Need help with a parametric equation question thanks (1 Viewer)

Bigboigood

New Member
Joined
Nov 19, 2019
Messages
2
Gender
Male
HSC
2020
This question has me pretty stumped. It's question 14 d) from the Sydney Girls High School 2020 Maths Extension 1 Paper:
Find the cartesian equation of the curve with the parametric equations:
x=4cos(theta)-sin(theta) y=5cos(theta)+sin(theta)

Sorry for the terrible formatting and thanks in advance.
 

Greninja340

Active Member
Joined
Apr 28, 2020
Messages
288
Gender
Male
HSC
2021
could you just solve both of them for sin(theta) and then equate them and solve for y
 

Bigboigood

New Member
Joined
Nov 19, 2019
Messages
2
Gender
Male
HSC
2020
There would still be a cos(theta) in the equation wouldn't there?
 

cossine

Well-Known Member
Joined
Jul 24, 2020
Messages
583
Gender
Male
HSC
2017
x = 4*cos(theta) - sin(theta)

Use the theorem:
cos(theta + phi) = cos(theta)*cos(phi) - sin(theta)*sin(phi)

=> r*cos(theta + phi) = r*cos(theta)*cos(phi) - r*sin(theta)*sin(phi)

Equate

x = r*cos(theta)*cos(phi) - r*sin(theta)*sin(phi)

=> r*cos(phi) = 4, r*sin(phi) =1

Therefore (r^2)*cos^2(phi) + (r^2)*sin^2(phi) = 17

Using theorem: sin^2(theta) + cos^2(theta) = 1

=> r = sqrt(17)

tan(phi) = (r*sin(phi))/ (r*cos(phi) ) = 1/4


Using theorem (f^-1( f(x) ) = x ):

=> phi = tan^-1(1/4)


Therefore x = sqrt(17)*cos(theta + tan^-1(1/4))

=> cos^-1( x/sqrt(17) ) - tan^-1(1/4) = theta

Substitute the expression for theta into y
 

fan96

617 pages
Joined
May 25, 2017
Messages
543
Location
NSW
Gender
Male
HSC
2018
Uni Grad
2024
Therefore x = sqrt(17)*cos(theta + tan^-1(1/4))

=> cos^-1( x/sqrt(17) ) - tan^-1(1/4) = theta

Substitute the expression for theta into y
If you plot this using software the graph you get is not the same as the original parametric equation.
(You only get the upper half of the graph.)

Using theorem (f^-1( f(x) ) = x ):
...
Therefore x = sqrt(17)*cos(theta + tan^-1(1/4))
=> cos^-1( x/sqrt(17) ) - tan^-1(1/4) = theta
This theorem requires to be invertible wherever is defined.
The input varies over an interval of length , where does not have an inverse.

Perhaps an easier way to state this is that if then the range of is an interval of length , yet to get the entire graph of the curve we want we must take over a range of length at least .

This problem is fixed by adding a in front, so

 
Last edited:

cossine

Well-Known Member
Joined
Jul 24, 2020
Messages
583
Gender
Male
HSC
2017
If you plot this using software the graph you get is not the same as the original parametric equation.
(You only get the upper half of the graph.)


This theorem requires to be invertible wherever is defined.
The input varies over an interval of length , where does not have an inverse.

Perhaps an easier way to state this is that if then the range of is an interval of length , yet to get the entire graph of the curve we want we must take over a range of length at least .

This problem is fixed by adding a in front, so

Thank you
 

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

Top