MATH6040: Spring 2019, Week 11

Test 2

Test 2, worth 15% and based on Chapter 3, will now take place Week 12, 30 April. There is a sample test [to give an indication of length and layout only] in the notes (marking scheme) and the test will be based on Chapter 3 only.

More Q. 1s (on the test) can be found on p.112; more Q. 2s on p. 117; more Q. 3s on p.125 and p.172, Q.1; more Q. 4s on p.136, and more Q. 5s on p. 143.

Chapter 3 Summary p. 144.

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

Homework

Once you are prepared for Test 2 you can start looking at Chapter 4:

Revision of Integration, p.161.
p.167, Q. 1-5
p.182

Week 11

We had some tutorial time from 18:00-19:00 for further differentiation (i.e. for Test 2, after Easter).

We completed our review of antidifferentiation before starting Chapter 4 proper.

We looked at Integration by Parts and centroids.

For those who could not make it here is some video and slides from what we did after the video died.

Week 12

We will have some tutorial time from 18:00-19:00 for further differentiation.

We will have Test 2 from 19:00-20:05.

Then we will look at completing the square, centres of gravity, and work.

Week 13

We will look the Winter 2018 paper at the back of your manual.

CIT Mathematics Exam Papers

These are not always found in your programme selection — most of the time you will have to look here.

Student Resources

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

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April 11, 2019 at 7:16 am

Student

Hi J.P.,

Stuck on p. 102, Q 1 (f)

I have attached my attempt. I’ve been checking out khan academy and it said I needed the change of base formula but I got confused from there.

Thanks.

April 11, 2019 at 7:23 am

J.P. McCarthy

The confusion here, largely my fault perhaps, is that when you see $\log x$ in the calculus context that it is actually the natural log. So that, in this context:

$\log x=\log_e x=\ln x$ .

This is the confusion.

So what you are differentiating is $y=\ln(x^2+1)$ . It isn’t $\ln x$ and so you have a chain rule where the ‘outside’ is $\ln v$ and the ‘inside’ is $x^2+1$ .

The chain rule says that

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

that is “differentiate the outside — evaluate that at the inside — and the multiply by the derivative of the inside’.

The derivative of the outside, $\ln v$ is in the tables and is $\frac{1}{v}$ . Evaluating this at the inside gives $\displaystyle \frac{1}{x^2+1}$ but we must multiply this by the derivative of the inside, the derivative of $x^2+1$ , which is $2x$ .

All in all we have:

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

You probably need to move beyond this revision and tackle the new material — it might be the case that you don’t have to calculate more difficult derivatives such as these to do the new material.

Regards,
J.P.

April 26, 2019 at 7:25 am

Just wondering if you could help with the differentiation of $y=xe^{2x}$ .

I understand to use the product rule to start, but then when I apply the chain rule to the $e^{2x}$ I get confused with the substitution after that. Could you help.

Regards and Thanks.

April 26, 2019 at 7:30 am

I am not altogether sure what you mean by “substitution after that”.

So, the chain rule says that

$\displaystyle \frac{d}{dx}(u(v))=u'(v)\cdot v'.$

In words, the derivative of a composition $u(v(x))$ is the derivative of the ‘outside’ – evaluated at the inside – times the derivative of the inside.

Here the outside is $e^v$ and the inside is $2x$ . The derivative of the outside is $e^v$ , evaluated at the inside is $e^{2x}$ , and the derivative of the inside is $2\cdot 1=2$ . Therefore we get

$\displaystyle \frac{d}{dx}(e^{2x})=e^{2x}\cdot 2=2e^{2x}$ .

As it happens this differentiation is in the tables:

$\displaystyle \frac{d}{dx}e^{ax}=a\cdot e^{ax}$ .

Altogether this gives, via the produce rule,

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

April 26, 2019 at 7:34 am

My question is about p. 112 Q.2.

When I differentiate $x(\theta)$ and $y(\theta)$ all is well. But when $\theta = 0$ shouldn’t the substitution everything equal to zero?
Because anything multiplied by zero is zero?

$x' = -\sin \theta + 2\cos 2 \theta$

So if I substitute 0 shouldn’t this all be 0?

$-\sin 0+2\cos2(0)=0$ ?

Regards.

April 26, 2019 at 7:37 am

No, because it isn’t sine multiplied by zero and cosine multiplied by zero — it is sine OF zero and cosine OF zero.

Sine and cosine are real-valued functions. They take in numbers as inputs and produce real number outputs.

When you input zero into sine the output is zero.

But when you input zero into cosine the output is one.

Therefore

$x'(0)=-\sin 0+2\cos 2(0)=-\sin 0+2\cos 0=-0+2\cdot 1=2$ .

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MATH6040: Spring 2019, Week 11

Test 2

Homework

Week 11

Week 12

Week 13

CIT Mathematics Exam Papers

Student Resources

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Categories

J.P. McCarthy Maths

6 comments

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MATH6040: Spring 2019, Week 11

Test 2

Homework

Week 11

Week 12

Week 13

CIT Mathematics Exam Papers

Student Resources

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