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File:Han's graph.jpg

2018-04-07T21:56:41Z

HANHONG: User created page with UploadWizard

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

2018-04-07T21:50:20Z

HANHONG:

[[File:12.jgp|thumbnail]]

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

2018-04-07T21:49:19Z

HANHONG: Blanked the page

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

2018-04-07T21:47:27Z

HANHONG:

<gallery>
File:Example.jpg|12.jpg

</gallery>

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

2018-04-07T21:47:01Z

HANHONG: Created page with "<gallery> File:Example.jpg|Caption1 File:Example.jpg|Caption2 </gallery>"

<gallery>
File:Example.jpg|Caption1
File:Example.jpg|Caption2
</gallery>

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

2018-04-07T21:33:10Z

HANHONG: Created page with "Based on what we attained in last few questions, e.g. increasing, decreasing intervals; concave up and concave down, inflection, etc. we can draw a rough graph."

Based on what we attained in last few questions, e.g. increasing, decreasing intervals; concave up and concave down, inflection, etc. we can draw a rough graph.

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

2018-04-07T20:14:36Z

HANHONG: Created page with "First, from previous question we know that <math display="block">f''(x)=\frac{2x-6}{x^4}.</math> The inflection point is <math display="inline">x=3</math> and <math display="i..."

First, from previous question we know that <math display="block">f''(x)=\frac{2x-6}{x^4}.</math> The inflection point is <math display="inline">x=3</math> and <math display="inline">f''(x)</math> has a not well defined point at <math display="inline">x=0</math>. Let’s consider three different intervals <math display="inline">(-\infty,0), (0,3), (3,\infty).</math>

(1) when <math display="inline">x\in (-\infty,0)</math>, <math display="inline">2x-6<0, x^4>0</math>, thus <math display="inline">f''(x)<0</math>;

(2) when <math display="inline">x\in (0,3)</math>, <math display="inline">2x-6<0, x^4>0</math>, thus <math display="inline">f''(x)<0</math>;

(3) when <math display="inline">x\in (3,\infty)</math>, <math display="inline">2x-6>0, x^4>0</math>, thus <math display="inline">f''(x)>0</math>;

In all, <math display="inline">f</math> is concave up in <math display="inline">\color{blue}(3,\infty);</math> while <math display="inline">f</math> is concave down in <math display="inline">\color{blue}(-\infty,0)\cup (0,3)</math>.

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

2018-04-07T20:03:34Z

HANHONG: Created page with "Recall that (1) <math display="inline">f</math> is concave up (convex) on the intervals where <math display="inline">f''(x)>0</math>; (2) <math display="inline">f</math> is..."

Recall that

(1) <math display="inline">f</math> is concave up (convex) on the intervals where <math display="inline">f''(x)>0</math>;

(2) <math display="inline">f</math> is concave down on the intervals where <math display="inline">f''(x)<0</math>;

(3) <math display="inline">f</math> has a inflection point at which <math display="inline">f''(x)=0</math>.

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

2018-04-07T19:52:11Z

HANHONG:

From part (d), <math> f'(x) = \frac{2 - x}{x^3} </math>. Critical points are <math display="inline">0,2</math>. Now let’s calculate the second derivative <math display="block">f''(x)=(f'(x))'=(\frac{2-x}{x^3})'=\frac{(2-x)'x^3-(x^3)'(2-x)}{x^6}=\frac{2x-6}{x^4}.</math>

(1) at <math display="inline">x=0</math>, <math display="inline">f''(x)</math> can not have value (not well defined or singularity);

(2) at <math display="inline">x=2</math>, <math display="inline">f''(2)=\frac{2\times 2-6}{2^4}=-\frac{1}{8}<0</math>, so it attains local maximum at this point.

In all, it has no local minimum, but has a local minimum at <math display="inline">x=2</math> and <math display="inline">f(2)=\frac{2-1}{x^2}=\frac{1}{4}</math>.

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

2018-04-07T19:42:56Z

HANHONG:

Recall that the local maxima and minima could happen at the critical points. Then we can use second derivative test:

(1) <math display="inline">f''(x)<0</math>, then <math display="inline">f</math> has a local maximum at <math display="inline">x</math>;

(2) <math display="inline">f''(x>0)</math>, then <math display="inline">f</math> has a local minimum at <math display="inline">x</math>;

(3) <math display="inline">f''(x)=0</math>, inclusive.

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

2018-04-07T19:39:20Z

HANHONG:

Let’s calculate the derivative first by quotient rule, <math display="block">f'(x)=\frac{(x-1)'x^2-(x^2)'(x-1)}{x^4}=\frac{x^2-2x^2+2x}{x^4}=\frac{(2-x)x}{x^4}=\frac{2-x}{x^3}.</math>

We can see that zero of <math display="inline">f'(x)</math> are <math display="inline">2</math> and it has a singularity at <math display="inline">x=0</math> (not well defined), these points are called critical points. At these points, <math display="inline">f'(x)</math> might change the sign. Now let’s consider three different intervals: <math display="inline">(-\infty,0), (0,2), (2,\infty)</math>.

(1) when <math display="inline">x\in (-\infty,0)</math>, <math display="inline">x^3<0,\ 2-x>0</math>, so <math display="inline">f'(x)<0</math>;

(2) when <math display="inline">x\in (0,2)</math>, <math display="inline">x^3>0, \ 2-x>0</math>, so <math display="inline">f'(x)>0</math>;

(3) when <math display="inline">x\in (2,\infty)</math>, <math display="inline">x^3>0,\ 2-x<0</math>, so <math display="inline">f'(x)<0</math>.

In all, <math display="inline">f</math> is increasing in the interval <math display="inline">\color{blue}(0,2)</math>, while it is decreasing in <math display="inline">\color{blue}(-\infty,0)\cup(2,\infty).</math>

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

2018-04-07T19:32:55Z

HANHONG: Created page with "Let’s calculate the derivative first by quotient rule, <math display="block">f'(x)=\frac{(x-1)'x^2-(x^2)'(x-1)}{x^4}=\frac{x^2-2x^2+2x}{x^4}=\frac{(2-x)x}{x^4}=\frac{2-x}{x^..."

Let’s calculate the derivative first by quotient rule, <math display="block">f'(x)=\frac{(x-1)'x^2-(x^2)'(x-1)}{x^4}=\frac{x^2-2x^2+2x}{x^4}=\frac{(2-x)x}{x^4}=\frac{2-x}{x^3}.</math>

We can see that zero of <math display="inline">f'(x)</math> are <math display="inline">2</math> and it has a singularity at <math display="inline">x=0</math> (not well defined). At these points, <math display="inline">f'(x)</math> might change the sign. Now let’s consider three different intervals: <math display="inline">(-\infty,0), (0,2), (2,\infty)</math>.

(1) when <math display="inline">x\in (-\infty,0)</math>, <math display="inline">x^3<0,\ 2-x>0</math>, so <math display="inline">f'(x)<0</math>;

(2) when <math display="inline">x\in (0,2)</math>, <math display="inline">x^3>0, \ 2-x>0</math>, so <math display="inline">f'(x)>0</math>;

(3) when <math display="inline">x\in (2,\infty)</math>, <math display="inline">x^3>0,\ 2-x<0</math>, so <math display="inline">f'(x)<0</math>.

In all, <math display="inline">f</math> is increasing in the interval <math display="inline">\color{blue}(0,2)</math>, while it is decreasing in <math display="inline">\color{blue}(-\infty,0)\cup(2,\infty).</math>

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

2018-04-07T18:35:45Z

HANHONG: Created page with "Calculate the derivative, and determine the intervals where <math display="inline">f'(x)>0</math> and <math display="inline">f'(x)<0</math>. When <math display="inline">f'(x..."

Calculate the derivative, and determine the intervals where <math display="inline">f'(x)>0</math> and <math display="inline">f'(x)<0</math>. When <math display="inline">f'(x)>0</math>, <math display="inline">f(x)</math> is increasing, while <math display="inline">f'(x)<0</math>, <math display="inline">f(x)</math> is decreasing.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 03 (c)/Hint 1

2018-04-07T18:25:56Z

HANHONG:

To show that this equation has zero is equivalent to show that the function <math display="inline">f(x)</math> has intersection with <math display="inline">e^{-(x-1)}</math>. Note that the intersection can only happen in the interval <math display="inline">(0,1)</math>, so we can apply intermediate value theorem to show that there exists at least one intersection in the interval <math display="inline">(0,1)</math>. Here we use intermediate value theorem: if function <math display="inline">f(x)</math> is continuous on <math display="inline">(a,b)</math>, and <math display="inline">f(a)\cdot f(b)<0</math>, then there at least exists one point <math display="inline">c\in (a,b)</math> such that <math display="inline">f(c)=0</math>.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 03 (c)/Solution 1

2018-04-07T18:22:12Z

HANHONG: Created page with "Before we apply intermediate value theorem, let’s discuss why the intersection can only happen in <math display="inline">(0,1)</math> rather than <math display="inline">(-\i..."

Before we apply intermediate value theorem, let’s discuss why the intersection can only happen in <math display="inline">(0,1)</math> rather than <math display="inline">(-\infty,0]\cup[1,\infty)</math>.

For the simplicity, we denote <math display="inline">g(x)=e^{-(x-1)}</math>, we know from the question that <math display="inline">f(x)=e^x</math> when <math display="inline">x\leq 0</math>, can we compare values of <math display="inline">f</math> and <math display="inline">g</math> in <math display="inline">(-\infty,0]</math>? The answer is yes, according to the mononicity of exponential function, we can easily see that when <math display="inline">x\in (-\infty,0]</math>, we have that <math display="block">f(x)\leq f(0)=e^0=1, \text{but} \ g(x)\geq g(0)=e^{-(0-1)}=e</math> i.e. <math display="inline">f(x)<g(x)</math> if <math display="inline">x\leq 0</math>.

In a similar manner, we can show that when <math display="inline">x\in [1,\infty)</math>, we have that <math display="block">f(x)>f(1)=1^2+1=2 \text{but} \ g(x)<g(1)=e^{-(1-1)}=1</math> i.e., <math display="inline">f(x)>g(x)</math> if <math display="inline">x\geq1</math>

The above analysis seems hopeless, but if you draw a piecewise graph for function <math display="inline">f</math> and <math display="inline">g</math>, you would directly find that the intersection can only happen in <math display="inline">(0,1)</math>.

In the following argument, for convenience denote <math display="block">F(x)=f(x)-g(x)=f(x)-e^{-(x-1)}.</math> Now that we know the intersection only shows up in <math display="inline">(0,1)</math>, the function <math display="inline">f(x)</math> is just <math display="inline">k(x-1)+2</math>, so <math display="block">F(x)=k(x-1)+2-e^{-(x-1)}.</math> It is time to apply intermediate value theorem, note that <math display="inline">F(x)</math> is continuous in the interval <math display="inline">(0,1)</math>, and <math display="block">F(0)=-k+2-e,\ F(1)=0+2-1=1.</math> If it has at least a zero in <math display="inline">(0,1)</math>, by the theorem, we need to make sure that <math display="inline">F(0)\cdot F(1)<0</math>, i.e., <math display="block">(-k+2-e)\cdot 1<0.</math> Solving this inequality gives us that <math display="block">k>2-e.</math> Thus, the solution is <math display="inline">\color{blue}\{k:k>2-e \}.</math>

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 03 (c)/Hint 1

2018-04-07T17:47:58Z

HANHONG:

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 03 (c)/Hint 1

2018-04-07T17:31:58Z

HANHONG: Created page with "If we can choose suitable <math display="inline">k</math> such that the function <math display="inline">f</math> is continuous, then we can apply intermediate value theorem to..."

If we can choose suitable <math display="inline">k</math> such that the function <math display="inline">f</math> is continuous, then we can apply intermediate value theorem to show that there exists at least one intersection.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 03 (b)/Solution 1

2018-04-07T17:25:56Z

HANHONG: Created page with "First note that the function is defined at <math display="inline">x=1</math> in the third piece of piecewise function, and <math display="inline">f(1)=1^2+1=2.</math> When <ma..."

First note that the function is defined at <math display="inline">x=1</math> in the third piece of piecewise function, and <math display="inline">f(1)=1^2+1=2.</math> When <math display="inline">k=2</math>, let’s see the left limit and right limit at point <math display="inline">x=1</math> as follows: <math display="block">f(1^+):=\lim_{x\rightarrow 1^+}f(x)=\lim_{x\rightarrow 1^+}x^2+1=1^2+1=2</math> and <math display="block">f(1^-):=\lim_{x\rightarrow 1^-}f(x)=\lim_{x\rightarrow 1^-}k(x-1)+2=2.</math> We get that the left limit is equal to right limit, thus <math display="block">\lim_{x\rightarrow 1}f(x)</math> exists. And note that <math display="block">\lim_{x\rightarrow 1}f(x)=2=f(1)</math> which is the reason that <math display="inline">f(x)</math> is continuous at <math display="inline">x=1</math>. So we choose <math display="inline">\color{blue} (i),(ii),(iii),(iv)</math>.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 03 (b)/Hint 1

2018-04-07T17:15:07Z

HANHONG: Created page with "On the contrary to the question (a), the definition of the continuity of function at one point is: the left limit equals the right limit, also equals the value at that point."

On the contrary to the question (a), the definition of the continuity of function at one point is: the left limit equals the right limit, also equals the value at that point.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 03 (a)/Solution 1

2018-04-07T17:13:20Z

HANHONG:

First, we note that the function is defined at <math display="inline">x=0</math> and <math display="inline">f(0)=e^0=1</math> since it is defined in the first piece, so (i) is not correct. Now let’s see the left limit and right limit as follows <math display="block">f(0^+):=\lim_{x\rightarrow 0^+}f(x)=\lim_{x\rightarrow 0^+}k(x-1)+2=-k+2=2</math> and <math display="block">f(0^-):=\lim_{x\rightarrow 0^-}f(x)=\lim_{x\rightarrow 0^-}e^x=e^0=1.</math> That is <math display="block">\lim_{x\rightarrow 0^+}f(x)\neq \lim_{x\rightarrow 0^-}f(x) .</math> Since the left limit is not equal to right limit at <math display="inline">x=0</math>, thus the limit <math display="block">\lim_{x\rightarrow 0}f(x)</math> does not exist. In all, we choose <math display="inline">\color{blue}(ii)</math>.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 03 (a)/Solution 1

2018-04-07T17:12:35Z

HANHONG: Created page with "First, we note that the function is defined at <math display="inline">x=0</math> and <math display="inline">f(0)=e^0=1</math> since it is defined in the first piece, so (a) is..."

First, we note that the function is defined at <math display="inline">x=0</math> and <math display="inline">f(0)=e^0=1</math> since it is defined in the first piece, so (a) is not correct. Now let’s see the left limit and right limit as follows <math display="block">f(0^+):=\lim_{x\rightarrow 0^+}f(x)=\lim_{x\rightarrow 0^+}k(x-1)+2=-k+2=2</math> and <math display="block">f(0^-):=\lim_{x\rightarrow 0^-}f(x)=\lim_{x\rightarrow 0^-}e^x=e^0=1.</math> That is <math display="block">\lim_{x\rightarrow 0^+}f(x)\neq \lim_{x\rightarrow 0^-}f(x) .</math> Since the left limit is not equal to right limit at <math display="inline">x=0</math>, thus the limit <math display="block">\lim_{x\rightarrow 0}f(x)</math> does not exist. In all, we choose <math display="inline">\color{blue}(ii)</math>.

Science:Math Exam Resources/Courses/MATH110/April 2016/Question 03 (a)/Hint 1

2018-04-07T17:01:34Z

HANHONG: Created page with "Recall the definition of discontinuity of a function at one point: 1, if the function is not defined at that point then it is not continuous at that point; 2, if the left lim..."

Recall the definition of discontinuity of a function at one point: 1, if the function is not defined at that point then it is not continuous at that point; 2, if the left limit is not equal to right limit at that point then it is not continuous at that point; 3 if the left limit is equal to right limit but not equal to the value of that point then it is not continuous.

Thread:Science talk:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (c)/Hint 1/Need a better explanation/reply (2)

2018-03-31T00:37:19Z

HANHONG: Reply to Need a better explanation

Dear Brian,

Never mind, I just expressed my opinion, it doesn't really matter if students don't get confused themselves.

Best

Han

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

2018-03-26T17:33:01Z

HANHONG:

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







{{MER Question page}}

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

2018-03-26T17:31:03Z

HANHONG:

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

2018-03-26T17:28:36Z

HANHONG:

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

2018-03-26T17:27:59Z

HANHONG:

Recall the following '''Quotient Rule''':

<br />
<math>D\left(\frac{h(x)}{g(x)}\right)=\frac{g(x)\cdot D(h(x))-h(x)\cdot D(g(x))}{(g(x))^2}.</math>

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

2018-03-26T17:26:48Z

HANHONG: Talk page autocreated when first thread was posted

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

2018-03-26T17:26:48Z

HANHONG: New thread: Need a better explanation

Hi Brian,

Your hint and your solution is definitely right telling students what to do. But to me, it didn't explain why they are doing that. I think it is better to explain the definition of vertical (undefined points) and horizontal asymptotes (existence of limit as x goes to infinity) , so they won't get lost when they encounter a more complicated question.
What do you think?

Best,

Han

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

2018-03-26T17:16:17Z

HANHONG:

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

2018-03-26T17:14:23Z

HANHONG:

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

2018-03-26T17:14:00Z

HANHONG:

Recall the definition of the domain of a function, in this question, for a fraction function.

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

2018-03-26T17:11:56Z

HANHONG:

The domain of <math>f(x)</math> is those x-values for the denominator of <math> f(x) </math> to be non-zero. That is, the domain is <math>\color{blue} \{x| \ x\neq 0\}.</math>

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 27/Solution 2

2018-03-26T17:07:45Z

HANHONG: Created page with "Since the solution is <math>x=[0,1,0,1]^{T} s+[7,0,1,0]^{T}t </math>, we can rewrite it as <math>x= \begin{bmatrix} 7t\\s\\t\\s \end{bmatrix}=\begin{bmatrix} 7x_3\\x_2\\x_3\\..."

Since the solution is <math>x=[0,1,0,1]^{T} s+[7,0,1,0]^{T}t </math>, we can rewrite it as
<math>x= \begin{bmatrix} 7t\\s\\t\\s \end{bmatrix}=\begin{bmatrix} 7x_3\\x_2\\x_3\\x_2 \end{bmatrix}</math> by changing variables.

It means that
when we transform the reduced echelon form to equations, we already have

<math>x_1=7x_3, \ \ \ \ x_4=x_2 </math>

Or equivalently

<math>x_1-7x_3=0, \ \ \ \ x_2-x_4=0 </math>

So the echelon form is easily seen to be

<math>\color{blue} \begin{bmatrix} 1 & 0 & -7 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}</math>

Since when we transform following matrix system to equations, we can attain above equations we want:

<math> \begin{bmatrix} 1 & 0 & -7 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}\cdot \begin{bmatrix} x_1\\x_2\\x_3\\x_4 \end{bmatrix}=\begin{bmatrix} 0\\0\\0\\0\end{bmatrix} </math>

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

2018-03-25T00:15:05Z

HANHONG:

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}
= \color{blue}t_1\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}+\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}.
</math>

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

2018-03-24T23:35:10Z

HANHONG:

According to definition of parametric form of plane, we will write <math> P_2 </math> as

:<math> \bold{x} = \bold{c} + s_1\bold{v_1} + s_2\bold{v_2} </math>

Our purpose is to find <math>c, \bold{v_1},\bold{v_2}</math>. As hinted, since <math> P_2</math> contains <math> l_1 </math>, we can set <math> \bold{v_1} = \bold{a}</math>. Now let's look for <math>\bold{ v_2} </math>, in fact, we know <math> P_2</math> is perpendicular to <math> P_1</math>, it not hard to see that the normal vector of <math> P_1 </math> is on the plane <math> P_1 </math>, this normal vector is going to be <math> \bold{ v_2} </math>.

We are already given the parametric form of <math> P_1 </math>, by the standard method of finding normal vector, we need to apply cross product to two nonparallel vectors <math> \bold{b}_1,\bold{b}_2 </math> in the plane <math> P_1 </math>.

The cross product <math> \bold{b}_1 \times \bold{b}_2 </math> is given by the ''formal'' determinant

:<math>

\det \begin{bmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & 1 & 0 \\ 1 & 2 & -1 \end{bmatrix} = \det \begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix} \hat{\mathbf{i}}\ -\ \det \begin{bmatrix} 0 & 0 \\ 1 & -1 \end{bmatrix} \hat{\mathbf{j}}\ +\ \det \begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix} \hat{\mathbf{k}} = -\hat{\mathbf{i}} - \hat{\mathbf{k}} = \begin{bmatrix} -1 \\ 0 \\ -1 \end{bmatrix},

</math>

so we set

:<math> \bold{v_2} = \begin{bmatrix} -1 \\ 0 \\ -1 \end{bmatrix}. </math>

It remains to compute <math>\bold{c} </math>. Notice that the origin is in <math> l_1 </math>, so it is in <math> P_2 </math> since <math> P_2 </math> contains <math> l_1</math>. Hence we can take <math> \bold{c}= 0 </math>. It follows that a parametric form of the equation for <math> P_2 </math> is

:<math>
\color{blue}

\mathbf{x} = s_1\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} + s_2 \begin{bmatrix} -1 \\ 0 \\ -1 \end{bmatrix}.

</math>

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

2018-03-24T22:51:01Z

HANHONG:

According to definition of parametric form of plane, we will write <math> P_2 </math> as

:<math> \bold{x} = \bold{c} + s_1\bold{v_1} + s_2\bold{v_2} </math>

Our purpose is to find <math>c, \bold{v_1},\bold{v_2}</math> in <math> P_2</math>. As hinted, since <math> P_2</math> contains <math> l_1 </math>, we can set <math> \bold{v_1} = \bold{a}</math>. In addition, we know <math> P_2</math> is perpendicular to <math> P_1</math> so it contains the vector perpendicular to <math> P_1</math>. We can compute this vector with a cross product.

The cross product <math> \bold{b}_1 \times \bold{b}_2 </math> is given by the ''formal'' determinant

:<math>

\det \begin{bmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & 1 & 0 \\ 1 & 2 & -1 \end{bmatrix} = \det \begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix} \hat{\mathbf{i}}\ -\ \det \begin{bmatrix} 0 & 0 \\ 1 & -1 \end{bmatrix} \hat{\mathbf{j}}\ +\ \det \begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix} \hat{\mathbf{k}} = -\hat{\mathbf{i}} - \hat{\mathbf{k}} = \begin{bmatrix} -1 \\ 0 \\ -1 \end{bmatrix},

</math>

so we set

:<math> \bold{v_2} = \begin{bmatrix} -1 \\ 0 \\ -1 \end{bmatrix}. </math>

It remains to compute <math>\bold{c} </math>. Notice that the origin is in <math> l_1 </math>, so it is in <math> P_2 </math>. Hence we can take <math> \bold{c}= 0 </math>. It follows that a parametric form of the equation for <math> P_2 </math> is

:<math>
\color{blue}

\mathbf{x} = s_1\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} + s_2 \begin{bmatrix} -1 \\ 0 \\ -1 \end{bmatrix}.

</math>

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

2018-03-24T22:49:08Z

HANHONG:

The form given in the question is actually vector form of a plane: given any point <math> a_0 </math> in the plane, and two nonparallel vectors <math> v,w</math> in the plane, then any other point <math> X</math> can be written as

<math display="block">X=a_0+t\cdot v+s\cdot w </math>.

Here, <math> t,s </math>are parameters. We already know one direction in the plane, i.e., the direction of line <math> l </math>, the target is to find one point and another direction in the plane.

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

2018-03-24T22:48:26Z

HANHONG:

We will write the parametric form of <math> P_2 </math> as

:<math> \bold{x} = \bold{c} + s_1\bold{v_1} + s_2\bold{v_2} </math>

We know <math>\bold{v_1},\bold{v_2}</math> are vectors in <math> P_2</math>. Since <math> P_2</math> contains <math> l_1 </math>, we can set <math> \bold{v_1} = \bold{a}</math>. In addition, we know <math> P_2</math> is perpendicular to <math> P_1</math> so it contains the vector perpendicular to <math> P_1</math>. We can compute this vector with a cross product.

The cross product <math> \bold{b}_1 \times \bold{b}_2 </math> is given by the ''formal'' determinant

:<math>

\det \begin{bmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & 1 & 0 \\ 1 & 2 & -1 \end{bmatrix} = \det \begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix} \hat{\mathbf{i}}\ -\ \det \begin{bmatrix} 0 & 0 \\ 1 & -1 \end{bmatrix} \hat{\mathbf{j}}\ +\ \det \begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix} \hat{\mathbf{k}} = -\hat{\mathbf{i}} - \hat{\mathbf{k}} = \begin{bmatrix} -1 \\ 0 \\ -1 \end{bmatrix},

</math>

so we set

:<math> \bold{v_2} = \begin{bmatrix} -1 \\ 0 \\ -1 \end{bmatrix}. </math>

It remains to compute <math>\bold{c} </math>. Notice that the origin is in <math> l_1 </math>, so it is in <math> P_2 </math>. Hence we can take <math> \bold{c}= 0 </math>. It follows that a parametric form of the equation for <math> P_2 </math> is

:<math>
\color{blue}

\mathbf{x} = s_1\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} + s_2 \begin{bmatrix} -1 \\ 0 \\ -1 \end{bmatrix}.

</math>

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

2018-03-24T22:47:29Z

HANHONG:

The form given in the question is actually vector form of a plane: given any point <math> a_0 </math> in the plane, and two nonparallel vectors <math> v,w</math> in the plane, then any other point <math> X</math> can be written as

<math display="block">X=a_0+t\cdot v+s\cdot w </math>.

We already know one direction in the plane, i.e., the direction of line <math> l </math>, the target is to find one point and another direction in the plane.

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 29

2018-03-19T03:20:00Z

HANHONG:

Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 29/Hint 1

2018-03-19T03:19:45Z

HANHONG:

Consider the eigenvector. The direction vector of the rotation is just the vector <math>v</math> unchanged after rotation, i.e., <math>Av=v.</math>

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 07 (b)

2018-03-19T03:09:51Z

HANHONG:

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 07 (a)

2018-03-19T03:05:18Z

HANHONG:

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 04 (b)

2018-03-19T03:02:51Z

HANHONG:

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 04 (a)

2018-03-19T02:23:22Z

HANHONG:

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{{MER Question page}}

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 04 (a)/Hint 1

2018-03-19T02:22:24Z

HANHONG:

A simple way of approximating <math>\sqrt{69}</math> is to construct the linear approximation of (i.e., ''tangent line'' to) <math>f(x) = \sqrt{x}</math> at some point <math>(a, f(a))</math> for which <math>f(a)</math> is known exactly. Therefore, the point needs to satisfy two things: have a pretty straightforward square root and also close to <math>\sqrt{69}</math>. Thus we choose <math>a = 64.</math>

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 04 (a)/Hint 1

2018-03-19T02:21:32Z

HANHONG:

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 03 (b)

2018-03-19T02:20:24Z

HANHONG:

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{{MER Question page}}

Science:Math Exam Resources/Courses/MATH104/December 2016/Question 03 (a)

2018-03-19T02:18:56Z

HANHONG: