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Squishy
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#61
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(Original post by infekt)
(1/8) arctan (4/4) - (arctan 0)
That's right, but it simplifies to pi/32.
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rockindemon
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#62
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anyone the answer to the intrinsics question?
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Llamas
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#63
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(Original post by beauford)
Hmm i don't think so, I(0) is arsinh 1.

So I(2) = root2/ 2 - arsinh 1 /2

So I(4) = root(2)/4 - 3/8 root(2) + 3/8 arsinh(1) = 3/8 arsinh(1) - root(2) /8
I(0) is the integral of 1dx, which is of course x.

I2 = (1/2)(root(2) - x)

I4 = (1/4)(root(2) - (3/2)(root(2) - x))

= (1/4)(root(2) - (3/2)root(2) + (3/2)x)
= (1/4)((3/2)x - (1/2)root(2))
= [(3/8)x - (1/8)root(2)] from 0 to arsinh(1)
= [(3/8)arsinh(1) - (1/8)root(2)] - [0 - (1/8)root(2)]
= (3/8)arsinh(1) - (1/8)root(2) + (1/8)root(2)
= (3/8)arsinh(1)
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Squishy
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#64
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(Original post by Llamas)
I(0) is the integral of 1dx, which is of course x.

I2 = (1/2)(root(2) - x)

I4 = (1/4)(root(2) - (3/2)(root(2) - x))

= (1/4)(root(2) - (3/2)root(2) + (3/2)x)
= (1/4)((3/2)x - (1/2)root(2))
= [(3/8)x - (1/8)root(2)] from 0 to arsinh(1)
= [(3/8)arsinh(1) - (1/8)root(2)] - [0 - (1/8)root(2)]
= (3/8)arsinh(1) - (1/8)root(2) + (1/8)root(2)
= (3/8)arsinh(1)
You used the limits in the wrong place. You have to do it for I0, which is x between the limits of arsinh1 and 0, so it's arsinh1.

I4 = ¼(sqrt2 - 1.5(sqrt2 - arsinh1))

= ¼(sqrt2 - 1.5sqrt2 + 1.5arsinh1)
= -(sqrt2)/8 + (3arsinh1)/8

Edited to add: I just used a graph-drawing program to approximate the answer, and I get a numerical value of 0.1537, which is -(sqrt2)/8 + (3arsinh1)/8 to 4 DP.
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Llamas
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#65
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(Original post by Squishy)
You used the limits in the wrong place. You have to do it for I0, which is x between the limits of arsinh1 and 0, so it's arsinh1.

I4 = ¼(sqrt2 - 1.5(sqrt2 - arsinh1))

= ¼(sqrt2 - 1.5sqrt2 + 1.5arsinh1)
= -(sqrt2)/8 + (3arsinh1)/8

Edited to add: I just used a graph-drawing program to approximate the answer, and I get a numerical value of 0.1537, which is -(sqrt2)/8 + (3arsinh1)/8 to 4 DP.
This does not bode well.
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IntegralAnomaly
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#66
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(Original post by rockindemon)
anyone the answer to the intrinsics question?
here:
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Squishy
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#67
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(Original post by Llamas)
This does not bode well.
Eh, it's only 4 marks, and you'll probably get 1 or 2 for what you did.
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Llamas
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#68
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(Original post by Squishy)
Eh, it's only 4 marks, and you'll probably get 1 or 2 for what you did.
Aye, but considering how many I've already missed I can ill afford to lose more marks! I'll just have to work to get a good A on P6. Thankfully I did P4 in January and got 87
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flyinghorse
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#69
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#69

YAY
Other people found it hard
Well, the first 4 questions were fine, and the intrinsic coordinates should have been easy but I screwed it up, then figured it out at the end when I ran out of time aarrrrrgggghhh.


Anyone who found it easy, please post in a different thread, I'd rather shield myself from the truth and eat Skittles mixed with ice-cream and chocolate bits (in homage to Homer Simpson's home-made prozac).
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jaffakidds
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#70
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#70
Ive not ever seen as hard a p5 paper. compared with this years paper last years was wee buns....nothin harder than a bit of wonky coordgmtry.

I found it quite an evil nasty tricky paper having to leave stuff out cos i hadnt enough time and stuff was rushed.. Question seven was a complete [insert suitable word here]. bound to have an algebra mistake in here...l And no mad integration question!!!! Was expecting a hard papaer this year after last years simplicity..

Hopefully p4 and m4 go allright...
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chandoug
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#71
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#71
isn't the answer to 5c is

(root 2 -1.5)/4
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hatton02
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#72
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Today's exam was horrendous. I think I had a maths mental block. From getting a spread of marks between 83% on P4 and 100% on P1, this pure module is by far the lowest...every question I struggled on. Even 1 b - not a good confidence boost. My calculator didnt agree with my answer, but i checked my answers over and over again...and didnt see an error so stuck with my answers, as I had the answer as ln2, and the calc showed -ln2. I was a bit worried when it said leave it as klnb or similar and I only came out with ln 2.

Here goes:

cosechx - 2coth x = 2

1/sinhx - 2coshx/sinhx = 2

1-2coshx = 2sinhx

At this point, my maths colleagues turned it into the e^x form, which is what I did all the time in revising...but not in the exam! I noticed a potential cosh²x-sinh²x = 1 coming up - so I squared everything.

1-4coshx+4cosh²x = 4sinh²x

So 4(cosh²x-sinh²x)=4coshx-1

4coshx-1=4

So 4coshx = 5
so cosh x = 5/4

So x = arcosh (5/4)

Using the form given in the formula booklet for arcosh, i.e. arcosh x = ln (x + root (x²-1)) I got x=ln2.

Subbing this back into the original equation (three times) gave me -2 instead of the required 2. Panic and confusion set in as I could not spot my error....can any of you? Surely squaring both sides is a valid move?! I never took any square roots either, so I haven't got to worry about negatives or positives?

Reduction formulae are easy! Fantastic! This question, YUK! I couldn't show the first bit, nor could I get the next bit with root 2 involved, but I managed to work out I4 I think!

The *only* thing I could do fully was the radius of curvature worth 9 marks, woohoo! 9/75, I'm well chuffed. No, seriously, about 40/75 is what I am expecting...a dire performance on my part and I actually felt a tear in my eye at the end of my exam as I am fully aware my university place could have already been lost thanks to one bad exam.

Kirk
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jaffakidds
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#73
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#73
Theres question 1b) unfort for some..
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Squishy
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#74
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(Original post by hatton02)
So 4coshx = 5
so cosh x = 5/4
You're fine up to there. However, cosh x = 5/4 has two solutions (-ln 2 and ln 2). If you plug both numbers into your calculator, you'll see they both work (just as every number always has two square roots).

However, the arcosh function is defined so that it always gives you the positive answer only (the same way that sqrt(x) always returns a positive answer). This is in order to make it a one-one function. However, in this case, the negative answer is the only solution to the equation...sorry to hear about the rest of your paper.
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ogs
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#75
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#75
(Original post by janetjanet)
hi,I got the radian of curveture like 25/9 root 3 ,something,did anyway get that???

I got that! well erm root 2
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ogs
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#76
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#76
It was the hardest p5 paper i have taken, adn i made really stupid mistakes, ie got question 2 wrong through carelessness! hopefully i'll get error carried forward marks! yelp! horrific!
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qqqwwweee
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#77
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(Original post by hatton02)
At this point, my maths colleagues turned it into the e^x form, which is what I did all the time in revising...but not in the exam! I noticed a potential cosh²x-sinh²x = 1 coming up - so I squared everything.
Every time you square something, an additional unexpected answer may appear.
Consider x = -2,
if you square it, x^2 = 4
then x = 2 or -2

So you should not square anything unless you cant do it without eliminating the square root. If you do so, you have to substitute the answers to the original equation and see whether it is correct.

In this case, you should get -ln2 and ln2. However, putting ln2 into the original equation gives you -2. So ln2 is the additional unexpected "answer".
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Llamas
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#78
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#78
(Original post by hatton02)
...
Everyone has already pointed out your mistake, but you should also read the question properly.

"Using the definitions of cosh x and sinh x in terms of exponentials"

Even if you had gotten the right answer you wouldn't have gained any method marks.
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Camford
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#79
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(Original post by Squishy)
You used the limits in the wrong place. You have to do it for I0, which is x between the limits of arsinh1 and 0, so it's arsinh1.

I4 = ¼(sqrt2 - 1.5(sqrt2 - arsinh1))

= ¼(sqrt2 - 1.5sqrt2 + 1.5arsinh1)
= -(sqrt2)/8 + (3arsinh1)/8

Edited to add: I just used a graph-drawing program to approximate the answer, and I get a numerical value of 0.1537, which is -(sqrt2)/8 + (3arsinh1)/8 to 4 DP.
I got that too, Only I took the 1/8 out and write the answer in the ln forms as well as arsinh1.
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qqqwwweee
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#80
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#80
I wonder if we should change arsinh1 to ln(1+root(2))
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