MATH6037: Week 1

February 5, 2015 in MATH6037

I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jpmccarthymaths@gmail.com and I will add you to the mailing list.

Note: We are supposed to be C212 rather than B212 as advertised by Conor.

Quiz 1 Question Bank

Please see here for your question bank for Quiz 1 (in Week 2). Here you can find the tables that will be allowed.

Quiz 1 runs from 19:00 to 19:15 sharp on Wednesday 11 February.

Continuous Assessment

The Continuous Assessment is broken into Weekly Quizzes (20%) and Maple (10%).

There will be ten weekly quizzes and your eight best results will count (so 2.5% per quiz from eight quizzes). You will receive an email (i.e. this one) on Thursday/Friday detailing the examinable exercises.

Maple consists of five labs and a Maple Test in the sixth lab. Satisfactory participation in labs gives you 1.5% and the Maple Test is worth 2.5%. More on this in the coming days.

Week 1

In Week 1 we explained the kind of thing that we would be looking at in this module. We did a quick review of integral calculus.

Week 2

In Week 2 we will look at $u$ -substitutions. Then we will study Integration by Parts: this is the start of the new material (i.e. not MATH6019 material).

If we have a Maple lab we will do some basic plotting, differentiation and integration.

Week 6

As briefly mentioned in class, I will be away in Week 6 (11 March). I am proposing that we have a class over the Easter Break. In particular, I am thinking 1 April. Please email me if this not possible for you giving your reasons.

Notes

I have given out 15 sets of the notes. I received the €11 from everyone who got a manual. Please bring €11 for next week as more manuals will be printed. If you know someone who is not registered and who was absent in Week 1 please get them to email me at jpmccarthymaths@gmail.com so that I can order them a manual.

A student was asking did I have MATH6019-type notes that revision could be done with. You could look at an old higher level maths book, look in the library for anything with “Calculus” in the title.

Study

Please feel free to ask me questions about the exercises via email or even better on this webpage. Anyone can give me exercises they have done and I will correct them.

Student Resources

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

14 comments

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February 9, 2015 at 11:52 am

Student

Hi J.P.,
Just a few questions on the quiz.

Q4 – I have worked out the answer just want to know why $0<t<20$ is quoted in the question as it seems to have no relevance to the question?

Q6 (b) – I have worked out the answer to be 2.083 and am just wondering why is there 2 answers to this question as shown as your answer ?
Q10 – am having trouble with dividing fractions , I will forward what I have worked out these to be if you could let me know where I'm going wrong?

Q13(b) – I have nothing! Are you looking for the area above the curve to the $y$ -axis and below to the $x$ -axis, I need more info please.

Q14 – (b) I have worked this to be the inverse of your answer. Again I might be going wrong with dividing fractions?

Q15 – I have nothing; am having trouble with the square root and the square on the function? Maybe you could give me part (a) and I could try the rest!

Thanks.

February 9, 2015 at 1:51 pm

J.P. McCarthy

Hi,

Q.4 – the formula is only ‘works’ for twenty seconds. For example, if you project an object upwards with an initial velocity of 10 m/s, then its height is given by

$h(t)=10t-4.9t^2.$

This is only valid for $0\leq t\leq \frac{1000}{49}$ because the object will hit the ground again after $\frac{1000}{49}$ s. Something similar happens in Q.4 — for $t>20$ , the height is negative — the tank is empty.

Q. 6 (b) Consider a trapezoid as follows: http://www.geometryexpressions.com/explore.php?p=04-Examples%2F02-Example_Book%2F02-Quadrilaterals&f=Example+034-Right_Trapezoid.htm

You can underestimate its area using $a\times h$ and you can overestimate the area using $b\times h$ . 2.083 is an overestimate of the area and 1.283 is the underestimate. Note that I said the answer is BETWEEN these two numbers.

Q. 10 Your answer to part (a) is fine as $\frac{\sqrt{3}}{3}=\frac{\sqrt{3}}{\sqrt{3}\cdot \sqrt{3}}=\frac{1}{\sqrt{3}}$ .

Your mistake in part (b) is in dealing with $\sqrt{x^3}$ . This is $\sqrt[3]{x^2}=(x^2)^{\frac{1}{3}}=x^{2\times\frac{1}{3}}=x^{\frac{2}{3}}$ .

Q.13 (b) The region in question is shown here: http://www5b.wolframalpha.com/Calculate/MSP/MSP11821h59e0h18f1ec03000010gdhbdif29eh0c8?MSPStoreType=image/gif&s=64&w=390.&h=300.&cdf=Animation

Q.14 (b) You are dealing with the fraction fine but $b=\frac{\pi}{2}$ not $\frac{2}{\pi}$ $\displaystyle\frac{1}{b-a}=\frac{1}{\frac{\pi}{2}-0}=\frac{1}{\frac{\pi}{2}}=\frac{2}{\pi}$ .

Q. 15 Do the square root LAST.

$(e^t)^2=e^{2\cdot t}=e^{2t}$ .

$\displaystyle\left(\frac{1}{x}\right)^2=\frac{1}{x}\frac{1}{x}=\frac{1}{x^2}=\cdots$

$(t^2)^2=t^{2\cdot 2}=t^4$

$(\sin r)^2=\sin^2r$ . Use the identity

$\sin^2A=\frac12\left(1-\cos 2A\right)$

Regards,
J.P.

February 9, 2015 at 1:55 pm

Student

Hi J.P.,

I’m a little lost on question 4. It says that $h = 4t-0.2t^2$ , where $t$ is greater or equal to zero and where $t$ is less than or equal to 20. Find the rate after 8 seconds…

If i plug in the 8 second value I get $19.2$ for $h$ …. I’m kind of lost after that….

Regards.

February 9, 2015 at 1:57 pm

J.P. McCarthy

Hi,

The key word is RATE.

19.2 m is just the height of the water after 8 s but you are interested in the rate at which the height is changing… 5 cm/s, 10 cm/s, etc?

To find the rate of change of a function we differentiate. So you are looking for the derivative $\frac{dh}{dt}$ evaluated at $t=8$ s.

Regards,
J.P.

February 9, 2015 at 3:44 pm

Student

Hi J.P.

Attached please find the homework assigned this week. I couldn’t solve some of the questions as you can see. would you be able to send me on the solutions so that i could see where i am going wrong.

Thanks a million.

February 9, 2015 at 4:04 pm

J.P. McCarthy

Student,

Q.3 for units that are products it is better to leave a space between the units. For example, $\text{m s}^{-1}$ rather than $\text{ms}^{-1}$ . The latter could be mistaken as “per millisecond” rather than “metres per second”.

Q. 4 There is a problem with your differentiation… namely the derivative of $t^2$ is $2t$ . In longer form:
$\displaystyle \frac{d}{dt}(4t-0.2t^2)=4\frac{d}{dt}t-0.2\frac{d}{dt}t^2=4(1)-0.2(2t)=4-0.4t$ . This differentiating of a sum by “differentiating term by term and pulling out of constants” is correct as differentiation is LINEAR.

Q. 5 You correctly got $6t^2+12t=8$ . This DOES imply that $6t(t+2)=8$ (which is slightly different to what you had) but this DOES NOT imply that $6t=8$ or $t+2=8$ . For example, consider $x=2$ and $y=4$ . Now $x\cdot y=8$ but neither of $x$ nor $y$ are equal to eight.

You are thinking of something called the zero divisors theorem… suppose I have two numbers $a$ and $b$ that multiply to give me zero:

$a\cdot b=0$ .

Now if they are both positive that $a\cdot b>0$ so not zero. If they are both negative we have the same situation. If they are different signs then we have $a\cdot b<0$ which ain't zero either. Therefore one of $a$ or $b$ MUST be zero:

$a\cdot b=0\Leftrightarrow a=0\text{ or }b=0$

So it is only when you have a product equal to zero can you conclude like you did; i.e. $6t(t+2)=0\Rightarrow 6t=0\text{ or }t+2=0$ .

The proper way to do this is to solve the quadratic equal to zero:

$6t^2+12t-8=0$ ,

divide both sides by two:

$3t^2+6t-4=0$

and attempt to factorise. In this case, there are no factors so we must use the $-b\pm\sqrt{b^2\cdots}$ formula.

Q.6 The graph of $y=1/x$ is not a straight line… see http://www.wolframalpha.com/input/?i=Plot%5B1%2Fx%2C%7Bx%2C0%2C6%7D%5D

Q.7 The graph of $y=25-x^2$ is not a straight line either… see http://www.wolframalpha.com/input/?i=Plot%5B25-x%5E2%2C%7Bx%2C-1%2C6%7D%5D

Q.8 (i) Your calculator was not in RADIAN mode.

Q. 8 (ii) I have a typo… it should be $(\tan t)'=\sec^2t$ .

Q.9 The anti-derivative is actually $e^2x$ so you should have

$\displaystyle \int_0^1e^2\,dx=\left[e^2x\right]_0^1=e^2(1)-e^2(0)=e^2\approx 7.389$ .

Q.10 Anti-differentiation is LINEAR. This means that

$\displaystyle \int (f(x)+k\cdot g(x))\,dx=\int f(x)\,dx+k\int g(x)\,dx$ for functions $f(x)$ and $g(x)$ and CONSTANT $k$ . In order words you can "attack a SUM of terms term-by-term" and pull out constants. So you should have

$\int \sqrt{3}\sqrt{t}\,dt=\sqrt{3}\int t^{1/2}\,dt=\cdots$ .

Note in general that $\displaystyle \int f(x)g(x)\,dx\neq \left(\int f(x)\,dx\right)\left(\int g(x)\,dx\right)$ and also that if you were calculating $\int\sqrt{3}\,dx=\sqrt{3}x+C$ as $\sqrt{3}$ is a constant.

Q. 13 should be $\int_1^2x^2\,dx$ .

Regards,
J.P.

February 11, 2015 at 10:15 am

Student

Hi J.P.,

Just on Q5 (ii), I thought quadratic equation is –

$\displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}\qquad ?$

Why do you divide $6t^2+12t-8$ by $2$ and where did you get $-b\pm\sqrt{b^2}$ ? Any help would be great.

February 11, 2015 at 10:21 am

J.P. McCarthy

You are correct in your formula. I actually meant to write $-b\pm\sqrt{b^2\cdots}$ rather than $-b\pm\sqrt{b^2}$ . The $\cdots$ signifies that I am not finished writing the formula. I could have referred to the quadratic formula or the “ $-b$ ” formula.

I did not divide $6t^2+12t-8$ by two! I divided both sides of an equation by two:

$6t^2+12t-8=0\Rightarrow 3t^2+6t-4=0.$

I did this because it makes it easier to factorise. In this case, it made no difference because the quadratic has no nice factors but suppose I had

$4t^2-4t-24=0.$

Now I can divide both sides by four:

$t^2-t-6=0$ ,

and I know how to factorise (and then solve) this:

$t^2-3t+2t-6=0$
$t(t-3)+2(t-3)=0$
$(t-3)(t+2)=0$
$(t-3)=0\,\text{or}\,(t+2)=0$
$t=3\,\text{or}\, t=-2$

Regards,
J.P.

February 10, 2015 at 2:09 pm

Student

Hi J.P.,

I found a few examples from MATH6019 and i tried following the steps through for the rms question. Could you look at Q15 b attached and tell me if I’m right or where I’m going wrong as my answer isn’t the same as yours.

Regards.

February 10, 2015 at 2:15 pm

J.P. McCarthy

Your approach doesn’t make much sense to me to be honest! Here we see a proper answer. Firstly $f(x)=\frac{1}{x}$ and so

$\displaystyle |f(x)|^2=(f(x))^2=\left(\frac{1}{x}\right)^2=\frac{1}{x}\cdot \frac{1}{x}=\frac{1}{x^2}$ .

Now we have an interval $[a,b]=[1,2]$ so we want to calculate:

$\displaystyle \int_1^2|f(x)|^2\,dx=\int_1^2\frac{1}{x^2}\,dx.$

We may write $\frac{1}{x^2}=x^{-2}$ so we have

$\displaystyle \int_1^2x^{-2}\,dx=\left[\frac{x^{-1}}{-1}\right]_1^2=\left[-\frac{1}{x}\right]_1^2=-\frac{1}{2}-\left(-\frac{1}{1}\right)=\frac12$

Now we have

$f_{\text{rms}}=\sqrt{\frac{1}{b-a}\int_a^b|f(x)|^2\,dx}=\sqrt{\frac{1}{2-1}\left(\frac12 \right)}=\sqrt{\frac{1}{2}}=\frac{\sqrt{1}}{\sqrt{2}}=\frac{1}{\sqrt{2}}$ .

Regards,
J.P.

February 11, 2015 at 9:39 am

Student

Hi JP,

I am having a few difficulties with some questions on quiz bank 1. Not sure how to go about these questions:

Questions: 4, 5(ii), 6, 7, 10, 11 & 13.

Would it be possible to get the solutions so I can revise them prior to tonight’s quiz?

Thanks

February 11, 2015 at 9:52 am

J.P. McCarthy

I am not giving out full solutions — this is all MATH6019 material. There are some hints in the above comments and I will make more comments here.

Q.4: The key word is RATE.

19.2 m is just the height of the water after 8 s but you are interested in the rate at which the height is changing… 5 cm/s, 10 cm/s, etc?

To find the rate of change of a function we differentiate. So you are looking for the derivative $\frac{dh}{dt}$ evaluated at $t=8$ s.

Q. 5 (ii) If you know how to find the velocity, you need to solve the equation

$v(t)=0$ .

That is you need to find the times, $t$ , when the velocity is equal to zero. This is discussed in this comment (https://jpmccarthymaths.com/2015/02/05/math6037-week-1/#comment-4173)

Q.6 & 7 You have to do something… can you not plot $f(x)=\frac{1}{x}$ just using some sample points $x=1,2,\dots$ ? Once this is done take inspiration from Figure 1.6 on P. 13 of the notes. For Q.6 the graph looks like http://www.wolframalpha.com/input/?i=Plot%5B1%2Fx%2C%7Bx%2C0%2C6%7D%5D and for Q.7 it looks like http://www.wolframalpha.com/input/?i=Plot%5B25-x%5E2%2C%7Bx%2C-1%2C6%7D%5D

Q.10 This is just antidifferentiation using $\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$ .
Anti-differentiation is LINEAR. This means that

$\displaystyle \int (f(x)+k\cdot g(x))\,dx=\int f(x)\,dx+k\int g(x)\,dx$

for functions $f(x)$ and $g(x)$ and CONSTANT $k$ , so you in part (a) you can take out the constant $\sqrt{3}$ . You need to be able to write $\sqrt{t}$ as $t^n$ and $\sqrt[3]{x^2}$ as a power of $x$ .

Q. 11 is tougher but is in the antidifferentiation tables… look for it here: https://jpmccarthymaths.files.wordpress.com/2012/02/newtables.pdf

For Q.13 you need to understand that “The area under a positive function is given by the integral”. That is what integration DOES.

Regards,
J.P.

February 11, 2015 at 9:55 am

Student

This is what I am getting for question 6 (b) can’t seem to get the way you are doing it. Is mine incorrect?

February 11, 2015 at 10:07 am

J.P. McCarthy

Let me say that

$6t^2+12t-8=0$

IS correct. From here what you did is you subtracted $12t$ from both sides and added $8$ to both sides:

$6t^2=-12t+8.$

This equation is still true but the question is, is the equation any easier to solve? What you attempt to do then is get rid of the the $t$ on the right-hand side. There is one big problem with what you did and one small problem.

The big problem can be seen as follows… you have two numbers that are equal to each other… namely $6t^2=(-12t+8)$ . Now if you DO THE SAME THING TO BOTH NUMBERS then they are still equal. Now to get rid of that $t$ on the left you have to divide both numbers by $t$ :

$\frac{6t^2}{t}=\frac{8-12t}{t}\Rightarrow 6t=\frac{8}{t}-12\neq 8-12$ .

Just to show you that the kind of thing you did doesn’t work. Suppose $t=3$ so that

$t^2=t+6$ .

Now according to what you did we can write:

$t=1+6=7$ … but $t=3$ so what we did is clearly wrong.

The small problem occurs if $t=0$ . Suppose you do have

$6t^2=-12t$ .

Now dividing both sides by $t$ gives
$6t=-12\Rightarrow t=\frac{-12}{6}=-2$ …

However clearly $t=0$ is another solution as

$6(0)^2=-12(0)\qquad\checkmark$

but we lost that solution by dividing by $t$ … the problem is if $t=0$ then we have one: YOU CAN’T DIVIDE BY ZERO.

To fix this factorise instead:

$6t^2=-12t\Rightarrow 6t^2+12t=0\Rightarrow 6t(t+2)=0$ .

Now, suppose I have two numbers $a$ and $b$ that multiply to give me zero:

$a\cdot b=0$ .

$a\cdot b=0\Leftrightarrow a=0\text{ or }b=0$

So it is only when you have a product equal to zero can you conclude; i.e. $6t(t+2)=0\Rightarrow 6t=0\text{ or }t+2=0\Rightarrow t=0\text{ or }t=-2$ .

Therefore, the proper way to solve the original problem is to set the quadratic equal to zero:

$6t^2+12t-8=0$ ,

divide both sides by two:

$3t^2+6t-4=0$

and attempt to factorise. In this case, there are no factors so we must use the $-b\pm\sqrt{b^2\cdot}$ formula.

Regards,
J.P.

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MATH6037: Week 1

Quiz 1 Question Bank

Continuous Assessment

Week 1

Week 2

Week 6

Notes

Study

Student Resources

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J.P. McCarthy Maths

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MATH6037: Week 1

Quiz 1 Question Bank

Continuous Assessment

Week 1

Week 2

Week 6

Notes

Study

Student Resources

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