This discussion is closed.
Barny
Badges: 15
Rep:
?
#1
Report Thread starter 16 years ago
#1
The curve C has parametric equations

x=4cos2t, y=3sint, -90 < t < 90

A is the point(2, 1 1/2) and lies on C.

a) Find the value of t at the point A

b) find dy/dx in terms of t.

c) Show that an equation of the normal to C at A is 6y - 16x + 23 = 0

The normal at A cuts C again at the point B.

d) Find the y-coordinate of the point B

Done a) (answer = 30) b) (answer = 3cost/-8sin2t) and c) but Im stuck on d)

Can anyone help? How do I get a cartesian equation from the curve C...
0
GeneralGrievous
Badges:
#2
Report 16 years ago
#2
stick your parametric values of x and y into the equation of the normal, you should get two values of t, one for the point A and one for the point B, and use that to find the y-coordinate for B
0
makesomenoise
Badges: 1
Rep:
?
#3
Report 16 years ago
#3
(Original post by imasillynarb)
The curve C has parametric equations

x=4cos2t, y=3sint, -90 < t < 90

A is the point(2, 1 1/2) and lies on C.

a) Find the value of t at the point A

b) find dy/dx in terms of t.

c) Show that an equation of the normal to C at A is 6y - 16x + 23 = 0

The normal at A cuts C again at the point B.

d) Find the y-coordinate of the point B

Done a) (answer = 30) b) (answer = 3cost/-8sin2t) and c) but Im stuck on d)

Can anyone help? How do I get a cartesian equation from the curve C...
a) 2=4cos2t
1/2=cos2t
arccos 1/2 = 2t = 60 => t = 30

b) dy/dt = 3cost
dx/dt = -8sin2t
dy/dx = -3cost/8sin2t

c) dy/dx at A = -3cos30/8sin60 = -3rt3/2/8rt3/2 = -3/8
m for the normal = 8/3 so y-1.5 = 8/3(x-2) = 8x/3-16/3
=> 6y-9 = 16x-32 => 6y-16x+23 = 0

Sorry, didn't notice you did these, oops...

And I'm also stuck on d. But this may be useful...
0
Barny
Badges: 15
Rep:
?
#4
Report Thread starter 16 years ago
#4
(Original post by GeneralGrievous)
stick your parametric values of x and y into the equation of the normal, you should get two values of t, one for the point A and one for the point B, and use that to find the y-coordinate for B
I didnt quite understand that, can you go into abit more detail?
0
GeneralGrievous
Badges:
#5
Report 16 years ago
#5
6(3sint) - 16(4cos2t) + 23 = 0

=> 18 sint -64(1-2sin²t) + 23 = 0

=> 128 sin²t + 18 sint - 41 = 0

sint = (-18 +- sqrt(324 + 4.128.41))/2.128

sint = -0.64, 0.5

[0.5 corresponds to t=30, ie point A]

y = 3 sint = 3*-0.64 = -1.92
0
Barny
Badges: 15
Rep:
?
#6
Report Thread starter 16 years ago
#6
Nice one, cheers
0
X
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

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

Personalise

Have you made your firm and insurance uni choices yet?

Yes (111)
54.95%
Yes, but I want to swap them (16)
7.92%
No, but I know who I want to choose (19)
9.41%
No, I still don't know who I want to choose (48)
23.76%
I have decided I don't want to go to uni anymore and will not be choosing (8)
3.96%

Watched Threads

View All