MATH6040: Spring 2020, Week 7

Tuesday 17 March 2020, Week 8

This class is postponed to the same time and room, but Wednesday 18 March.

Week 7

In Week 7 we did a matrices Concept MCQ and then did a quick revision of differentiation:

Then we looked at Parametric Differentiation.

We had no tutorial time

Homework/Study

How much time you put into homework is up to you: of course the more time you put in the better but we all have competing interests. I strongly recommend that you do some exercises every week. Please feel free to ask me questions about the exercises via email, or even better on this webpage.

I recommend strongly that everyone completes P.111, Q.1.

I would invite you to look at the following exercises:

P.111, Q. 1
P.123, Q. 1-3

Revision:

P.111, Q. 2
P.124, Q. 4-8

In addition, you can hand up your work in class and I will have it corrected for the week after.

Week 8

We will look at Related Rates and then look at Implicit Differentiation. If you are interested in a very “mathsy” approach to curves you can look at this.

I have video of the above material here.

Week 9

We will look at partial differentiation and its applications to error analysis.

Looking further ahead, a good revision of integration/antidifferentiation may be found here.

Test 2

Test 2, worth 15% and based on Chapter 3, will probably take place Week 11.

Study

Please feel free to ask me questions about the exercises via email or even better on this webpage.

Academic Learning Centre

If you are a little worried about your maths this semester, perhaps after the Diagnostic Test or in general, I would just like to remind you about the Academic Learning Centre. Next week, some students will receive emails detailing areas of maths that they should brush up on. The timetable is here: there is some availability after 17:00 on Mondays and Tuesdays.

Student Resources

Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc.

2 comments

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March 19, 2020 at 7:31 am

Student

Hello J.P.,

Struggling with a couple of exercises: 1(d) and 2(a) on p.111; Q. 3,7 and 8 on p123.

Thanks

March 19, 2020 at 7:57 am

J.P. McCarthy

Your work in general is very good. I will focus on the questions you mentioned here.

P. 111, Q. 1 (d)
My advice would be to differentiate the two parts, $(2x+1)^3$ , and $x+2$ , separately before using the product rule.

You do the chain rule slightly different to how I do the chain rule. I will do it your way and then my way, and then explain why I possibly prefer my way.

Give or take, your method is to identify $y$ and $u$ and then say that:

$\displaystyle \frac{d}{dx}(y(u(x)))=\frac{dy}{du}\cdot \frac{du}{dx}$ .

This is perfectly OK. My only concern is that I am not sure how you would get on if $u$ also needs a chain rule. Say if you had to differentiate:

$y=\ln(\sin(x^2))$ .

Well anyway, using your method:

$y=(2x+1)^3=u^3$ where $u=2x+1$ . Using now your formula:

$\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}=3u^2\cdot 2=6u^2=6(2x+1)^2$ .

How do I do it? I use:

$\displaystyle\frac{d}{dx}o(i(x))=o'(i(x))\cdot i(x)$ .

I do not necessarily label/name $o$ (outside function), and $i$ (inside function). Here $o'(i(x))$ is “the derivative of the outside function evaluated at the inside function”.

Here the outside function is $x^3$ and the inside function is $2x+1$ so I get:

$\displaystyle\frac{d}{dx}(2x+1)^3=3(2x+1)^2\cdot 2=6(2x+1)^2$ .

Both my method and your method are valid.

O.K., back to:

$y=(2x+1)^3(x+2)$ . Now we identify, for a product rule:

$y=\underbrace{(2x+1)^3}_{u}\underbrace{(x+2)}_{v}$ and we use the product rule (but we have already found the derivative of $(2x+1)^3$ separately. So we get:

$\displaystyle\frac{dy}{dx}=(2x+1)^3\cdot 1+(x+2)\cdot 6(2x+1)^2=(2x+1)^3+6(x+2)(2x+1)^2$ ,

which looks different to the answer in the book but is equal to it.

I think you just went a little wrong by not taking your time.
________________________________________________

p.111, Q. 2 (a)
First of all your

$m=x+2x\cdot \ln x$ is correct.

This gives the slope of the tangent to the point at ANY point. But you are interested in the point $(e,e^2)$ , i.e. $x=e$ . So you need to evaluate this at $e$ :

$m=e+2e\cdot (\ln e)$ .

You can put this in the calculator, or note that $\ln x$ and $e^x$ are an inverse pair so that:

$\ln(e)=\ln(e^1)=1$ .

At any rate $m=3e$ .

Now you have

$\begin{aligned}y-y_1&=m(x-x_1)\\ \Rightarrow y-e^2&=3e\cdot (x-e)\\ \Rightarrow y&=3e\cdot x-3e^2+e^2\\ \Rightarrow y&=3e\cdot x-2e^2\end{aligned}$
________________________________________________________

p. 123, Q. 3
You have done all the hard work to get to

$\displaystyle\frac{3t^2-12}{-2t}=0$ .

There is two approaches here.

1. Note that a fraction $\displaystyle\frac{a}{b}=0$ if and only if $a=0$ (and $b\neq 0$ .

Proof: Suppose $b\neq 0$ . Suppose $\frac{a}{b}=0$ . Multiply both sides by $b$ :

$\begin{aligned}b\cdot \frac{a}{b}&=b\cdot 0\\\Rightarrow\frac{ab}{b}&=0\\\Rightarrow a\frac{b}{b}&=0\\\Rightarrow a\cdot 1&=0\\\Rightarrow a&=0\qquad \bullet\end{aligned}$

That is doing to the dog on the proof a little.

2. Or argue like this directly. Look at

$\displaystyle\frac{3t^2-12}{-2t}=0$

What is making this difficult? The divides by $-2t$ . What is the inverse of dividing by $-2t$ ? Multiplying by $-2t$ . Do this to both sides to get:

$3t^2-12=0$

(because $-2t\cdot 0=0$ ).

Now solve $3t^2-12=0$ . I will let you take it from here.
_____________________________________________________
P.124, Q. 7

I can’t see where you have tried this so not sure where you are having difficulty.
______________________________________________________
P. 124, Q. 8

You don’t have to calculate the second derivative. It won’t be examinable.

I can’t see how you would have a problem with the first derivatives.
_____________________________________________________

Please get back to me if you have any questions.

Regards,
J.P.

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MATH6040: Spring 2020, Week 7

Tuesday 17 March 2020, Week 8

Week 7

Homework/Study

Week 8

Week 9

Test 2

Study

Academic Learning Centre

Student Resources

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J.P. McCarthy Maths

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MATH6040: Spring 2020, Week 7

Tuesday 17 March 2020, Week 8

Week 7

Homework/Study

Week 8

Week 9

Test 2

Study

Academic Learning Centre

Student Resources

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J.P. McCarthy Maths

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