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MATH6037: Properties of the Laplace Transform — Selected Exercise Solutions

March 31, 2011 in MATH6037 | 2 comments

Question 1

Find the Laplace transform of the following functions.

(i)

6\cos 4t+t^3

Solution

Using linearity:

\mathcal{L}\{6\cos 4t+t^3\}=6\mathcal{L}\{\cos 4t\}+\mathcal{L}\{t^3\}.

Now using the tables:

=6\frac{s}{s^2+4^2}+\frac{3!}{s^4}=\frac{6s}{s^2+16}+\frac{6}{s^4}.

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MATH6037: The Laplace Transform of Basic Functions — Selected Exercise Solutions

March 31, 2011 in MATH6037 | Leave a comment

Question 3

Find, from first principles, \mathcal{L}\{t^2\}.

Solution

Using the definition:

\mathcal{L}\{t^2\}=\int_0^\infty t^2e^{-st}\,dt

To ease the steps, first we will evaluate the indefinite integral:

I= \int t^2e^{-st}\,dt.

This integral needs to be integrated by parts (this hint would be proffered in an exam situation). Using the LIATE rule, choose u=t^2 and dv=e^{-st}\,dt. Now

\frac{du}{dt}=2t\Rightarrow du=2t\,dt, and

v=\int dv=\int e^{-st}\,dt=\frac{e^{-st}}{-s}=-\frac{1}{s}e^{-st}.

In this context -1/s is a constant and it will be handy to have it out the front so it can be easily taken out of an integral (\int kf(x)=k\int f(x).) Hence using the integration by parts formula:

I=t^2\left(-\frac{1}{s}e^{-st}\right)-\int \left(-\frac{1}{s}e^{-st}2t\,dt\right)=-\frac{1}{s}t^2e^{-st}+\frac{2}{s}\underbrace{\int te^{-st}}_{=:J}.

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MATH6037: Definition of the Laplace Transform Selected Exercise Solutions

March 31, 2011 in MATH6037 | Leave a comment

Question 1

Determine whether the integrals are convergent or divergent. Evaluate those that are convergent.

This exercise is more to get us used to evaluating infinite integrals — none of these bar (iii) are necessarily related to Laplace transforms.

(i)

\int_1^\infty \frac{1}{(3x+1)^2}\,dx

Solution

We first try and evaluate

\int \frac{1}{(3x+1)^2}\,dx.

Try the substitution u=3x+1:

\frac{du}{dx}=3\Rightarrow dx =\frac{du}{3}.

I=\int\frac{1}{u^2}\cdot\frac{du}{3}=\frac{1}{3}\int u^{-2}\,du

=\frac{1}{3}\frac{u^{-1}}{-1}=-\frac{1}{3u}=-\frac{1}{3(3x+1)}.

Now evaluating I between the limits x=1 and x=R:

\left(-\frac{1}{3(3R+1)}\right)-\left(-\frac{1}{3(3+1)}\right)=\frac{1}{12}-\frac{1}{9R+3}.

Now taking the limit as R\rightarrow\infty:

I=\lim_{R\rightarrow\infty}\left(\frac{1}{12}-\underbrace{\frac{1}{9R+3}}_{\rightarrow 0}\right)=\frac{1}{12}.

The integral is convergent with value 1/12.

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MATH6037: Maple Test

March 25, 2011 in MATH6037 | Leave a comment

Due to Miguel being away next Wednesday I have decided to postpone the Maple test until the week after.

MATH6037 Test Results

March 24, 2011 in MATH6037 | Leave a comment

The results are down the bottom. You are identified by the last three digits of your student number. If there is a blank it means that you did not write down a student number. If you want email me on jippo@campus.ie and I can confirm your score — otherwise you’ll be waiting until Wednesday.

The scores are itemized as you can see. At the bottom there are some average scores. Finally the last column displays your Continuous Assessment mark for the Test (out of 15).

If you would like to see your paper or have it discussed please email me at jippo@campus.ie

 

St No Q1 Q2 Q3 Q4 Q5 Q6 Q7 Total % CA
219 10 10 10 10 10 10 10 70 100 15
4 7 10 10 10 9 8 58 83 13
993 8 9 10 6 8 9 1 51 73 11
102 9 4 8 10 6 10 3 50 71 11
960 6 7 10 10 2 10 4 49 70 11
787 4 4 10 5 1 7 10 41 59 9
4 6 10 10 6 2 1 39 56 9
675 10 4 4 8 3 2 3 34 49 8
6 0 9 4 2 2 4 27 39 6
5 0 1 7 5 1 7 26 37 6
271 3 0 4 8 4 2 2 23 33 5
4 0 2 5 4 2 2 19 27 5
Ave 6.08 4.25 7.33 7.75 5.08 5.50 4.58 40.58 57.98 9.08

 

MATH6037: Approximate Integration Selected Exercise Solutions

March 14, 2011 in MATH6037 | Leave a comment

Remarks in italics are by me for extra explanation. These comments would not be necessary for full marks in an exam situation. Exercises taken from p.87 of these notes.

Question 1

Estimate \int_0^1\cos(x^2)\,dx using  (a) the Trapezoidal Rule and (b) the Midpoint Rule, each with n=4.

Solution

Part (a)

For the Trapezoidal Rule, we have \Delta x=0.25 and we have the points x_0=0, x_1=0.25, x_2=0.5, x_3=0.75, x_4=1. Hence using the formula:

\int_0^1\cos(x^2)\,dx\approx\frac{0.25}{2}[\cos(0)+2[\cos(0.25^2)+\cos(0.5^2)+\cos(0.75^2)]

+\cos(1)]\approx 0.895795.

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MATH6037: Root Approximation Exercise Selected Solutions

March 14, 2011 in MATH6037 | Leave a comment

Remarks in italics are by me for extra explanation. These comments would not be necessary for full marks in an exam situation. Exercises taken from p.75 of these notes.

Question 1

If f(x)=x^3-x^2+x, show that there is a number c such that f(c)=10.

Solution

Solving the equation f(x)=10 is equivalent to finding a root of the continuous function g(x)=f(x)-10=x^3-x^2+x-10. Note that this function is continuous hence we can use the Intermediate Value Theorem to find an interval with a root.

g(0)=-10<0,

g(1)=(1)^3-(1)^2+(1)-10=-9<0,

g(2)=(2)^3-(2)^2+(2)-10=-4<0,

g(3)=(3)^3-(3)^2+(3)-10=11>0.

Hence g changes sign between 2 and 3. Hence g has a root in (2,3), say at c. Hence g(c)=0\Leftrightarrow f(c)=10 \bullet

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MATH6037: Applications to Error Analysis Exercise Solutions

March 14, 2011 in MATH6037 | 2 comments

Remarks in italics are by me for extra explanation. These comments would not be necessary for full marks in an exam situation. Exercises taken from p.63 of these notes.

Question 1

Use differentials to estimate the amount of tin in a closed tin closed tin can with diameter 8 cm and height 12 cm if the can is 0.04 cm thick.

Solution

Assuming that the measurements of 8 cm and 12 cm are taken from the outside of the can, then we could estimate the change in volume of a cylinder if the radius were increased by 0.04 cm to 4 cm and the height increased by 0.08 cm to 12 cm (convince yourself of this with a picture.). Now the tin in the can comprises the difference between a (r,h)=(3.96,11.92) cylinder and a (r,h)=(3,12) cylinder.

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MATH6037: Sample Test

March 8, 2011 in MATH6037 | Leave a comment

The MATH6037 test will be held at 7 p.m. Wednesday 16/03/11. The test is worth 15% of your final mark. The test will be 60 minutes long and you must answer all questions. All questions carry equal marks and I will attach a set of tables. You will get a copy of these tables tomorrow night.

Please find a sample here.

 

MATH6037: Partial Derivatives Exercise Solutions

February 19, 2011 in MATH6037 | 1 comment

Remarks in italics are by me for extra explanation. These comments would not be necessary for full marks in an exam situation. Exercises taken from p.58 of these notes.

Question 1

(i)

\frac{\partial f}{\partial x}=3x^2-4y^2(1)+0.

————————————————————————————————————————————————–

\frac{\partial f}{\partial y}=-4x(2y)+4y^3=4y^4-8xy.

(ii)

\frac{\partial f}{\partial x}=e^y(2x)+0=2xe^y.

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