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- Thread starter Needhelp
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- #1

- Feb 1, 2012

- 20

What is the amplitude of this curve? (p)

After how many radians does the cycle start repeating itself? (determines q)

About what line does the curve oscillate? (determines r)

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- #3

surely the amplitude of the curve is 4? (1+7/2), and perhaps the period is π? although im not too sure what you mean by the line the curve oscillates? thank you so much for the help by the way!

according to the markscheme this is the method to work it out-

1=p-r

-7=-p-r

and therefore p=4 and r=3

Not trying to be a pain, and im so grateful for the help, but do possibly understand how this works and could explain it to me? My exam board marks very strictly..

according to the markscheme this is the method to work it out-

1=p-r

-7=-p-r

and therefore p=4 and r=3

Not trying to be a pain, and im so grateful for the help, but do possibly understand how this works and could explain it to me? My exam board marks very strictly..

Last edited:

- Mar 1, 2012

- 249

Is this for a graded assessment?

- Feb 1, 2012

- 20

The period is indeed [tex]\pi[/tex] and from this, one can deduce that the value of q would be 2.surely the amplitude of the curve is 4? (1+7/2), and perhaps the period is π? although im not too sure what you mean by the line the curve oscillates? thank you so much for the help by the way!

according to the markscheme this is the method to work it out-

1=p-r

-7=-p-r

and therefore p=4 and r=3

Not trying to be a pain, and im so grateful for the help, but do possibly understand how this works and could explain it to me? My exam board marks very strictly..

Usually, the period of the standard cos curve is [tex]2\pi[/tex]. The curve that you got completes 2 cycles in [tex]2\pi[/tex], hence q = 2.

For r, I forgot the technical name for it, sorry about that... but yes, that's one way that you can find it. If you join all the troughs together, you get a line. Joining the crests together gives you another line. The 'axis of oscillation' will be the line exactly between those two lines. You will notice that the curve never goes further than the amplitude from this axis.

For the second part, you simply need to solve simultaneously. You have two intersecting graphs (the x-axis and the curve you now know the equation). I hope you know how to solve equations involving trigonometry

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- #6

And in response to SuperSonic4, no, just revision for my c2 exam that is in two weeks

- Jan 26, 2012

- 890

Since this is a sinusoid the period is the interval between alternate turning pointts and so is \( \pi \), and so \(q \pi =2 \pi\), \(q=2\).I was wondering if anyone could help with this- I've never seen a question like this before and don't know exactly how to tackle it... any help would be greatly appreciated!!!

The maximum of \(y=p \cos(q x) +r\) is \(p+r\) and the minimum is \(-p+r\), you are given these which gives you a pair of simultaneous equations for \(p\) and \(q\) to solve.

CB

- Feb 1, 2012

- 20

Yes, the line is y = -3, which means, r = 3, since in the given equation, you are given -r. Substituting r = -3 would result in y = pcos(qx) - (-3) = pcos(qx) + 3 which is not what the diagram is showing.

And in response to SuperSonic4, no, just revision for my c2 exam that is in two weeks

If you had been given [tex]y = p\cos(qx)+r[/tex], then r would be equal to -3.

In other words, whatever constant coming after the trigonometry part makes the entire curve go up or down. It goes up if the constant is positive and it goes down if negative.

For part 2, it is a little less intuitive. Could you show us how you would solve the simultaneous equations formed? Don't worry if it's wrong, you learn better through mistakes

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- #9

- Feb 1, 2012

- 20

Yup! You just need to know which ones are the ones shown in the diagram