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C3 Trig Question, Help? (attached)

I know how to do every part, apart from part C) where it asks to find t
Screen Shot 2017-10-21 at 18.05.12.png149.9KB
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Avatar for Notnek
Notnek
21
Original post by chrlhyms
I know how to do every part, apart from part C) where it asks to find t

Ah the ferris wheel question - a very common question in the maths forum :smile:

You say you can't find t but does that mean you've already found the maximum value of H? Please post all your working / thoughts.
Try rewriting the H equastion in th Rcos ("") form and then think about ways you can maximise your H value
Original post by chrlhyms
I know how to do every part, apart from part C) where it asks to find t


what a classic, i remember i struggled on this , now if i remember.

The max value of H means to make H as big as possible.

So what can you do to the cos thing you now got from part a to make H as big as possible?
Set it equal to 0? -1? or 1? what gives you the biggest value of H
Original post by Notnek
Ah the ferris wheel question - a very common question in the maths forum :smile:

You say you can't find t but does that mean you've already found the maximum value of H? Please post all your working / thoughts.


So i've found H as 10 + root(85)
and the equation of H as

H=10-root85 cos(pit/5 +0.2187)
but cant think how I would find the maximum value of t as surely this would be 1 but the MS says -1?
Reply 5
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Notnek
21
Original post by chrlhyms
So i've found H as 10 + root(85)
and the equation of H as

H=10-root85 cos(pit/5 +0.2187)
but cant think how I would find the maximum value of t as surely this would be 1 but the MS says -1?

If cos(pit/5 +0.2187) = 1 then

10-root85 cos(pit/5 +0.2187) = 10 - root85

But you said the maximum was 10 + root85 so cos(pit/5 +0.2187) must be -1.
Original post by Notnek
If cos(pit/5 +0.2187) = 1 then

10-root85 cos(pit/5 +0.2187) = 10 - root85

But you said the maximum was 10 + root85 so cos(pit/5 +0.2187) must be -1.


Sorry, I still dont understand!
Reply 7
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Notnek
21
Original post by chrlhyms
Sorry, I still dont understand!

How did you find that the maximum value of H is 10 + root(85)? Please explain fully.
Original post by Notnek
How did you find that the maximum value of H is 10 + root(85)? Please explain fully.


well since the original equation was stretched by a factor of root 85, the new equation was +10 onto that, therefore the new maximum was 10 + root85 and minimum 10-root85
Original post by Notnek
If cos(pit/5 +0.2187) = 1 then

10-root85 cos(pit/5 +0.2187) = 10 - root85

But you said the maximum was 10 + root85 so cos(pit/5 +0.2187) must be -1.


if cos(pit/5) +0.2187)=-1
then surely 10-root85 cos(pit/5) +0.2187)= -10+root85
which isnt the max?
Reply 10
Avatar for Notnek
Notnek
21
Original post by chrlhyms
if cos(pit/5) +0.2187)=-1
then surely 10-root85 cos(pit/5) +0.2187)= -10+root85
which isnt the max?

I'll write it using brackets to make it clearer:

H=10−[85cos⁡(πt5+0.2187)]\displaystyle H = 10-\left[\sqrt{85} \cos\left(\frac{\pi t}{5} +0.2187\right)\right]

So when

cos⁡(πt5+0.2187)=−1\displaystyle \cos\left(\frac{\pi t}{5} +0.2187\right) = -1

you get

H=10−[85×−1]=10−[−85]=10+85\displaystyle H = 10 - \left[\sqrt{85} \times -1\right] = 10 - [-\sqrt{85}] = 10 + \sqrt{85}
Original post by Notnek
I'll write it using brackets to make it clearer:

H=10−[85cos⁡(πt5+0.2187)]\displaystyle H = 10-\left[\sqrt{85} \cos\left(\frac{\pi t}{5} +0.2187\right)\right]

So when

cos⁡(πt5+0.2187)=−1\displaystyle \cos\left(\frac{\pi t}{5} +0.2187\right) = -1

you get

H=10−[85×−1]=10−[−85]=10+85\displaystyle H = 10 - \left[\sqrt{85} \times -1\right] = 10 - [-\sqrt{85}] = 10 + \sqrt{85}


perfect! I understand now thank you! that question was horrible
Reply 12
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Notnek
21
Original post by chrlhyms
perfect! I understand now thank you! that question was horrible

Yes it was a bit nasty. They've asked a ferris wheel type question a few times in C3 exams so it's worth practicing them :smile:
Original post by Notnek
Yes it was a bit nasty. They've asked a ferris wheel type question a few times in C3 exams so it's worth practicing them :smile:


at least now i will know how to solve it! thank you!

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