MATH7021: Spring 2018, Week 10

April 19, 2018 in MATH7021

Assignment 2

Assignment 2 has a hand-in date of 17:00 23 April: the Monday of Week 11. Assignment 2 is in the manual, P. 149..

Week 10

The Monday lecture was another tutorial on the P.116 exercises. In the Wednesday lectures we worked on systems of differential equations. In the Thursday lecture we worked on systems of differential equations exercises.

In the Wednesday tutorial we continued with the full Laplace Transform questions.

Week 11

In the Monday and Wednesday lectures we will make a start on the final chapter by looking at double integrals.

In the Wednesday tutorial we will work on the p. 163 and p. 182 exercises. This work will continue on Thursday.

Week 12

On Monday and Wednesday we will finish looking at double integrals and then triple integrals.

In the Wednesday tutorial we will work on the p.182, p.192, and p.186 exercises. This work will continue on Thursday.

Week 13

We will go through last year’s exam on the board and then I will answer your questions if there are any. If there are none I will help one-to-one.

Study

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

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 19, 2018 at 12:15 pm

Student

Hey J.P.,

Still stuck on the same question from the other day. It’s the only question I have left and I’m so confused. Question 3.5.1 c ii.

Is there an example of this question I can look at?

Thanks.

April 19, 2018 at 12:27 pm

J.P.

It is first year engineering material rather than third year material.

You have to show that $\displaystyle \tilde{x}(t)=-7\cdot \cos\left(\frac{\sqrt{5}}{2}\cdot t\right)$ satisfies the differential equation.

So you have to calculate

$\displaystyle m\cdot \frac{d^2\tilde{x}}{dt^2} + k\cdot \tilde{x}(t)$ (*)

and see do you get zero.

To calculate the second derivative you differentiate twice… so you start by differentiating once.

The thing in green is not finished, it hasn’t used the Chain Rule.

The chain rule says if you have a function of a function, $y=u(v(x))$ (where $u$ ‘outside’ and $v$ ‘inside’), like here $-7\cos(t)$ is the outside and $\displaystyle \frac{\sqrt{5}}{2}\cdot t$ is the inside, that the derivative is given by:

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

That is you differentiate the outside, evaluate at the inside, and then multiply by the derivative of what is inside.

With $\tilde{x}(t)$ , fixing the constant $-7$

$\displaystyle \frac{d\tilde{x}}{dt}=-7\cdot \left(-\sin\left(\frac{\sqrt{5}}{2}\cdot t\right)\cdot \frac{d}{dt}\left(\frac{\sqrt{5}}{2}\cdot t\right)\right)$ .

That is, after fixing the constant $-7$ , the derivative of $\cos(x)$ , which is $-\sin(x)$ , evaluated at the inside $\displaystyle \frac{\sqrt{5}}{2}\cdot t$ (so this instead of $x$ ), multiplied by the derivative of what is inside.

The last day you were tired and stuck on

$\displaystyle \frac{d}{dt}\left(\frac{\sqrt{5}}{2}\cdot t\right)$ .

This is just like differentiating say $y=2x$ … the derivative of a constant times $x$ is just the constant; for a constant $a$ :

$\displaystyle \frac{d}{dx}(a\cdot x)=a$ .

Therefore

$\displaystyle \frac{d}{dt}\left(\frac{\sqrt{5}}{2}\cdot t\right)=\frac{\sqrt{5}}{2}$ ,

$\displaystyle \frac{d\tilde{x}}{dt}=-7\cdot \left(-\sin\left(\frac{\sqrt{5}}{2}\cdot t\right)\cdot \frac{\sqrt{5}}{2}\right)=\frac{7\sqrt{5}}{2}\sin\left(\frac{\sqrt{5}}{2}\cdot t\right)$ .

Now you need to differentiate a second time to find the second derivative… then see does (*) give you zero.

Regards,
J.P.

April 19, 2018 at 3:29 pm

Student

One last thing as I’m revising, I believe I missed a step.

April 19, 2018 at 3:37 pm

J.P.

Where you have “redo but change missed a step” you do have an issue. So you have

$\displaystyle Y(s)[s^2+16s+64] =\frac{8}{s}+2s+32$ .

This is not equal to

$\displaystyle \frac{8+16s+32}{s(s^2+16s+32)}$ .

You have four options.

1. Divide both sides by $(s^2+16s+64)$ to get

$\displaystyle Y(s)=\frac{\frac{8}{s}+2s+32}{s^2+16s+64}$ .

Now the issue is your fraction in your fraction. This is not good. To get rid of it you need to multiply above and below by $s$ :

$\displaystyle Y(s)=\frac{\frac{8}{s}+2s+32}{s^2+16s+64}\cdot \underbrace{\frac{s}{s}}_{=1}=?$ .

2. Divide each of the three terms by $(s^2+16s+64)$ separately:

$\displaystyle Y(s)=\frac{8}{s(s^2+16s+64)}+\frac{2s}{s^2+16s+64}+\frac{32}{s^2+16s+64}$ .

You can send these all back separately or…

3. Write this as a single fraction:

$\displaystyle Y(s)=\frac{8}{s(s^2+16s+64)}+\frac{2s}{s^2+16s+64}\cdot \frac{s}{s}+\frac{32}{s^2+16s+64}\cdot \frac{s}{s}$
$\displaystyle \Rightarrow Y(s)=?$

4. The other thing you might have done was write the right hand side as a single fraction here:

$\displaystyle Y(s)[s^2+16s+64] =\frac{8}{s}+2s+32$
$\displaystyle =\frac{8}{s}+\frac{2s^2}{s}+\frac{32}{s}.=\frac{8+2s^2+32s}{s}$ .
$\displaystyle \Rightarrow Y(s)=?$

Regards,
J.P.

April 19, 2018 at 4:50 pm

Student

Clearly I’m just not doing this right… I still can’t get the right answer.

April 19, 2018 at 4:56 pm

J.P.

You have, correctly,

$\displaystyle Y(s)=\frac{2s^2+32s+8}{s(s+8)^2}$ .

You have the correct partial fraction decomposition:

$\displaystyle \frac{2s^2+32s+8}{s(s+8)^2}=\frac{A}{s}+\frac{B}{s+8}+\frac{C}{(s+8)^2}$ .

It is true that

$\displaystyle \frac{A}{s}+\frac{B}{s+8}+\frac{C}{(s+8)^2}=\frac{A(s+8)^3+Bs(s+8)^2+Cs(s+8)}{s(s+8)(s+8)^2}$

Note this is equal to

$\displaystyle \frac{A(s+8)^3+Bs(s+8)^2+Cs(s+8)}{s(s+8)^3}$ .

and you are correct to force

$\displaystyle \frac{2s^2+32s+8}{s(s+8)^2}\overset{!}{=}\frac{A(s+8)^3+Bs(s+8)^2+Cs(s+8)}{s(s+8)^3}$ .

You conclude that the tops are the same? Why do you say that?

The bottoms aren’t the same… there is no good reason why the tops are the same (top means numerator and bottom denominator).

We are trying to get

$\displaystyle \frac{u(s)}{q(s)}=\frac{v(s)}{q(s)}$ so we can conclude the tops are equal:

$u(s)=v(s)$ .

But we can only conclude this if the bottoms are equal too.

You need to write

$\displaystyle \frac{A}{s}+\frac{B}{s+8}+\frac{C}{(s+8)^2}$

as a single fraction with the same bottom as

$\displaystyle \frac{2s^2+32s+8}{s(s+8)^2}$

Then you can force them equal to each other and conclude that

$2s^2+32s+8=v(s,A,B,C)$ and find $A,\, B,\,C$ .

Regards,
J.P.

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MATH7021: Spring 2018, Week 10

Assignment 2

Week 10

Week 11

Week 12

Week 13

Study

Student Resources

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J.P. McCarthy Maths

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MATH7021: Spring 2018, Week 10

Assignment 2

Week 10

Week 11

Week 12

Week 13

Study

Student Resources

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J.P. McCarthy Maths

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