You are correct in your formula. I actually meant to write rather than . The signifies that I am not finished writing the formula. I could have referred to the quadratic formula or the “” formula.

I did not divide by two! I divided both sides of an equation by two:

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

Now I can divide both sides by four:

,

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

Regards,

J.P.

Hi J.P.,

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

Why do you divide by and where did you get ? Any help would be great.

]]>Let me say that

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

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 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 . Now if you DO THE SAME THING TO BOTH NUMBERS then they are still equal. Now to get rid of that on the left you have to divide both numbers by :

.

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

.

Now according to what you did we can write:

… but so what we did is clearly wrong.

The small problem occurs if . Suppose you do have

.

Now dividing both sides by gives

…

However clearly is another solution as

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

To fix this factorise instead:

.

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

.

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

So it is only when you have a product equal to zero can you conclude; i.e. .

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

,

divide both sides by two:

and attempt to factorise. In this case, there are no factors so we must use the formula.

Regards,

J.P.

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 evaluated at s.

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

.

That is you need to find the times, , 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 just using some sample points ? 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 .

Anti-differentiation is LINEAR. This means that

for functions and and CONSTANT , so you in part (a) you can take out the constant . You need to be able to write as and as a power of .

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.

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

]]>Your approach doesn’t make much sense to me to be honest! Here we see a proper answer. Firstly and so

.

Now we have an interval so we want to calculate:

We may write so we have

Now we have

.

Regards,

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.

]]>Student,

Q.3 for units that are products it is better to leave a space between the units. For example, rather than . 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 is . In longer form:

. 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 . This DOES imply that (which is slightly different to what you had) but this DOES NOT imply that or . For example, consider and . Now but neither of nor are equal to eight.

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

.

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

So it is only when you have a product equal to zero can you conclude like you did; i.e. .

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

,

divide both sides by two:

and attempt to factorise. In this case, there are no factors so we must use the formula.

Q.6 The graph of 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 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 .

Q.9 The anti-derivative is actually so you should have

.

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

for functions and and CONSTANT . In order words you can "attack a SUM of terms term-by-term" and pull out constants. So you should have

.

Note in general that and also that if you were calculating as is a constant.

Q. 13 should be .

Regards,

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.

]]>