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#1 2009-02-07 02:19:29
- ganesh
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Question Bank : Age Group 16-17 : I
1. Find the angle between the asymptotes to the hyperbola 3x² - 5xy -2y² + 17x + y + 14 = 0.
2. Two sides of a triangle are 4 meters and 5 meters in length and the angle between them is increasing at the rate of 0.06 radians/second. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length if
3. If ,
find.
4. Solve: .
5. Find the equation of the ellipse whose foci are (4,0) and (-4,0) and .
6. For what values of x is the rate of increase of
twice the rate of increase of x?
Character is who you are when no one is looking.
#2 2009-05-07 07:06:17
- quittyqat
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Re: Question Bank : Age Group 16-17 : I
Ganesh, can you make a question bank for younger kids?
I'll be here at least once every month. XP
#3 2009-08-16 00:32:40
- rzaidan
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Re: Question Bank : Age Group 16-17 : I
Hi ganesh
For the second problem:
Assume that the angle between the sides of fixed length is θ, so
A = ½(4)(5) sin θ=10 sin θ where A is the area of triangle at any time , so
dA/dt =10 cos θ (dθ/dt) when θ=pi/3 and dθ/dt = 0.06
dA/dt =10 cos (pi/3) * 0.06
=10 (½)(0.06)=0.3
Best Regards
Riad Zaidan
#4 2009-08-16 20:03:58
- rzaidan
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Re: Question Bank : Age Group 16-17 : I
Hi ganesh
For the third problem:
w=x+2y+z^2 , x= cos(t) , y = sin(t) , z=t
dw/dt 
dw/dx)(dx/dt)+ (dw/dy)(dy/dt)+( dw/dz)(dz/dt)
=1(-sin (t) + 2(cos (t) + 2 z (1)
= -sin(t)+2 cos(t) + 2z
Best Regards
Riad Zaidan
#5 2009-08-16 20:11:24
- rzaidan
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Re: Question Bank : Age Group 16-17 : I
Hi ganesh
For the forth problem:
dy/dx + xy =x
dy/dx=x-xy=x(1-y)
dy/(1-y )=x dx
∫dy/(1-y )=∫x dx
-ln(1-y) x²) /2 + c
ln(1/(1-y)) x²) /2 + c ⇒
1/(1-y)=e^(( x²) /2 + c) ⇒
1-y = 1/(e^(( x²) /2 + c))
y=1-1/(e^(( x²) /2 + c)) and you can simplify more
Best Regards
Riad Zaidan
#6 2009-08-16 20:24:04
- bobbym
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Re: Question Bank : Age Group 16-17 : I
Hi ganesh;
We were working on this one at the same time but Riad was faster.
For #4;
Use separation of variables.
Integrate both sides:
Raise both sides to the power of e.
c= e^c
Invert both sides and solve for y:
Solution is:
Last edited by bobbym (2009-08-16 20:26:56)
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#7 2009-08-17 06:00:57
- rzaidan
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Re: Question Bank : Age Group 16-17 : I
Hi ganesh;
For the sixth problem:
Assume that
y= x^3-5x^2+5x+8 and differentiate both sides w.r.t (t) we get:
dy/dt=(3 x^2 - 10 x + 5) (dx/dt) ...........(1)
but dy/dt =2 * (dx/dt) so substitute in (1) we have the following:
2 (dx/dt) 3 x^2 - 10 x + 5) (dx/dt) so if dx/dt ≠0 we get
2 = 3 x^2 - 10 x + 5 therfore
3 x^2 - 10 x + 3 = 0 so
(3x-1)(x-3)=0 so
either x= 1/3 or x= 3 Q.E.D
Best Regards
Riad Zaidan
#8 2009-08-17 06:09:07
- rzaidan
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Re: Question Bank : Age Group 16-17 : I
Hi ganesh;
For the fifth problem:
The foci are (4,0) and (-4,0) and e=1/3 the foci are on the x-axis with center on (0,0)
c = 4 but e=c/a so c/a= 1/3 and we have
4/a=1/3 so a =12 but c^2 = a^2 - b^2 or b^2=a^2-c^2=144-16=128
so the requiered equation is
(x^2)/144 + (y^2)/128 = 1
Best Regards
Riad Zaidan
#9 2009-09-11 17:52:38
- Fruityloop
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Re: Question Bank : Age Group 16-17 : I
1. Find the angle between the asymptotes to the hyperbola 3x² - 5xy -2y² + 17x + y + 14 = 0.
This is honestly the most lengthy, involved problem I've ever worked on. I never studied the general conic equation while in school, so this is new for me. Hopefully I've done everything correctly.
First we need to get rid of the xy term by doing a rotation of axes..
using the formulae for the axes of rotation
We then substitute these into the given equation and after some work we get..
The x'y' terms cancel and after much, much more work we end up with...
(I changed x' and y' to simply x and y)
Now here I had a little help from the computer and amazingly...
continuing further along we eventually end up with...
So the slope of the asymptotes are
The hyperbola is oriented parallel to the y-axis and to get the angle between the asymptote and x-axis we use
So the angle between the asymptotes is
or on the other side the angle is
I hope I did everything right. I wouldn't wish this problem upon my worst enemy.
Last edited by Fruityloop (2009-09-11 19:36:40)
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