Need help in maths questions; angle between two lines (1 Viewer)

4025808 · Apr 9, 2010

This is for my extension 1 maths homework to be due once i get back to school....

and here are the questions I need help on...

2. Find the acute angle (to the nearest minute) between the curves y=(e^2x) and y=(e^x) + 2 at their point of intersection

3. The curves xy=2e and y=ln(x^2) intersect at P (e, 2). If (θ) is the acute angle between these curves, show that tan(θ) = 4e/(e^2)-4

4. Find the acute angle (to the nearest minute) between the tangents to the curves y=sinx and y=cosx at the point of intersection.

5. Find the size of the obtuse angle (to the nearest degree) between the two curves y=ln((x^2)+1) and y=2ln(x+2) at their point of intersection.

6. The angle between the line y=x/A and the tangent to the curve y=A(x^2) at x=1 is 45*. Find the value(s) of A.

If you can help me it would be greatly appreciated...

4025808 · Apr 9, 2010

Btw if you can't see it clearly then

2. $y=e^2x$ and $y=e^x +2$ (e^2x is what it should be...)

3. $xy=2e$ , $y=lnx^2$ , $tan(θ) = 4e/(e^2)-4$

5. $y=ln(x^2+1)$ , $y=2ln(x+2)$

6. $y=A(x^2)$

Drongoski · Apr 9, 2010

I'll try the 1st one

$\100dpi \small y\ =\ e^{2x}\ \ and\ \ \ y\ =\ e^{x} + 2\ \ and\\ \\ y'\ =\ 2e^{2x}\ \ \ and\ \ \ y'\ =\ e^{x}\ \ resp.\\ \\ where\ they\ intersect:\ \ \ e^{2x}\ =\ e^{x}+2\ \ \Rightarrow \ \ e^{2x}-e^{x}-2\ =\ 0\\ \\ \Rightarrow (e^{x}-2)(e^{x}+1)\ =\ 0\ \ \Rightarrow \ \ e^{x}-2\ =\ 0\ \ \Rightarrow x\ =\ ln\ 2\ \ \ \ [e^{x}+1\ \neq \ 0]\\ \\ The\ corresponding\ gradients\ of\ the\ 2\ tangents\ at\ point\ of\ intersection\ are:\\ \\ m_{1}\ =\ 2e^{2ln\ 2}\ =\ 2e^{ln\ 4}\ =\ 2\times 4\ =\ 8\\ \\ m_{2}\ =\ e^{ln\ 2}\ =\ 2\\ \\ If\ \theta \ is\ angle \ between\ the\ 2\ tangents:\ \ tan\ \theta \ =\ \frac{m_{2}-m_{1}}{1+m_{1}m_{2}}\ \ =\ \ \frac{8-2}{1+2\cdot 8}\ \ =\ \ \frac{6}{17}\\ \\ \therefore \ \ \theta \ =\ tan^{-1}\frac{6}{17}\ \ =\ \ 19{}^{\circ}\ 26' \ 24' ' \ \ =\ \ 19^{\circ}\ 26'\ \ \ (nearest\ deg)$

captainvagina · Apr 9, 2010

quick question
why are you doing logs and exponentials in year 11?

4025808 · Apr 9, 2010

captainvagina said:
quick question
why are you doing logs and exponentials in year 11?

It's because we have to, it's part of our syllabus...

AAEldar · Apr 9, 2010

captainvagina said:
quick question
why are you doing logs and exponentials in year 11?

I'm in year 11 and my class will be doing it in either Term 2 or 3.

Drongoski · Apr 9, 2010

in yr 11 most would not even have done calculus yet at this stage.

hscishard · Apr 9, 2010

He goes to a selective school.
Logs and e are actually pretty simple, I bet thats why its in their syllabus.

Edit: They don't do integration in year 11 though

bouncing · Apr 9, 2010

Okay so for question 5 i got this..

$y=ln(x^2+1)----[1]\\ y=2ln(x+2)-----[2]\\ x^2+1=(x+2)^2\\ 4x+3=0\\ x=-3/4\\ From [1] y'=2x/x^2+1\\ when x=-3/4\\ y'=-24/25\\ From [2] y'=2/x+2\\ y'=8/5\\$

Now you use the angle between two lines formula (sorry i dotn know how to put it into latex...) but i got an answer of 320/67
Therefore ...
$\Theta =79^{0}27' (acute angle)\\ Theta = 100^{0}32' (obtuse angle)$

For question 6:
$y=x/A -----[1]\\ y=Ax^2-----[2]\\ From [1] y'=1/A\\ From [2] y'=2A\\$
Then using tantheta =1 solve using the angle between two lines formula again

you end up with
$2A^2-1/3A=1\\ 2A^2-1=3A\\$

using the quadratic formula...

$A=\frac {-b\pm \sqrt {b^2-4ac}}{2a}\\ A=\frac {3\pm\sqrt{9+8}}{4}\\ A=\frac {3\pm\sqrt{17}}{4}\\$

NOTE: YOU SHOUDL PROBABLY CHECK THESE ANSWERS

hscishard · Apr 9, 2010

I knew the second one wasn't rational.

bouncing · Apr 9, 2010

i cannot prove that 3rd one

4025808 · Apr 9, 2010

Alright thanks guys...
I've got it now =D....
and if i've mistaken anyone, then i should have said angle between two curves...

had to use calculus... never learnt calc before
i guess i'll go and learn it sometime soon

Need help in maths questions; angle between two lines (1 Viewer)

4025808

Well-Known Member

4025808

Well-Known Member

Drongoski

Well-Known Member

captainvagina

Member

4025808

Well-Known Member

AAEldar

Premium Member

Drongoski

Well-Known Member

hscishard

Active Member

bouncing

Member

hscishard

Active Member

bouncing

Member

4025808

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)