How Do I Tackle an Unfamiliar Challenge?

[Needhelp2]

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...(Sweating) any help would be greatly appreciated!

 

[Unknown008]

Well, you should be familiar with cosine curves (Smile)

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)

 

[Needhelp2]

surely the amplitude of the curve is 4? (1+7/2), and perhaps the period is π? although I am 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 I am so grateful for the help, but do possibly understand how this works and could explain it to me? My exam board marks very strictly.. (Worried)

 

[SuperSonic4]

Is this for a graded assessment?

 

[Unknown008]

surely the amplitude of the curve is 4? (1+7/2), and perhaps the period is π? although I am 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 I am so grateful for the help, but do possibly understand how this works and could explain it to me? My exam board marks very strictly.. (Worried)

The period is indeed \pi and from this, one can deduce that the value of q would be 2.

Usually, the period of the standard cos curve is 2\pi. The curve that you got completes 2 cycles in 2\pi, 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 :)

 

[Needhelp2]

Haha yes it was really part 2 that was the problem :) If the line joining the peaks together is y=1 and the one joining the troughs together is y=-7, would that not mean that the axis of oscillation was y=-3? Sorry for all the questions- I am just quite confused!

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

 

[CaptainBlack]

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...(Sweating) any help would be greatly appreciated!

Since this is a sinusoid the period is the interval between alternate turning pointts and so is \( \pi \), and so \(q \pi =2 \pi\), \(q=2\).

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

 

[Unknown008]

Haha yes it was really part 2 that was the problem :) If the line joining the peaks together is y=1 and the one joining the troughs together is y=-7, would that not mean that the axis of oscillation was y=-3? Sorry for all the questions- I am just quite confused!

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

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.

If you had been given y = p\cos(qx)+r, 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 (Wink)

 

[Needhelp2]

I think it, once the values for p,r and q are found you can sub them into the equation,so it is now 4cos2x-3=0. cos2x= 3/4, so you get cos^-1(3/4)=2x. therefore 2x= + or - 0.723,5.560 and 7.006. Hence x would equal half of each of these values? It seems a lot easier now I have all the values!

 

[Unknown008]

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