# Cartesian equation

1. Nov 28, 2009

### Slimsta

1. The problem statement, all variables and given/known data
eliminate the parameter to find cartesian equation of the curve

2. Relevant equations

3. 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 doesnt make any sense..
so i make x=y in some way.. or what?

2. Nov 28, 2009

### rock.freak667

do you know any trigonometric identities involving cosδ and sinδ?

3. Nov 28, 2009

### Slimsta

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

4. Nov 28, 2009

### lanedance

so which trig identity would you use here?

5. Nov 29, 2009

### Slimsta

cos2x + sin2x = 1
but then what do i do about the 4 and 5 that are in there?

6. Nov 29, 2009

### rock.freak667

Right so

cos2δ+sin2δ=1

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

7. Nov 29, 2009

### Slimsta

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

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

8. Nov 29, 2009

### lanedance

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

9. Nov 29, 2009

### HallsofIvy

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

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

10. Nov 29, 2009

### Slimsta

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?

11. Nov 29, 2009

### lanedance

$x/4= cos(\delta)$ is true

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

12. Nov 29, 2009

### Slimsta

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?

13. Nov 30, 2009

### lanedance

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

14. Dec 1, 2009

### Slimsta

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

15. Dec 2, 2009

### 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

16. Dec 2, 2009

### Slimsta

well cos6t just means it gonna 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 :/