MATH6015: Week 10

I am emailing a link of this to everyone on the class list every Friday afternoon. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jpmccarthymaths@gmail.ie and I will add you to the mailing list.

Test 2

The second test will take place at 9 am next Friday 30 November. It is another 15% test that could arguably take 48 minutes but I’ll give ye from 9.05 — 10 am. Please find a sample.

Additional Tutorial

An additional tutorial for MATH6015 has been arranged for Tuesdays 17:00 – 18:00 in PF45.

To keep the numbers low at tutorials (so I can spend more time with ye) we have the following arrangement.

The BioEng students who have their lab in Week 10 (19 November $\rightarrow$ 23 November) are known as GROUP I. Those who are doing their lab in Week 11 are known as GROUP II.

The following groups may attend the relevant tutorials:

Week 10

Tuesday 20 November: ALL

Week 11

Monday 26 November: Group I and BIS ONLY

Tuesday 27 November: ALL

Week 12

Monday 3 December: Group II and BIS ONLY

Tuesday 4 December: ALL

Lectures

We evaluated areas under and between curves using integration. On Friday we showed how to evaluate volumes of revolution using integration.

Next Week

On Monday & Tuesday we will have tutorials where we will work more on the stuff for Test 2. On Tuesday and Thursday we will finish off the applications to volumes and introduce two more applications of integration.

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December 3, 2012 at 9:52 am

Student 26

I’m a student in your technological maths class, Tech Maths 2.

I’m currently revising by doing past exam papers, and have encountered a question I can’t seem to crack.

The question is from the 2011 Semester 1 exam, Q1, (b).

”Differentiate by rule: $\displaystyle y(x) = \frac{x}{\sqrt{1-x^2}}$ ”

Immediately I try the quotient rule, but by my 3rd line I’m lost, is there a chain rule inside this question also?

Appreciate your time,

December 3, 2012 at 3:41 pm

J.P. McCarthy

First of all you certainly need the quotient rule as your function is a quotient. We have that

$\displaystyle \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right)=\frac{vu'-uv'}{v^2}$

where $u=u(x)$ , $u'=u'(x)$ , etc.

So we have
$\displaystyle \frac{d}{dx}\left(\frac{x}{\sqrt{1-x}^2}\right)$

$\displaystyle= \frac{\sqrt{1-x^2}\frac{d}{dx}x-x\frac{d}{dx}\sqrt{1-x^2}}{(\sqrt{1-x^2})^2}$ .

To differentiate $x$ is no problem but we will need the Chain Rule to differentiate $\sqrt{1-x^2}$ . We are good at differentiating powers so we write $\sqrt{1-x^2}=(1-x^2)^{1/2}$ . Now this is a composition, a function of a function so we need to use the Chain Rule.

Roughly the Chain Rule says “Differentiate the outside function at the inside function and multiply the by the derivative of the inside function”:

$\displaystyle \frac{d}{dx}u(v(x))=u'(v(x))\times v'(x)$

$\displaystyle \Rightarrow \frac{d}{dx}(1-x^2)^{1/2}=\frac{1}{2}(1-x^2)^{-1/2}\times(-2x)$

where $-2x$ is the ‘derivative of the inside’ $1-x^2$ . Putting things back together (using $\displaystyle\frac{d}{dx}x=1$ ):

$\displaystyle \frac{dy}{dx}= \frac{\sqrt{1-x^2}-x\left(\frac{1}{2}(1-x^2)^{-1/2}(-2x)\right)}{(\sqrt{1-x^2})^2}$ .

We have got quite a lot of the marks at this point but we can do more:

1. By definition $(\sqrt{1-x^2})^2=1-x^2$ (well if $1-x^2$ is positive which we assume that it is.)
2. $x^{-1/2}=\frac{1}{x^{1/2}}=\frac{1}{\sqrt{x}}$ .
3. There is a two by a half term (which is equal to one) in the second term of the numerator.
4. $-(-2x)=2x$ .

Using all of these we have:

$\displaystyle \frac{dy}{dx}= \frac{\sqrt{1-x^2}+x^2\frac{1}{\sqrt{1-x^2}}}{1-x^2}$ .

Now we probably have all the marks but I don’t like the fraction in the numerator. If I multiply by $\displaystyle \frac{\sqrt{1-x^2}}{\sqrt{1-x^2}}=1$ this will do the trick:

$\displaystyle \frac{dy}{dx}= \frac{\sqrt{1-x^2}+x^2\frac{1}{\sqrt{1-x^2}}}{1-x^2}\cdot\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}}=\frac{(\sqrt{1-x^2})^2+x^2}{(1-x^2)\sqrt{1-x^2}}$

$\displaystyle = \frac{1}{(1-x^2)\sqrt{1-x^2}}$ .

Alternatively you could start by writing $y(x)=x(1-x^2)^{-1/2}$ and use a Product Rule.

Regards,
J.P.

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MATH6015: Week 10

Test 2

Additional Tutorial

Lectures

Next Week

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J.P. McCarthy Maths

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MATH6015: Week 10

Test 2

Additional Tutorial

Lectures

Next Week

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J.P. McCarthy Maths

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