

Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)./Hint 3

2019-04-10T04:18:49Z

HyunjuKwon: HyunjuKwon moved page Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)./Hint 3 to Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Hint 3 over redirect

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Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)./Hint 2

2019-04-10T04:18:48Z

HyunjuKwon: HyunjuKwon moved page Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)./Hint 2 to Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Hint 2 over redirect

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Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Hint 2

2019-04-10T04:18:48Z

(Solution 2) Think about the derivative of <math>\sqrt{x^ 2-3}</math>. Then, use the definition of anti-derivative.

Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)./Hint 1

2019-04-10T04:18:48Z

HyunjuKwon: HyunjuKwon moved page Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)./Hint 1 to Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Hint 1 over redirect

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Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Hint 1

2019-04-10T04:18:48Z

(Solution 1) Use Integration by Substitution; to answer this question, find the right substitution.

Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b).

2019-04-10T04:18:48Z

HyunjuKwon: HyunjuKwon moved page Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b). to Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b) over redirect

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Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Statement

2019-04-10T04:17:40Z

HyunjuKwon: HyunjuKwon moved page Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Statement to Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)./Statement

Evaluate <math>\int \frac{3x}{\sqrt{x^2-3}} dx.</math>

Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Solution 2

2019-04-10T04:17:40Z

HyunjuKwon: HyunjuKwon moved page Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Solution 2 to Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)./Solution 2

Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Solution 1

2019-04-10T04:17:40Z

HyunjuKwon: HyunjuKwon moved page Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Solution 1 to Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)./Solution 1

Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Hint 2

2019-04-10T04:17:40Z

HyunjuKwon: HyunjuKwon moved page Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Hint 2 to Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)./Hint 2

(Solution 2) Think about the derivative of <math>\sqrt{x^ 2-3}</math>. Then, use the definition of anti-derivative.

Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Hint 1

2019-04-10T04:17:40Z

HyunjuKwon: HyunjuKwon moved page Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)/Hint 1 to Science:Math Exam Resources/Courses/MATH101/April 2017/Question 01 (b)./Hint 1

(Solution 1) Use Integration by Substitution; to answer this question, find the right substitution.

Science:Math Exam Resources/Courses/MATH101/April 2017

2019-04-10T04:13:53Z

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Science:Math Exam Resources/Courses/MATH101/April 2017

2019-04-10T04:12:54Z

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Science:Math Exam Resources/Courses/MATH101/April 2017/Question 02 (a)

2019-04-10T04:12:31Z

HyunjuKwon:

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Science:Math Exam Resources/Courses/MATH101/April 2017/Question 02 (a)/Solution 1

2019-04-10T04:12:06Z

HyunjuKwon:

We consider the region between two curves <math>y^2 = 1-x</math> and <math>2y^2 = 5 -x</math>.

First, we find the intersection points. Since two curves can be written as <math>x=1-y^2</math> and <math>x=5-2y^2</math>, plugging the first one into the second one, we get
<math display = "block">
1-y^2 = 5-2y^2 \iff y^2 =4 \iff y=\pm2.</math>
Therefore, the intersection points are <math>(-3,2)</math> and <math>(-3,-2)</math>.

On the other hand, according to the question, the curve <math>x=5-2y^2</math> lies on the right side of <math>x=1-y^2</math>, so that <math>1-y^2 \leq 5-2y^2</math> when <math>y\in (-2,2)</math>.

To summarize, the area between the two curves has to be an integral of the form

<math display = "block>
\int_{-2}^{2} |(5 - 2y^2) - (1 - y^2)| \,dy
= \int_{-2}^{2} (5 - 2y^2) - (1 - y^2) \,dy
= \int_{-2}^{2} 4-y^2 \,dy
= 2\int_0^2 4-y^2 \,dy.
</math>

The last equality follows from the fact that <math>4-y^2</math> is even. Therefore, the answer is <math>a</math>.

'''Answer:''' <math>\color{blue} a </math>.

Science:Math Exam Resources/Courses/MATH105/April 2018

2019-04-09T17:53:42Z

HyunjuKwon:

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Science:Math Exam Resources/Courses/MATH101/April 2017/Question 08/Solution 1

2019-04-09T04:57:06Z

HyunjuKwon:

First note that the denominator of the integrand factorizes as :

<math> \frac{x-8}{x^ 3 - 4x^ 2 + 4x} = \frac{x-8}{x} \frac{1}{(x-2)^ 2} = \frac{1}{(x-2)^ 2} - \frac{8}{x(x-2)^ 2}</math>

So now that we know the the factors, we want to convert it to an expression of the following form:

<math> -\frac{8}{x(x-2)^ 2} = \frac{A}{x} \frac{Bx+C}{(x-2)^ 2}</math>

Equate the numerators
<math> A(x-2)^ 2 + Bx^2 +Cx = -8</math>

and obtain the equations
<math> A+B = 0,\, -4A +C = 0,\, 4A= -8</math> which yields

<math> A = -2,\,B= 2, C = -8 </math>

and thus we rewrite our integral as

<math> \int_{-3}^{-4} \frac{1}{(x-2)^ 2} - \frac{2}{x} + \frac{2x-8}{(x-2)^ 2} dx </math>

which we rewrite once more, so that the numerator of the 3rd term is the derivative of denominator

<math> \int_{-3}^{-4} \frac{-3}{(x-2)^ 2} - \frac{2}{x} + \frac{2(x-2)}{(x-2)^ 2} dx
=\frac{3}{x-2}\bigg|_{-4}^{-3} + \ln x^ {-2}\bigg|_{-4}^{-3} + \ln (x-2)^2\bigg|_{-4}^{-3}
=-\frac 35 + \frac 12 +\ln\left(\frac {16}9\right) + \ln \left(\frac{25}{36} \right)
</math>

'''Answer:''' <math>\color{blue} -\frac 35 + \frac 12 +\ln\left(\frac {100}{81}\right)</math>

Science:Math Exam Resources/Courses/MATH101/April 2017/Question 10

2019-04-09T04:46:41Z

HyunjuKwon:

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Science:Math Exam Resources/Courses/MATH105/April 2018/Question 03/Hint 1

2019-04-09T04:44:55Z

HyunjuKwon: Created page with "By the method of Lagrange multipliers, find three equations that a local extremum for the function <math>f</math> on the ellipse should satisfy."

By the method of Lagrange multipliers, find three equations that a local extremum for the function <math>f</math> on the ellipse should satisfy.