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-
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..
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!!!
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.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- im just quite confused!!
And in response to SuperSonic4, no, just revision for my c2 exam that is in two weeks