c3 25th January 2013 exam

Well, we wanted to find the maximum value of p(θ)

p(θ) = 4 / (12 + (6cosθ + 8sinθ))

In the previous question we found out that 6cosθ + 8sinθ = 10cos(θ - 0.927)

p(θ) = 4 / (12 + 10cos(θ - 0.927))

If we want to maximise p(θ), we want the denominator of the function to be as small as possible, therefore giving a higher value of p(θ)

So we can find the minimum of 10cos(θ - 0.927) which will be 10(-1) = -10

So now we get p(θ) = 4 / (12-10) = 4 / 2 = 2

The maximum value for p(θ) = 2

I got the same value as you :smile:

Reply 181

PlutoMars

Q7 AHRRRRHHGGGHHHHHHHHHHHHHHHHH ! Why didnt i realise it was quadratic :confused:

Reply 182

A*'s

Original post by Cosectant

Well, we wanted to find the maximum value of p(θ)

p(θ) = 4 / (12 + (6cosθ + 8sinθ))

In the previous question we found out that 6cosθ + 8sinθ = 10cos(θ - 0.927)

p(θ) = 4 / (12 + 10cos(θ - 0.927))

If we want to maximise p(θ), we want the denominator of the function to be as small as possible, therefore giving a higher value of p(θ)

So we can find the minimum of 10cos(θ - 0.927) which will be 10(-1) = -10

So now we get p(θ) = 4 / (12-10) = 4 / 2 = 2

The maximum value for p(θ) = 2

So how did you finds part ii?

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Reply 183

Nia_Beth

9

2/11

..... does anyone have a hunch about how high the grade boundaries will be (particularly for A/A*)??

(edited 11 years ago)

Reply 184

Cosectant

Original post by Nia_Beth

.... I thought cos(θ - 0.927)=+1 at a maximum point.....

Indeed, it does - but in this case, we want to find the minimum of cos(θ - 0.927) since we're trying to make the denominator of p(θ) as small as possible. You're correct about the maximum of 10cos(θ - 0.927), but that's not what we're trying to find. We're trying to find the maximum of P(θ)

Reply 185

A*'s

Arsey where are you?

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Reply 186

PlutoMars

A* shld be 69 >.< or below ... plz :colondollar:

Reply 187

StrawHat_Shifaz

anyone here who did the 7th question ( i think) where you had to find the value of t?

Reply 188

shumen9523

6

When does Arsey normally post the UMS ??

Reply 189

Cosectant

Original post by A*'s

So how did you finds part ii?

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p(θ) = 4 / (12 + (6cosθ + 8sinθ))

Well, since we know that the maximum value of p(θ) = 2, we can say

p(θ) = 4 / (12 + (6cosθ + 8sinθ)) = 2

4 = 2(12+(6cosθ + 8sinθ))

2 = 12 + (6cosθ + 8sinθ)
-10 = 6cosθ + 8sinθ
-10 = 10cos(θ-0.927)

Cos(θ-0.927) = -1
(θ-0.927) = π
θ = π + 0.927 = 4.069

Reply 190

Nia_Beth

9

Original post by StrawHat_Shifaz

anyone here who did the 7th question ( i think) where you had to find the value of t?

that was 8(b) - I remember it cus i flopped that part :/

Reply 191

kvohra

I am so annoyed with myself! I worked out the Rcos(alpha + theta) in the last question correctly but didn't see the connection between the two questions and I knew how to do it!

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Reply 192

A*'s

Original post by Cosectant

p(θ) = 4 / (12 + (6cosθ + 8sinθ))

Well, since we know that the maximum value of p(θ) = 2, we can say

p(θ) = 4 / (12 + (6cosθ + 8sinθ)) = 2

4 = 2(12+(6cosθ + 8sinθ))

2 = 12 + (6cosθ + 8sinθ)
-10 = 6cosθ + 8sinθ
-10 = 10cos(θ-0.927)

Cos(θ-0.927) = -1
(θ-0.927) = π
θ = π + 0.927 = 4.069

You speak/text like my American friend...it's creepy

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Reply 193

shumen9523

6

Original post by StrawHat_Shifaz

anyone here who did the 7th question ( i think) where you had to find the value of t?

Check the previous page.

Reply 194

Cosectant

Original post by A*'s

You speak/text like my American friend...it's creepy

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Is that a compliment? If so, why thank you! :colondollar:

Reply 195

Cosectant

Original post by shumen9523

It's quadratic. I think the question was something like this:
9500= 17000e^-0.25t + 2000e^-0.5t + 500
9000= 17000e^-0.25t + 2000e^-0.5t
17000e^-0.25t + 2000e^-0.5t -9000=
2^-0.5t + 17e^-0.25t -9 =
2(e^-0.25t)^2 + 17e^-0.25t -9 =
e^-0.25t = X
2X^2 +17X - 9 =
(2X - 1 ) ( X + 9)
X =0.5 X = -9 (rejected)
e^-0.25t =0.5
-0.25t = ln 0.5
t = 4 ln(2) if you substitutethis in the equation in the place of t, you'll get 9500 so it is likely to be correct.
Hope this helps

Hey, I got the same value as you :smile:

- although, I put my answer as ln(16)

Reply 196

A*'s

Original post by Cosectant

Is that a compliment? If so, why thank you! :colondollar:

Lol
John is that you?
I'm not even joking anymore XD
You're meant to be at university... How the hell did you get hold of a C3 paper :smile:

?

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Reply 197

SuziieB

11

Original post by Cosectant

Hey, I got the same value as you :smile:

- although, I put my answer as ln(16)

I didn't tidy up my answer like you guys...I left it as ln(0.5)/-0.25 :awkward: will it still be ok?

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