Eliminate Parameter to Find Cartesian Equation of Curve

[Slimsta]

Homework Statement

eliminate the parameter to find cartesian equation of the curve

Homework Equations

The Attempt at a Solution

@ means delta

x = 4cos@, y=5sin@, -pi/2 <= @ <= pi/2

i have no idea how to do it.. i read the whole chapter and it doesn't make any sense..
so i make x=y in some way.. or what?

 

[rock.freak667]

do you know any trigonometric identities involving cosδ and sinδ?

 

[Slimsta]

do you know any trigonometric identities involving cosδ and sinδ?

yeah.. obv.. i just didnt know how to put this delta symbol on here

 

[lanedance]

so which trig identity would you use here?

 

[Slimsta]

cos^2x + sin^2x = 1
but then what do i do about the 4 and 5 that are in there?

 

[rock.freak667]

cos^2x + sin^2x = 1
but then what do i do about the 4 and 5 that are in there?

Right so

cos²δ+sin²δ=1

4 and 5 are constants, so you can just divide by them to get sinδ and cosδ

 

[Slimsta]

Right so

cos²δ+sin²δ=1

4 and 5 are constants, so you can just divide by them to get sinδ and cosδ

so that would mean, cos²δ = x² = (4cosδ)²?

x²+y² = (4cosδ)² + (5sinδ)² = 1
is this correct?

 

[lanedance]

so that would mean, cos²δ = x² = (4cosδ)²?

x²+y² = (4cosδ)² + (5sinδ)² = 1
is this correct?

no, that doesn't make sense, the identity is:
cos²δ + sin²δ = 1

start with that form and think about what you have to divide x & y by to substitute into it

 

[HallsofIvy]

cos^2x + sin^2x = 1
but then what do i do about the 4 and 5 that are in there?

Right so

cos²δ+sin²δ=1

4 and 5 are constants, so you can just divide by them to get sinδ and cosδ

(4sin(\delta))^2+ (5cos(\delta))^2= 16sin^2(\delta)+ 25cos^2(\delta) is NOT equal to 1. sin^2(\delta)+ cos^2(\delta) is equal to 1.

If x= 4 cos(\delta) then cos(\delta)= ?

 

[Slimsta]

(4sin(\delta))^2+ (5cos(\delta))^2= 16sin^2(\delta)+ 25cos^2(\delta) is NOT equal to 1. sin^2(\delta)+ cos^2(\delta) is equal to 1.

If x= 4 cos(\delta) then cos(\delta)= ?

oh so x/4= cos(\delta)
and then same thing for x/5= sin(\delta)

then plug it in the equation..
(x/5)^2+ (x/4)^2 = 1

and solve for x or how do i find cartesian equation of the curve?

 

[lanedance]

oh so x/4= cos(\delta)
and then same thing for x/5= sin(\delta)

then plug it in the equation..
(x/5)^2+ (x/4)^2 = 1

and solve for x or how do i find cartesian equation of the curve?

x/4= cos(\delta) is true

x/5= sin(\delta) is definitely not, where did you get that?

 

[Slimsta]

x/4= cos(\delta) is true

x/5= sin(\delta) is definitely not, where did you get that?

that was my mistake... i meant y/5= sin(\delta)

so from there (y/5)^2+ (x/4)^2 = 1
==> (y^2/25)+ (x^2/16) = 1

what i did next on my paper is, solved for y, and this cartesian equation of the curve..
is that right?

 

[lanedance]

be careful though, there is two solutions for y... the one you should use is based on the orginal domainof \delta

 

[Slimsta]

how do you know what the graph going to look like though?
i mean, what if you have a question like this:
x = 3cos6t
y = 4sin2t

I can do the same thing like x/3=cos6t.. because then what about this 6 in there?

 

[lanedance]

yeah, that is a bit more complicated, the first one simplfies a lot because it is an ellipse

what do you think will happen? think about cycles

 

[Slimsta]

well cos6t just means it going to be horizontally compressed (by 1/6 from normal)
same for sin2t just by 1/2

the only thing i can come up with is: cos2x = 2sinxcosx
so would cos6x = 6sinxcosx ? my guess is no..
then i have no idea what to do :/