This discussion is closed.
13 years ago
#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?
0
13 years ago
#62
thanx dvs and syncman.. feel free to do the rest of the question!! (sorry for spamming syncman )
0
13 years ago
#63
hehe np. Ill let dvs do the honours. im gonna go watch tv
0
13 years ago
#64
(Original post by lgs98jonee)
ok thanks for that :-)....i got dy/dx=x^x(1+ln x) is this correct?

So yes, looks good to me
0
13 years ago
#65
haha don't watch tv go and study young man!!
0
13 years ago
#66
(Original post by Lucky Penny)
feel free to do the rest of the question!!

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
0
13 years ago
#67
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 :'(
0
13 years ago
#68
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.
0
13 years ago
#69
(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}
0
13 years ago
#70
(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?
0
13 years ago
#71
(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 =
\log =

This only works for the common ones.
0
#72
everyone ready for the big day tomorrow? I've finsihed revising - went threough a few papers

pk
0
13 years ago
#73
Oh, btw Gaz, did you know you could get better looking functions in tex? Just add a backslash before the function, e.g.
\cosh =
\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.

0
13 years ago
#74
thank you thank you thank you dvs... don't know why i couldn't do that question!
0
13 years ago
#75
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!!!
0
13 years ago
#76
(Original post by dvs)
2.
sinht/sqrt(1+sinh²t) = tant/sqrt(1+tan²t)
eh? it does ?
0
13 years ago
#77
Help some one :'(

why does

sinht/sqrt(1+sinh²t) = tant/sqrt(1+tan²t)
0
13 years ago
#78
(Original post by lgs98jonee)
Help some one :'(

why does

sinht/sqrt(1+sinh²t) = tant/sqrt(1+tan²t)
It does in this question as though it's not an identity.
The identity we used to move from to the left hand side is
0
X
new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Cranfield University
Cranfield Forensic MSc Programme Open Day Postgraduate
Thu, 25 Apr '19
• University of the Arts London
Open day: MA Footwear and MA Fashion Artefact Postgraduate
Thu, 25 Apr '19
• Cardiff Metropolitan University
Sat, 27 Apr '19

### Poll

Join the discussion

#### Have you registered to vote?

Yes! (344)
37.84%
No - but I will (72)
7.92%
No - I don't want to (65)
7.15%
No - I can't vote (<18, not in UK, etc) (428)
47.08%