UBC Wiki - User contributions [en]

Template:Alfred Xing

2015-10-26T18:52:33Z

Alfred: Blanked the page

User:Alfred

2015-10-26T18:48:17Z

Alfred: Blanked the page

User:Alfred

2013-04-19T04:29:28Z

Alfred:

<div style="width:100%;background:#0098a0">
<div style="width:50%;margin:0 auto; padding:2em 0">[[File:Alfred Xing Logo.png|link=http://alfredxing.com]]</div>
</div>
<div style="width:50%;margin: 2em auto; font: 700 2em 'Lato', 'Open Sans', 'Helvetica', 'Segoe UI';padding-left:10px;color:#123">
Web designer. Student. Minimalist.
</div>

File:Alfred Xing Logo.png

2013-04-19T04:19:11Z

Alfred:

=={{int:filedesc}}==
{{Information
|description={{en|1=Logo for Alfred's user page}}
|date=2013-04-18
|source={{own}}
|author=[[User:Alfred|Alfred Xing]]
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File:MATH101 April 2010 Question 4 Cone.png

2013-04-07T17:15:58Z

Alfred: Alfred uploaded a new version of "File:MATH101 April 2010 Question 4 Cone.png"

=={{int:filedesc}}==
{{Information
|description={{en|1=A render of the anthill in question 4 of the April 2010 exam for MATH 101.}}
|date=2013-04-07
|source={{own}}
|author=[[User:Alfred|Alfred Xing]]
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File:MATH101 April 2010 Question 4 Cone.png

2013-04-07T17:11:58Z

Alfred: Alfred uploaded a new version of "File:MATH101 April 2010 Question 4 Cone.png": Add Δx and r

Science:Math Exam Resources/Courses/MATH101/April 2010/Question 04/Solution 1

2013-04-07T17:05:54Z

Alfred: Rerender the image

As the hint suggests, we split the cone into cylinders of height say <math> \Delta x </math> which are a height of <math> 1-x_{i}^{*} </math> from the ground (where the <math> x_{i}^{*} </math> is an arbitrary sample point)[[File:MATH101_April_2010_Question_4_Cone.png|500px|thumbnail|right]]. We would like to know the radius of each of these cylinders depending upon how high up the cone they are, then we can calculate the volume of these cylinders and add them all up. In order to find the radius ''r'' of each cylinder with respect to its height up the cone, we use ratios and the fact that the radius of the base of the cone is 1/2. I.e.: Since the base of the cone is 1/2 and that's 1 ft away from the tip of the cone, we know that (via similar triangles in the cone - see the diagram)
:<math>
\frac{1}{1/2} = \frac{x_{i}^{*}}{r}
</math>
and so
:<math>
r = \frac{x_{i}^{*}}{2}
</math>

In our diagram below, we have set it up so that the origin is at the apex (another way to solve this problem is to set up the 0 value at the bottom of the anthill). We now have, since the volume of a cylinder is <math> \pi r^2 h </math> where <math> r </math> is the radius of the base and <math> h </math> is the height of the cylinder, that
:<math>
V=\pi \frac{(x_{i}^{*})^2}{4} \Delta x
</math>

Since

work done = weight <math> \times </math> distance

and

weight = density <math> \times </math> volume

we have

work = density <math> \times </math> volume <math> \times </math> distance

where the distance is <math> 1-x_{i}^{*} </math>. Hence we can get the total volume by adding the contributions from each of the cylinders. In the limit as <math>\Delta{x}</math> goes to zero this becomes and integral and we get

:<math>\begin{align}
\text{work} &= \int_0^1 150\, \pi \frac{x^2}4\,(1-x) dx\\
&= \left[\frac{150\pi x^3}{12} - \frac{150\pi x^4}{16}\right]_0^1 \\
&= \frac{150 \pi}{12} -\frac{150\pi }{16} \\
&= \frac{25\pi}{8}
\end{align}</math>

File:MATH101 April 2010 Question 4 Cone.png

2013-04-07T17:03:59Z

Alfred:

Science:Math Exam Resources/Courses/MATH101/April 2012/Question 02 (a)

2013-03-26T16:45:19Z

Alfred:

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







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH101/April 2012/Question 02 (a)/Hint 1

2013-03-26T16:44:33Z

Alfred: Created page with "The area between the curves ''f''(x) and ''g''(x) is :: <math>\int_a^b(f(x)-g(x))dx</math> where ''a'' and ''b'' are the intersects of the two curves and ''f''(x) is the upper..."

The area between the curves ''f''(x) and ''g''(x) is
:: <math>\int_a^b(f(x)-g(x))dx</math>
where ''a'' and ''b'' are the intersects of the two curves and ''f''(x) is the upper curve.

File:101AR12Q6 - PNG.png

2013-01-25T05:36:36Z

Alfred:

=={{int:filedesc}}==
{{Information
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|date=2013-01-24
|source={{own}}
|author=[[User:Alfred|Alfred Xing]]
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{{self|cc-by-sa-2.5-ca}}

Science:Math Exam Resources/Courses/MATH101/April 2012/Question 06

2013-01-25T05:33:01Z

Alfred:

[[Category:MER RQ flag]][[Category:MER CH flag]][[Category:MER CS flag]][[Category:MER CT flag]]







{{MER Question page}}

File:101AR12Q6 - PNG.png

2013-01-25T05:31:57Z

Alfred: Alfred uploaded a new version of "File:101AR12Q6 - PNG.png": Fixed added margins (last time didn't add)

=={{int:filedesc}}==
{{Information
|description={{en|1=PNG version of the SVG file.}}
|date=2013-01-24
|source={{own}}
|author=[[User:Alfred|Alfred Xing]]
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|other_versions=
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=={{int:license-header}}==
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File:101AR12Q6 - PNG.png

2013-01-25T05:29:55Z

Alfred: Alfred uploaded a new version of "File:101AR12Q6 - PNG.png": Added margins

Science:Math Exam Resources/Courses/MATH101/April 2012/Question 06/Statement

2013-01-25T05:28:30Z

Alfred:

A tank in the shape of a hemispherical bowl of radius 3m, with an outlet that rises 2m above its top (see the diagram below), is full of water. Using the fact that the density of water is <math> 1000\text{kg/m}^{3}</math>, find the work (in joules) required to pump all the water out of the outlet. You may use the value <math> g=9.8\text{m/s}^{2} </math> for the acceleration due to gravity.You do not need to simplify your answer, but you must completely evaluate any integral(s) that arise.

[[File:101AR12Q6 - PNG.png|frame|Diagram for the above question]]

Science:Math Exam Resources/Courses/MATH101/April 2012/Question 06/Statement

2013-01-25T05:27:32Z

Alfred:

Science:Math Exam Resources/Courses/MATH101/April 2012/Question 06/Statement

2013-01-25T05:27:11Z

Alfred:

File:101AR12Q6 - PNG.png

2013-01-25T05:26:35Z

Alfred:

Science:Math Exam Resources/Courses/MATH101/April 2012/Question 06/Statement

2013-01-25T05:25:06Z

Alfred:

File:101AR12Q6.svg

2013-01-25T05:22:06Z

Alfred:

=={{int:filedesc}}==
{{Information
|description={{en|1=Figure for Question 6 of the April 2012 exam for MATH 101}}
|date=2013-01-24
|source={{own}}
|author=[[User:Alfred|Alfred Xing]]
|permission=
|other_versions=
|other_fields=
}}

=={{int:license-header}}==
{{self|cc-by-sa-2.5-ca}}

Science:Math Exam Resources/Courses/MATH101/April 2012/Question 06/Statement

2013-01-25T05:20:44Z

Alfred:

Science:Math Exam Resources/Courses/MATH101/April 2012/Question 06/Statement

2013-01-25T04:41:56Z

Alfred:

Thread:Science talk:MER/Solutions to MATH 100/180 December 2011 and December 2010 final exams/reply (2)

2012-12-04T04:34:25Z

Alfred:

I'll work on that once I'm done my finals.
I sent you an email regarding the latter (there doesn't seem to be a private messaging system here).

Also, here are the solutions to December 2006 (they're not from UBC but I checked the answers and they should be fine):
http://yukon-mathematics-assessment.wikispaces.com/file/view/UBC+Math+100+%26+180+Final+Exam+December+2006+Solutions.pdf

Thread:Science talk:MER/Solutions to MATH 100/180 December 2011 and December 2010 final exams/reply (2)

2012-12-04T04:12:17Z

Alfred: Reply to Solutions to MATH 100/180 December 2011 and December 2010 final exams

I'll work on that once I'm done my finals.
I sent you an email regarding the latter (there doesn't seem to be a private messaging system here).

Thread:Science talk:MER/Challenge: wolframalpha plugin to display stats/reply (5)

2012-12-04T04:06:12Z

Alfred: Reply to Challenge: wolframalpha plugin to display stats

Both can plot equations. Integrating the WolframAlpha one will be a bit hard though because the MediaWiki widget doesn't allow other input parameters.

User:Alfred

2012-12-01T04:23:50Z

Alfred:

<div style="font-family: 'Helvetica Neue Light', 'Segoe UI Light', 'Batik Regular'; font-size: 4em; float: right; padding-right: 1em; text-align: right;background: #003D51; color: #FFFFFF; height: 65px;">
<div style="position: relative; right: -11px;">xing,</div><br />
<div style="position: relative;">alfred</div>
</div>

{| style="font-family:'Segoe UI Light', 'Batik Regular'; font-size: 1.25em;"
|style="border: 1px solid #003D51; width: 20%;"|<div style="padding: 5px;">You are currently visiting my user page.</div>
|style="width: 50%;"|
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|-
|style="width: 20%;"|
|style="width: 50%; border: 1px solid #003D51;"|<div style="padding: 5px;">I don't know why, but most likely you just want to know who I am. Sadly, I'm not going to say much about myself here but I will state that I am a first-year student in the Faculty of Science.</div>
|style="width: 30%;"|
|-
|style="width: 20%;"|
|style="width: 50%;"|
|style="width: 30%; border: 1px solid #003D51;"|<div style="padding: 5px;">I also enjoy coding, particularly web design. You can visit my website at [http://alfredxing.com alfredxing.com].</div>
|}

[[Category:MER Participant]]

Thread:Science talk:MER/Solutions to MATH 100/180 December 2011 and December 2010 final exams

2012-12-01T03:50:29Z

Alfred: New thread: Solutions to MATH 100/180 December 2011 and December 2010 final exams

I found the solutions to these exams:
2011: http://www.math.ubc.ca/~gupta/m100-180_common/M100-180Final201112_solns.pdf
2010: http://www.math.ubc.ca/~gupta/m100-180_common/M100-180Final201012_solns.pdf

This saves us a lot of work (especially for 2010).

Thread:Science talk:MER/Challenge: wolframalpha plugin to display stats/reply (3)

2012-12-01T02:52:00Z

Alfred:

Not until we get the admin to install the iframe module.
One thing that's not good about graph.tk is that it can't plot points...

I found a graphing widget, but if it's not good enough we can make our ourselves; it isn't too hard:
'''<nowiki>{{#widget:WolframAlpha|id=45a99effbf17c0286917f132ab9d3595}}</nowiki>'''

Thread:Science talk:MER/Challenge: wolframalpha plugin to display stats/reply (3)

2012-12-01T02:51:12Z

Alfred:

Not until we get the admin to install the iframe module.
One thing that's not good about graph.tk is that it can't plot points...

I found a graphing widget, but if it's not good enough we can make our ourselves; it isn't too hard:
<nowiki>{{#widget:WolframAlpha|id=45a99effbf17c0286917f132ab9d3595}}</nowiki>

Thread:Science talk:MER/Challenge: wolframalpha plugin to display stats/reply (3)

2012-12-01T02:50:54Z

Alfred: Reply to Challenge: wolframalpha plugin to display stats

Not until we get the admin to install the iframe module.
One thing that's not good about graph.tk is that it can't plot points...

I found a graphing widget, but if it's not good enough we can make our ourselves; it isn't too hard:
{{#widget:WolframAlpha|id=45a99effbf17c0286917f132ab9d3595}}

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 02 (a)/Solution 1

2012-11-29T23:37:39Z

Alfred: Created page with "== Draft == Newton's Law of Cooling states that :<math>\displaystyle T(t)-T_s=(T(0)-T_s)e^{kt} </math> We know that the initial temperature, ''T''(0), is equal to 3; ''T<sub..."

== Draft ==

Newton's Law of Cooling states that
:<math>\displaystyle
T(t)-T_s=(T(0)-T_s)e^{kt}
</math>

We know that the initial temperature, ''T''(0), is equal to 3; ''T<sub>s</sub>'' , the temperature of the surroundings, is 19°; and that ''T''(30) is equal to 11°.

Then we know that
:<math>\displaystyle
T(t)=3-16e^{kt}
</math>

And since ''k'' is a constant:
:<math>\begin{align}
T(30)=11&=3-16e^{30k}\\
8&=-16e^{30k}\\
-\frac{1}{2}&=e^{30k}\\
ln(-\frac{1}{2})&=30k\\
\frac{ln(-1/2)}{30}&=k
\end{align}
</math>

Thread:Science talk:MER/Challenge: wolframalpha plugin to display stats/reply

2012-11-29T22:48:03Z

Alfred: Reply to Challenge: wolframalpha plugin to display stats

A very useful application, graph.tk, has API we can use to implement graphs easily using LaTeX.
Unlike WolframAlpha, it's open source so we can use it basically how we like.

http://graph.tk/about/api.html

However, this involves iframes, which can be implemented with the iframe widget: http://www.mediawikiwidgets.org/Iframe

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (a)

2012-11-23T06:45:32Z

Alfred:

[[Category:MER RQ flag]][[Category:MER CH flag]][[Category:MER RS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (a)/Solution 1

2012-11-23T06:45:17Z

Alfred: Created page with "Directly plugging in ''x''=1 into the limit gives 0/0. Since this is an indeterminate form, we can use L'Hospital's rule: :<math> \displaystyle \begin{align} &\lim_{x\to 1}{\..."

Directly plugging in ''x''=1 into the limit gives 0/0. Since this is an indeterminate form, we can use L'Hospital's rule:

:<math>
\displaystyle
\begin{align}
&\lim_{x\to 1}{\frac{x^2+2x-3}{x-1}}\\
=&\lim_{x\to 1}{\frac{(x^2+2x-3)'}{(x-1)'}}\\
=&\lim_{x\to 1}{\frac{2x+2}{1}}\\
=&\lim_{x\to 1}{2x+2}\\
=&3(1)+2\\
=&5
\end{align}
</math>

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (e)

2012-11-23T06:38:08Z

Alfred:

[[Category:MER QGQ flag]][[Category:MER CH flag]][[Category:MER CS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (d)

2012-11-23T06:37:40Z

Alfred:

[[Category:MER RQ flag]][[Category:MER CH flag]][[Category:MER CS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (j)/Solution 1

2012-11-06T00:19:21Z

Alfred: Created page with "To find the second degree Taylor polynomial, we use the equation :<math> T_2(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2 </math> We know that ''a'' =4 and that &fnof;(x)=&radic..."

To find the second degree Taylor polynomial, we use the equation
:<math>
T_2(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2
</math>
We know that ''a'' =4 and that &fnof;(x)=√{{overline|x}}. We can easily compute the derivatives to get
:<math>
\begin{align}
f'(x)&=\frac{1}{2}x^{-1/2}\\
f''(x)&=-\frac{1}{4x^{3/2}}
\end{align}
</math>
Plugging this all in to the Taylor equation, we get:
:<math>\displaystyle
\begin{align}
T_2(x)&=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2\\
&=f(4)+f'(4)(x-a)+\frac{f''(4)}{2}(x-4)^2\\
&=\sqrt{4}+\frac{1}{2}4^{-1/2}(x-a)+\frac{-\frac{1}{4\times 4^{3/2}}}{2}(x-4)^2\\
&=2+\frac{1}{4}(x-a)-\frac{1}{64}(x-4)^2\\
&=2+\frac{x-a}{4}-\frac{(x-4)^2}{64}\\
\end{align}
</math>

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (j)/Hint 1

2012-11-06T00:05:58Z

Alfred: Created page with "Recall that the general equation for a second degree Taylor polynomial is given by: :<math> T_2(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2 </math>"

Recall that the general equation for a second degree Taylor polynomial is given by:
:<math>
T_2(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2
</math>

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (j)/Statement

2012-11-06T00:04:51Z

Alfred: Created page with "{{Math Exam Resources/Courses/MATH100/December 2010/Question 1/Statement/Header}} Find the second degree Taylor polynomial T<sub>2</sub>(x) for :<math>\displaystyle f(x)=\sqr..."

{{Math Exam Resources/Courses/MATH100/December 2010/Question 1/Statement/Header}}

Find the second degree Taylor polynomial T<sub>2</sub>(x) for
:<math>\displaystyle
f(x)=\sqrt{x}
</math>
At 4 (or about ''x'' =4)

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (j)

2012-11-06T00:03:11Z

Alfred:

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







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (i)/Solution 1

2012-11-06T00:02:06Z

Alfred:

Let ''&fnof;'' (x)=x<sup>1/3</sup>.
First we need a number ''a''  as close to 30 as possible but whose cube root we know. A good candidate would be 3<sup>3</sup>=27.
We know that the general formula for a linear approximation is given by
:<math>\displaystyle
L(x)=f(a)+f'(a)(x-a)
</math>
We can then find the derivative:
:<math>\displaystyle
f'(x)=\frac{x^{-2/3}}{3}
</math>

Knowing these values, we can plug them in to the linear approximation equation and find the approximation for x=30:
:<math>\displaystyle
\begin{align}
L(x)&=f(a)+\frac{27^{-2/3}}{3}(x-27)\\
L(30)&=f(27)+\frac{27^{-2/3}}{3}(30-27)\\
&=3+\frac{1/9}{3}(3)\\
&=3+\frac{1}{9}\\
&=\frac{28}{9}
\end{align}
</math>

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (i)/Solution 1

2012-11-05T23:53:15Z

Alfred: Created page with "Let ''&fnof;'' (x)=x<sup>1/3</sup>. First we need a number as close to 30 as possible but whose cube root we know. A good candidate would be 3<sup>3</sup>=27."

Let ''&fnof;'' (x)=x<sup>1/3</sup>.
First we need a number as close to 30 as possible but whose cube root we know. A good candidate would be 3<sup>3</sup>=27.

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (i)

2012-11-05T23:46:35Z

Alfred:

[[Category:MER RQ flag]][[Category:MER CH flag]][[Category:MER RS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (i)/Statement

2012-11-05T23:46:11Z

Alfred: Created page with "{{Math Exam Resources/Courses/MATH100/December 2010/Question 1/Statement/Header}} Use a suitable linear approximation to estimate (30)<sup>1/3</sup>. Give your answer as a fr..."

{{Math Exam Resources/Courses/MATH100/December 2010/Question 1/Statement/Header}}

Use a suitable linear approximation to estimate (30)<sup>1/3</sup>. Give your answer as a fraction with integer numerator and denominator.

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (h)/Solution 1

2012-11-05T23:44:20Z

Alfred: Created page with "To differentiate this, we should use logarithmic differentiation: :<math>\displaystyle \begin{align} ln(f(x))&=\ln{((\tan{x})^{\cos{x}})}\\ ln(f(x))&=\cos{x}\ln{(\tan{x})}\\ (..."

To differentiate this, we should use logarithmic differentiation:
:<math>\displaystyle
\begin{align}
ln(f(x))&=\ln{((\tan{x})^{\cos{x}})}\\
ln(f(x))&=\cos{x}\ln{(\tan{x})}\\
(ln(f(x)))'&=(\cos{x}\ln{(\tan{x})})'\\
\frac{f'(x)}{f(x)}&=\cos{x}(\ln{(\tan{x})})'-\sin{x}\ln{(\tan{x})}\\
&=\cos{x}\frac{\sec^2(x)}{\tan{x}}-\sin{x}\ln{(\tan{x})}\\
f'(x)&=f(x)\left(\frac{\cos{x}\sec^2(x)}{\tan{x}}-\sin{x}\ln{(\tan{x})}\right)\\
&=(\tan{x})^{\cos{x}}\left(\frac{\cos{x}\sec^2(x)}{\tan{x}}-\sin{x}\ln{(\tan{x})}\right)\\
\end{align}

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (h)

2012-11-05T23:37:04Z

Alfred:

[[Category:MER RQ flag]][[Category:MER CH flag]][[Category:MER RS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (h)/Statement

2012-11-05T23:36:49Z

Alfred: Created page with "{{Math Exam Resources/Courses/MATH100/December 2010/Question 1/Statement/Header}} Find ''&fnof;'' ’(x), if :<math>\displaystyle f(x)=(\tan{x})^{\cos{x}} </math>"

{{Math Exam Resources/Courses/MATH100/December 2010/Question 1/Statement/Header}}

Find ''&fnof;'' ’(x), if
:<math>\displaystyle
f(x)=(\tan{x})^{\cos{x}}
</math>

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (g)/Solution 1

2012-11-05T23:20:41Z

Alfred: Created page with "First, we must find the derivative of the curve using implicit differentiation: :<math>\displaystyle \begin{align} x^4-x^2y+y^4 & =1\\ (x^4-x^2y+y^4)'&=(1)'\\ 4x^3-(x^2y)'+4y^..."

First, we must find the derivative of the curve using implicit differentiation:
:<math>\displaystyle
\begin{align}
x^4-x^2y+y^4 & =1\\
(x^4-x^2y+y^4)'&=(1)'\\
4x^3-(x^2y)'+4y^3y'&=0\\
4x^3-x^2y'-2xy+4y^3y'&=0\\
y'(-x^2+4y^3)&=2xy-4x^3\\
y'&=\frac{2xy-4x^3}{4y^3-x^2}
\end{align}
</math>
This gives us the equation for the slope. The question is asking us for the slope at the point (-1,1) so we plug in x=-1 and y=1:
:<math>\displaystyle
\begin{align}
y'&=\frac{2xy-4x^3}{4y^3-x^2}\\
&=\frac{2(-1)(1)-4(-1)^3}{4(1)^3-(-1)^2}\\
&=\frac{-2+4}{4-1}\\
&=\frac{2}{3}\\
\end{align}
</math>
Therefore, the slope of the tangent line to the curve at (-1,1) is 2/3.

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (g)

2012-11-05T23:06:18Z

Alfred:

[[Category:MER RQ flag]][[Category:MER CH flag]][[Category:MER RS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (g)/Statement

2012-11-05T23:06:03Z

Alfred: Created page with "{{Math Exam Resources/Courses/MATH100/December 2010/Question 1/Statement/Header}} Find the slope of the tangent line to the curve :<math>\displaystyle x^4-x^2y+y^4=1 </math> ..."

{{Math Exam Resources/Courses/MATH100/December 2010/Question 1/Statement/Header}}

Find the slope of the tangent line to the curve
:<math>\displaystyle
x^4-x^2y+y^4=1
</math>
At the point (-1,1).