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IF you would like help with any math problems, feel free to post them

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Reply 20
Avatar for f1mad
f1mad
13
Original post by raheem94
Why differentiate twice?

The question clearly states that it is not to be proved that the value is a maximum.


I didn't bother reading that.

Anywho, no big deal.
The point P (cp, c/p) and the point Q(cq, c/q) lie on a rectangular parabola, equation xy=c(squared). The gradient of the chord PQ is -1/pq . The points P, Q and R lie on the rectangular parabola and angle QPR being a right angle. Show that the angle between QR and the tangent at P is also a right angle. HELP???
Original post by sulexk
Just note this is a maths thread! :smile:

Nevertheless, through your pretty clever words, you intended to mean something.
I would think you were saying that a particular way of life, puts one entity in higher status than the other..

Tell me what were you thinking?

Thanks :smile:


Was a joke about the political correctness of exam papers, dude. :tongue:
Lmao

http://www.guardian.co.uk/education/2011/jun/02/exam-board-apologises-impossible-maths-question
Reply 24
Avatar for sulexk
sulexk
OP
11
Original post by Brit_Miller
Was a joke about the political correctness of exam papers, dude. :tongue:


Aha! :smile:
Hi I was wondering if someone could help me with the following C2 integration question.


Sketch the following and find the area of the finite region or regions bounded by the curve and the x-axis:

y=x(x+2)

I did the question but got -4/3 instead of the 4/3 (which is the answer) and I just wondering if someone could explain to me why it's 4/3 and not -4/3.

Thanks!
Reply 26
Avatar for raheem94
raheem94
13
Original post by starfish232
Hi I was wondering if someone could help me with the following C2 integration question.


Sketch the following and find the area of the finite region or regions bounded by the curve and the x-axis:

y=x(x+2)

I did the question but got -4/3 instead of the 4/3 (which is the answer) and I just wondering if someone could explain to me why it's 4/3 and not -4/3.

Thanks!


No problem here, area is always positive, so write the answer as 4/3.

You get the answer as -4/3, because the area is below the x-axis. So when you use integration to find the area below the x-axis, you will always get a negative answer.

Below image shows the graph:



From the graph, you can see that the area is below the x-axis. But remember that area is always positive(can we say anything to have negative area?), so write it as positive.
Original post by raheem94
No problem here, area is always positive, so write the answer as 4/3.

You get the answer as -4/3, because the area is below the x-axis. So when you use integration to find the area below the x-axis, you will always get a negative answer.

Below image shows the graph:



From the graph, you can see that the area is below the x-axis. But remember that area is always positive(can we say anything to have negative area?), so write it as positive.


Oh ok so if I get a question like this then get the answer as negative then I should turn into positive as the area is always positive.

But, why is the area is always positive? That sort of confuses me.
Thank you very much for your help!
Original post by starfish232
Oh ok so if I get a question like this then get the answer as negative then I should turn into positive as the area is always positive.

But, why is the area is always positive? That sort of confuses me.
Thank you very much for your help!


How can you have a negative area?
You are reminded to read the Guide to Posting, and in particular what it has to say about posting full solutions.
Reply 30
Avatar for raheem94
raheem94
13
Original post by starfish232
Oh ok so if I get a question like this then get the answer as negative then I should turn into positive as the area is always positive.

But, why is the area is always positive? That sort of confuses me.
Thank you very much for your help!


Can you say a length of a side to -2cm?, we can say a length exists as 2cm but -2cm can't be length, the case is similar with area, area can't be negative, can you say the area of a thing to be -20cm^2, it will mean nothing.
Original post by Brit_Miller
How can you have a negative area?


Oh you can't. My mistake!
I have another question. I'm sorry but I'm having to do some chapters myself since my teacher won't teach them till next term and I find it better to finish it myself.


The diagram shows part of the curve with equation y=x2+2 and the line with equation y=6. The line cuts the curve at the points A and B.


(a) Find the coordinates of the points A and B.

(b) Find the area of the finite region bounded by AB and the curve.
Hint:
Put line = curve to find A and B then integrate ‘line − curve’.

I did the question and got 12 1/3 but the actual answer is 10 2/3 and I don't understand why it's 10 2/3 instead of 12 1/3?

Thank you very much!
Original post by raheem94
Can you say a length of a side to -2cm?, we can say a length exists as 2cm but -2cm can't be length, the case is similar with area, area can't be negative, can you say the area of a thing to be -20cm^2, it will mean nothing.


I understand what you mean. Thanks!
Original post by starfish232



I did the question and got 12 1/3 but the actual answer is 10 2/3 and I don't understand why it's 10 2/3 instead of 12 1/3?




Because that is what you get when you integrate and put in the limits

What do you have as the limits
Reply 35
Avatar for raheem94
raheem94
13
Original post by starfish232
I have another question. I'm sorry but I'm having to do some chapters myself since my teacher won't teach them till next term and I find it better to finish it myself.


The diagram shows part of the curve with equation y=x2+2 and the line with equation y=6. The line cuts the curve at the points A and B.


(a) Find the coordinates of the points A and B.

(b) Find the area of the finite region bounded by AB and the curve.
Hint:
Put line = curve to find A and B then integrate ‘line − curve’.

I did the question and got 12 1/3 but the actual answer is 10 2/3 and I don't understand why it's 10 2/3 instead of 12 1/3?

Thank you very much!


See the graph below,




The shaded region is the area which we require.

We know that the line AB cuts the curve at x=-2 and x=2.

So we need to intergrate using the formula,
∫x1x2(y1−y2)dx \displaystyle \int^{x_2}_{x_1} (y_1-y_2)dx

So our expression will be,
∫−22(6−(x2+2))dx \displaystyle \int^{2}_{-2} (6-(x^2+2))dx

Hope it helps :smile:.
(edited 12 years ago)
Reply 36
Avatar for Azland
Azland
8
The complex number u = -1 - i

Sketch an Argand diagram showing the points representing the complex numbers u and u^2.

Shade the region whose points represent the complex numbers z which satisfy both the inequalities |z| < 2
and |z − u^2 | < | z - u |

Just lost on what to do in this question regarding the inequality.
Original post by raheem94
See the graph below,




The shaded region is the area which we require.

We know that the line AB cuts the curve at x=-2 and x=2.

So we need to intergrate using the formula,
∫x1x2(y1−y2)dx \displaystyle \int^{x_2}_{x_1} (y_1-y_2)dx

So our expression will be,
∫−22(6−(x2+2))dx \displaystyle \int^{2}_{-2} (6-(x^2+2))dx

Hope it helps :smile:.


It does! Thanks.
Reply 38
Avatar for Dog4444
Dog4444
10
I found a good problem, but I can't manage to solve it. It supposed to be pretty easy one:

Prove that there are infinitely many positive integers nn such that n2+1n^2+1 has a prime divisor greater than 2n+2n 2n+\sqrt{2n}

Help!
(edited 12 years ago)
Reply 39
Avatar for f1mad
f1mad
13
Original post by Azland
The complex number u = -1 - i

Sketch an Argand diagram showing the points representing the complex numbers u and u^2.

Shade the region whose points represent the complex numbers z which satisfy both the inequalities |z| < 2
and |z − u^2 | < | z - u |

Just lost on what to do in this question regarding the inequality.


Plug in a couple of complex numbers (z= x+iy after all...). This should give you some basis on which part to shade in.

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