P5 questions help, thread:) Watch

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lgs98jonee
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#61
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#61
(Original post by dvs)
If that's from the sample exam at the end of the P5 book, then it should be y=x^x, not x^sinhx.
ok thanks for that :-)....i got dy/dx=x^x(1+ln x) is this correct?
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Lucky Penny
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#62
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thanx dvs and syncman.. feel free to do the rest of the question!! (sorry for spamming syncman )
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Syncman
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#63
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hehe np. Ill let dvs do the honours. im gonna go watch tv
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Gaz031
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#64
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(Original post by lgs98jonee)
ok thanks for that :-)....i got dy/dx=x^x(1+ln x) is this correct?
y=x^{x}\\

lny=xlnx\\

\frac{1}{y} \frac{dy}{dx} = 1+lnx\\

\frac{dy}{dx}=x^{x}(1+lnx)
So yes, looks good to me
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Lucky Penny
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#65
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#65
haha don't watch tv go and study young man!!
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dvs
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#66
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(Original post by Lucky Penny)
feel free to do the rest of the question!!
:rolleyes::p:

b) Substitute the equation of the normal in xy=c² and solve for the other pt of intersection. You might need to use the quadratic formula. You should get Q (-c/t³, -ct³).
c) This should be straight-forward once you get the coords of Q.
d)
From the coords of the midpoit:
x = (c/2)(t - 1/t³)
y = (c/2)(1/t - t³) = (c/2t)(1 - t4)

You know that t²=-y/x from the previous part. So:
y² = (c²/4t²)(1 - t4
4y³/x + c²(1 - y²/x²)² = 0

Multiply by x²/y²:
4xy + c²(x/y - y/x)² = 0
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lgs98jonee
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#67
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Ok I am really struggling to get the correct answer of q4 of the paper at the back of the P5 book. I get ln1/2 :/.....I have no idea how to do question 2 either :'(
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dvs
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#68
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2.
a) tanht = sinht/cosht = sinht/sqrt(1+sinh²t) = tant/sqrt(1+tan²t) = tant/sect = sint
b)
sinhx = 0.5(e^x - e^(-x))
coshx = 0.5(e^x + e^(-x))
sinhx + coshx = e^x
=> e^x = tant + sqrt(1+tan²t) = tant + sect

4.
y = mx + c
Sub this in x²/a² - y²/b² = 1, and you should get a quadratic. This line is a tangent iff there is only one point of intersection, i.e. b²-4ac=0.
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Gaz031
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#69
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(Original post by lgs98jonee)
Ok I am really struggling to get the correct answer of q4 of the paper at the back of the P5 book. I get ln1/2 :/.....I have no idea how to do question 2 either :'(
Q2:
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
(a)\\

tanhx=\frac{sinhx}{coshx}\\

=\frac{tant}{\sqrt{1+sinh^{2}x}\ \

=\frac{tant}{\sqrt{1+tan^{2}t}\\

=\frac{tant}{sect}\\

=\frac{sint}{cost}cost

=sint\\

(b)\\

sect+tant=\sqrt{1+sinh^{2}x}+sin hx\\

=coshx+sinhx

=e^{x}
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lgs98jonee
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#70
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(Original post by dvs)
2.
a) tanht = sinht/cosht = sinht/sqrt(1+sinh²t) = tant/sqrt(1+tan²t) = tant/sect = sint
b)
sinhx = 0.5(e^x - e^(-x))
coshx = 0.5(e^x + e^(-x))
sinhx + coshx = e^x
=> e^x = tant + sqrt(1+tan²t) = tant + sect

4.
y = mx + c
Sub this in x²/a² - y²/b² = 1, and you should get a quadratic. This line is a tangent iff there is only one point of intersection, i.e. b²-4ac=0.
couldnt there be intersection on teh right hand side by a tangent coming from the right? or would this be ignore :/...also how is 3b doen?
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dvs
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#71
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(Original post by lgs98jonee)
couldnt there be intersection on teh right hand side by a tangent coming from the right? or would this be ignore :/...also how is 3b doen?
If there's more than one point of intersection then the line isn't a tangent.

For 3b, did you try plugging in the limits?
arsinh(3/4) - arsinh(5/12), and use the log form of arsinh.

Oh, btw Gaz, did you know you could get better looking functions in tex? Just add a backslash before the function, e.g.
\cosh = \cosh
\log = \log

This only works for the common ones.
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Font
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#72
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#72
everyone ready for the big day tomorrow? I've finsihed revising - went threough a few papers

pk
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Gaz031
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#73
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Oh, btw Gaz, did you know you could get better looking functions in tex? Just add a backslash before the function, e.g.
\cosh = \cosh
\log = \log
This only works for the common ones.
Thanks for the advice! I only started learning yesterday so i'm still kinda new to it. A lot of the commands seem fairly intuitive once you know what they are but there's so many. Of course, the TeXnicCenter menus are a useful reference.
\cosh x \geq \int_0^x \cosh x dx, \forall x \in \Re .
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Lucky Penny
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#74
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thank you thank you thank you dvs... don't know why i couldn't do that question!
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Lucky Penny
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#75
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I'm so nervous about tmrw.. have this nasty feeling they're gonna throw a really hard question at us and can't help feeling that i might not know something that will come up.. ahh!!!
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lgs98jonee
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#76
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#76
(Original post by dvs)
2.
sinht/sqrt(1+sinh²t) = tant/sqrt(1+tan²t)
eh? it does ?
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lgs98jonee
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#77
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Help some one :'(

why does

sinht/sqrt(1+sinh²t) = tant/sqrt(1+tan²t)
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Gaz031
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#78
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(Original post by lgs98jonee)
Help some one :'(

why does

sinht/sqrt(1+sinh²t) = tant/sqrt(1+tan²t)
It does in this question as \sinh t = \tan t though it's not an identity.
The identity we used to move from \tanh t to the left hand side is  \cosh ^{2}t - \sinh ^{2}t=1
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