MATH7021: Week 13

December 10, 2012 in MATH7021

Exam Layout

Answer four questions: each worth the same marks.

1. Multiple Topics Question

(a) Interpolation (b) Linear Algebra (c) Laplace Methods

2. Interpolation

Finite Difference Methods, Least Squares, Lagrangian Interpolation

3. Linear Algebra

Gaussian Elimination with and without partial pivoting, Cramer’s Rule, Jacobi’s Method, Gauss-Siedel Method

4. Laplace Methods

Three differential equations

5. Multiple Integration

line integrals, double integrals with applications including second moment of area, triple integrals.

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December 14, 2012 at 11:43 am

Student 28

Just saw your email with the triple integral example (pg 142) and on the very last integral I was wondering should $z$ not be from 0 to 3 rather than 0 to 2?

Sorry for annoying you again but I wanted to check the answers to Q3(a) on the sample paper you gave us (2009/10). I’ve tried it out and the answers I have are (i) 264 (ii) 14 and (iii) 196. If you have answers for this question could you let me know if I’ve done it right.

December 14, 2012 at 12:07 pm

J.P. McCarthy

You are absolutely correct that $z:0\rightarrow 3$ — corrected.

Firstly what we have is

$\displaystyle \int_{\mathcal{C}}(30y^2\mathbf{i}+12x\mathbf{j})\cdot d\mathbf{r}$ .

We will not have an integral presented to us in this form. Rather the $d\mathbf{r}=dx\,\mathbf{i}+dy\,\mathbf{j}$ so that the integral is

$\displaystyle \int_{\mathcal{C}}(30y^2\mathbf{i}+12x\mathbf{j})\cdot (dx\,\mathbf{i}+dy\,\mathbf{j})=\int_{\mathcal{C}}30y^2\,dx+12x\,dy$

The first line integral, without a picture. The line segment from $(0,1)\rightarrow (1,3)$ is a line which cuts the $y$ -axis at $1$ and has a slope of 2. Hence the curve may be parameterised by $x=t$ and $y=2x+1=2t+1$ with $t:0\rightarrow 1$ . Also $dx=dt$ and $dy=2\,dy$ . Hence we have

$\displaystyle \int_{\mathcal{C}}30y^2\,dx+12x\,dy=\int_{0}^130(2t+1)^2\,dt+12t(2dt)$

$\displaystyle=\int_0^1(30(4t^2+4t+1)+24t)dt$

$\displaystyle=\int_0^1(120t^2+120t+30+24t)dt=\int_0^1(120t^2+144t+30)\,dt$

$\displaystyle=\left[120\frac{t^3}{3}+144\frac{t^2}{2}+30t\right]^1_0=\left[40t^3+72t^2+30t\right]^1_0$

$\displaystyle=(40(1^3)+72(1)+30(1))-(40(0)^3+72(0)^2+30(0))=142$ .

Your answer to the second integral is correct.

For the third integral, without a picture we have $x=y^2+1$ . Hence we can use the parameterisation $y=t$ and $x=y^2+1=t^2+1$ where $t$ — morryah $y$ — runs from $t:0\rightarrow 2$ . We have $dx=2t\,dt$ and $dy=dt$. Hence our integral is

$\displaystyle \int_{\mathcal{C}}30y^2\,dx+12x\,dy=\int_{1}^230(t)^2\,(2t\,dt)+12(t^2+1)(dt)$

$\displaystyle=\int_1^2(60t^3+12t^2+12)dt$

$\displaystyle=\left[60\frac{t^4}{4}+12\frac{t^3}{3}+12t\right]^2_1=\left[15t^4+4t^3+12t\right]^2_1$

$\displaystyle=(15(2)^4+4(2)^3+12(2))-(15(1)^4+4(1)^3+12(1))=(240+32+24)-(15+4+12)=265$ .

Regards and Happy Christmas,
J.P.

December 14, 2012 at 12:12 pm

Student 29

Can you take a look over my answer to

$\displaystyle \frac{d^2y}{dt^2}+4\frac{dy}{dt}+20y(t)=80$ , $y(0)=y'(0)=0$ .

December 14, 2012 at 12:24 pm

J.P. McCarthy

The first problem in your solution is that you completely forgot about $20y=20y(t)$ when you took the Laplace Transform.

After that things are good up until the point where correctly you infer that $A+B\overset{!}{=}0$ and $4A+C\overset{!}{=}0$ … which is O.K. except the same equation

$\displaystyle\frac{80}{s(s^2+4)}=\frac{(A+B)s^2+(4A+C)s}{s(s^2+4)}$

also implies that $0=80$ (i.e. comparing the constant coefficients)… which is a bit of a problem.

Here I present a worked solution with (overly?) extensive explanations:

The first thing to do is the take the Laplace Transform of the differential equation. This transforms a differential equation with solution $y=f(t)$ to an algebraic solution with solution $Y=F(s)$ — and algebraic equations are easier to solve.

$\displaystyle \frac{d^2y}{dt^2}+4\frac{dy}{dt}+20y(t)=80$

Note that the Laplace Transform is linear, which means that we can transform term-by-term and pull out constants:

$\displaystyle \Rightarrow \mathcal{L}\left\{\frac{d^2y}{dt^2}\right\}+4\mathcal{L}\left\{\frac{dy}{dt}\right\}+20\mathcal{L}\left\{y(t)\right\}=\mathcal{L}\left\{80\right\}$

Now we have expressions for the Laplace Transform of the first and second derivatives in the tables. They are given in terms of the Laplace Transform of the solution, $Y(s)=\mathcal{L}\{y(t)\}$ and the initial conditions (at time, $t=0$ ) of $y(t)$ and $y'(t)$ . The Laplace Transform of a constant is also in the tables: $\displaystyle A\overset{\mathcal{L}}{\rightsquigarrow} \frac{A}{s}$ . Hence we have

$\displaystyle \ s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0))+20Y(s)=\frac{80}{s}$ .

Now we apply the initial/ boundary condition that $y(0)=y'(0)=0$ :

$\displaystyle s^2Y(s)+4sY(s)+20Y(s)=\frac{80}{s}$ .

Now we solve for $Y(s)$ :

$\displaystyle Y(s)(s^2+4s+20)=\frac{80}{s}$

$\displaystyle \Rightarrow Y(s)=\frac{80}{s(s^2+4s+20)}$ .

We want to apply the inverse Laplace Transform $\mathcal{L}^{-1}$ to go from $Y(s)\overset{\mathcal{L}^{-1}}{\rightsquigarrow} y(t)$ . Now the idea is that nothing like this is in the tables. We need to write this as a sum of simpler terms… partial fractions. The first thing to do is factorise the denominator $s(s^2+4s+20)$ inasmuch as possible. When you try and factor the quadratic it doesn’t seem to factor so we check $b^2-4ac=(4)^2-4(1)(20)<0$ . This means that the roots of the quadratic are complex. However the Factor Theorem says that

$k$ a root a polynomial $p(x)$ if and only if $(x-k)$ is a factor.

We have complex roots and hence cannot factor without complex numbers. Such a quadratic is not factorisable (such a quadratic is called irreducible over the real numbers).

O.K. we cannot factorise the denominator any further so what we must do now is associate to each term in the facorisation of the denominator a term in the partial fraction expansion. $s$ is a linear factor so begets a term of the form

$\displaystyle s\rightsquigarrow \frac{A}{s}$ for some constant $A$ .

$s^2+4s+20$ is a quadratic factor so begets a term of the form

$\displaystyle s^2+4s+20\rightsquigarrow \frac{Bs+C}{s^2+4s+20}$ for some constants $B$ and $C$ .

Note that we may not use the cover-up method as we do not have a product of distinctive linear terms. Hence what we will do is write

$\displaystyle Y(s)=\frac{80}{s(s^2+4s+20)}=\frac{A}{s}+\frac{Bs+C}{s^2+4s+20}$ ,

write the partial fraction expansion as a single fraction again and compare.

$\displaystyle \frac{A}{s}+\frac{Bs+C}{s^2+4s+20}=\frac{A}{s}\cdot\frac{s^2+4s+20}{s^2+4s+20}+\frac{Bs+C}{s^2+4s+20}\cdot\frac{s}{s}$

$\displaystyle=\frac{A(s^2+4s+20)+(Bs+C)s}{s(s^2+4s+20)}$ .

Hence we want, for all values of $s$ ,

$\displaystyle \frac{80}{s(s^2+4s+20)}=\frac{A(s^2+4s+20)+(Bs+C)s}{s(s^2+4s+20)}$ .

The denominators of these are the same so we need the numerators to be equal:

$80=A(s^2+4s+20)+(Bs+C)(s)$ .

Note that these must be equal for all values of $s$ . Hence we could put $s=0,1,-1$ and generate three equations in the three unknowns $A,\,B,\,C$ and hence find their values (note that the partial fraction expansion is unique — e.g. as opposed to fractions which can be expanded in different ways $\frac{1}{20}=\frac{1}{4}-\frac{1}{5}=\frac{5}{4}-\frac{6}{5}$ for example).

Alternatively we can do what you do which was multiply out and compare coefficients (this works because if $i(s)=j(s)$ for all $s$ then $i'(s)=j'(s)$ ; e.g. $1\overset{!}{=}As+B\Rightarrow 0=A$ after differentiating both sides with respect to $s$ ).

$80=As^2+4As+20A+Bs^2+Cs$

$\Rightarrow 80=(A+B)s^2+(4A+C)s+20A$

Hence we have $A+B\overset{!}{=}0$ , $4A+C\overset{!}{=}0$ and — what we were missing — $20A=80$ . Thence we have $A=4$ , $B=-4$ and $C=-16$ . Thus we have

$\displaystyle Y(s)=\frac{80}{s(s^2+4s+20)}=\frac{4}{s}+\frac{-4s-16}{s^2+4s+20}$ .

Now we know how to deal with $\displaystyle \frac{4}{s}\overset{\mathcal{L}^{-1}}{\rightsquigarrow}4$ . What about $\displaystyle \frac{-4s-16}{s^2+4s+20}$ . All we can handle is, in terms of quadratics,

$\displaystyle \mathcal{L}^{-1}\left\{\frac{\omega}{s^2+\omega^2}\right\}=\sin(\omega t)$ , and

$\displaystyle \mathcal{L}^{-1}\left\{\frac{s}{s^2+\omega^2}\right\}=\cos(\omega t)$ .

Hence all we can really say is that $s^2+4s+20$ ‘kind of looks like’ $(s-q)^2+\omega^2$ . Could we write $s^2+4s+20$ in this form? Let’s try

$s^2+4s+20=(s-q)^2+\omega^2$

$\Rightarrow s^2+4s+20=s^2-2qs+q^2+\omega^2$

Now this will be equal, again for all values of $s$ (cf. partial fraction expansion), if the coefficients are equal on both sides; i.e.

$4=-2q$ and $20=q^2+\omega^2$ .

The first equation yields $q=-2$ and hence the second reads $20=4+\omega^2\Rightarrow \omega^2=16\Rightarrow \omega=\pm4$ . You can take $\omega=+4$ . Hence we have

$\displaystyle Y(s)=\frac{80}{s(s^2+4s+20)}=\frac{4}{s}+\frac{-4s-16}{(s+2)^2+4^2}$ .

Now if we could write the second term here in the form

$\displaystyle \frac{-4s-16}{(s+2)^2+4^2}=m\cdot \frac{4}{(s+2)^2+4^2}+n\cdot\frac{s+2}{(s+2)^2+4^2}$ ,

for some constants $m$ and $n$ then we are well set. The first term here is a $s\rightarrow s+2$ shifted sine, while the second is a $s\rightarrow s+2$ shifted cosine. We have in the tables that

$\mathcal{L}\left\{e^{at}\cdot f(t)\right\}=F(s-a)$ ,

where $F(s)=\mathcal{L}\{f(t)\}$ . On the other hand we have that a $s\rightarrow s-a$ shifted function is multiplied by $e^{at}$ by the inverse Laplace Transform:

$\mathcal{L}^{-1}\left\{F(s-a)\right\}=e^{at}\cdot f(t)$ .

To actually write

$\displaystyle\frac{-4s-16}{(s+2)^2+4^2}=m\cdot \frac{4}{(s+2)^2+4^2}+n\cdot\frac{s+2}{(s+2)^2+4^2}$

is not that easy and I don’t have a good way of doing it apart from playing around with the thing:

$\displaystyle\frac{-4s-16}{(s+2)^2+4^2}=-4\cdot\frac{s+4}{(s+2)^2+4^2}$

$\displaystyle =-4\cdot\frac{(s+2)+2}{(s+2)^2+4^2}=-4\cdot \frac{s+2}{(s+2)^2+4^2}-4\cdot \frac{2}{(s+2)^2+4^2}$

$\displaystyle =-4\cdot \frac{s+2}{(s+2)^2+4^2}-2\cdot \frac{4}{(s+2)^2+4^2}$ ,

which is exactly the form we need. The first term is a shifted cosine while the second is a shifted sine. Now we send the whole thing back (note that the inverse Laplace Transform is also linear)

$\displaystyle y(t)=\mathcal{L}^{-1}\left\{Y(s)\right\}$

$\displaystyle=\mathcal{L}^{-1}\left\{\frac{4}{s}\right\}-4\mathcal{L}^{-1}\left\{\frac{s+2}{(s+2)^2+4^2}\right\}-2\mathcal{L}^{-1}\left\{\frac{4}{(s+2)^2+4^2}\right\}$

$\displaystyle \Rightarrow y(t) =4-4e^{-2t}\cos(4t)-2e^{-2t}\sin(4t)$ .

Sure enough $y(0)=y'(0)=0$ (a good idea has a check to see is the initial condition satisfied) and also this function satisfies the differential equation.

Regards and Happy Christmas,
J.P.

December 14, 2012 at 1:42 pm

Student 30

Can you give a quick run through about the rules that are applied in the formula which I have outlined in green here. What are $m$ & $n$ ?

$\displaystyle \frac{Cs+D}{(s-q)^2+\omega^2}=m\cdot \frac{\omega}{(s-q)^2+\omega^2}+n\cdot\frac{s-q}{(s-q)^2+\omega^2}$ .

December 14, 2012 at 1:45 pm

J.P. McCarthy

Suppose you get down to a point where your Laplace Transform looks like

$\displaystyle Y(s)=\frac{Cs+D}{s^2+bs+c}$

where $C,\,D,\,b,\,c$ are constants but $s^2+bs+c$ cannot be factorised any more because the roots are complex: $b^2-4(1)c<0$ .

All we can handle is, in terms of quadratics,

$\displaystyle \mathcal{L}^{-1}\left\{\frac{\omega}{s^2+\omega^2}\right\}=\sin(\omega t)$ , and

$\displaystyle \mathcal{L}^{-1}\left\{\frac{s}{s^2+\omega^2}\right\}=\cos(\omega t)$ .

However all we can really say is that $s^2+bs+c$ can be written as $(s-q)^2+\omega^2$ — once we complete the square. However if we could write

$\displaystyle \frac{Cs+D}{(s-q)^2+\omega^2}=m\cdot \frac{\omega}{(s-q)^2+\omega^2}+n\cdot\frac{s-q}{(s-q)^2+\omega^2}$ ,

for some constants $m$ and $n$ then we are well set. The first term here is a $s\rightarrow s-q$ shifted sine, while the second is a $s\rightarrow s-q$ shifted cosine. We have in the tables that

$\mathcal{L}\left\{e^{at}\cdot f(t)\right\}=F(s-a)$ ,

where $F(s)=\mathcal{L}\{f(t)\}$ . On the other hand we have that a $s\rightarrow s-a$ shifted function is multiplied by $e^{at}$ by the inverse Laplace Transform:

$\mathcal{L}^{-1}\left\{F(s-a)\right\}=e^{at}\cdot f(t)$ .

To actually write

$\displaystyle \frac{Cs+D}{(s-q)^2+\omega^2}=m\cdot \frac{\omega}{(s-q)^2+\omega^2}+n\cdot\frac{s-q}{(s-q)^2+\omega^2}$ .

is not that easy and I don’t have a good way of doing it apart from playing around with the thing. The first term is a shifted cosine while the second is a shifted sine. Now we send the whole thing back (note that the inverse Laplace Transform is linear which means you can transform term-by-term and pull out constants):

$\displaystyle y(t)=\mathcal{L}^{-1}\left\{Y(s)\right\}$

$\displaystyle=m\cdot\mathcal{L}^{-1}\left\{\frac{\omega}{(s-q)^2+\omega^2}\right\}+n\cdot\mathcal{L}^{-1}\left\{\frac{s-q}{(s-q)^2+\omega^2}\right\}$

$\displaystyle \Rightarrow y(t) =me^{qt}\cos(\omega t)+ne^{qt}\sin(\omega t)$ .

There is an example of a question like this in the above comment.

Regards and Happy Christmas,
J.P.

December 14, 2012 at 3:02 pm

Nick

Ok thanks JP! Happy Christmas to you too! Thanks for all your help.

Nick

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MATH7021: Week 13

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MATH7021: Week 13

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