Parametric equations.. Please help. (1 Viewer)

[Michaelmoo]

Hi. Lately Ive been transforming a pair of parametrics into Cartesians. I've been a little confused when it comes to a qustion that has restriction on y-values instead of x. I work out the solution but when I look at the answers, the answer is usually a different equation. For example:

Solve the following pair of parametrics:

x=2cosT and y=sinT for 0 Degrees < T < 180 degrees

Im finding that sometimes my circles/elipses are actually semi circles/elipses with these types of restrictions. Is this just using common sense or is there a way to derive this throughout the calculations?

Also if I write the equation for the complete circle/elipse with the y restriction near it (e.g. for -1 < y < 1), would I still get it correct? Or is it expected that I should manipulate the eqn itself?

Or in other words, please do the above equation with working.

Please help.

Thanks.

 

[elliots]

not sure if this will answer your question or not as i dont 100% know wat ur asking:

so you have
x=2cosT and y = sinT... 2y = 2sinT

square both

x^2 = 4(cosT)^2
(2y)^2 = 4(sinT)^2

add them

x^2 + (2y)^2 = 4(cosT^2 + sinT^2)
x^2 + (2y)^2 = 4

ok so this is going to be the equation of your circle/elipse (0< T <360)
now look at the boundaries.. it will either be
0< T <180 or -180< T <0

for 0< T <180
2y = root(4 - (x)^2)

for -180< T <0
2y = -root(4 - (x)^2)

 

[Michaelmoo]

not sure if this will answer your question or not as i dont 100% know wat ur asking:

so you have
x=2cosT and y = sinT... 2y = 2sinT

square both

x^2 = 4(cosT)^2
(2y)^2 = 4(sinT)^2

add them

x^2 + (2y)^2 = 4(cosT^2 + sinT^2)
x^2 + (2y)^2 = 4

ok so this is going to be the equation of your circle/elipse (0< T <360)
now look at the boundaries.. it will either be
0< T <180 or -180< T <0

for 0< T <180
x = root(4 - (2y)^2)

for -180< T <0
x = -root(4 - (2y)^2)

Ok I get everything up to the boundaries of 0< t< 180 parts. Can someone explain further please? dont understand how you getting those forumulae from 0< t< 180?.

 

[12o9]

Ok I get everything up to the boundaries of 0< t< 180 parts. Can someone explain further please? dont understand how you getting those forumulae from 0< t< 180?.

if 0< t <180 , then y>0, hence you take the positive half of the curve.

 

[elliots]

make y the subject and when u square root your going to get + and -

+ for top half of circle (positive y values)
- for bottom half of circle (negative y values)