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Parametric equations question

The question is, find a cartesian equation from

x = 1 + sin t
y = 2 - sec t

I made it so

x - 1 = sin t
2 - y = sec t,

I cant figure out a trig formula to help me get this into an equation? any ideas would be apprienciated!
Reply 1
Avatar for Zacken
Zacken
22
Original post by SunDun111
The question is, find a cartesian equation from

x = 1 + sin t
y = 2 - sec t

I made it so

x - 1 = sin t
2 - y = sec t,

I cant figure out a trig formula to help me get this into an equation? any ideas would be apprienciated!


The usual:

sin⁡2t=(x−1)2\sin^2 t = (x-1)^2

cos⁡2t=(12−y)2\cos^2 t = \left(\frac{1}{2-y}\right)^2

You know that sin⁡2t+cos⁡2t=1\sin^2 t+ \cos^2 t = 1
Reply 2
Avatar for SunDun111
SunDun111
OP
11
Original post by Zacken
The usual:

sin⁡2t=(x−1)2\sin^2 t = (x-1)^2

cos⁡2t=(12−y)2\cos^2 t = \left(\frac{1}{2-y}\right)^2

You know that sin⁡2t+cos⁡2t=1\sin^2 t+ \cos^2 t = 1

oh crap yeah, thanks!
Reply 3
Avatar for Zacken
Zacken
22
Original post by SunDun111
oh crap yeah, thanks!


No problem. :tongue:
Reply 4
Avatar for SunDun111
SunDun111
OP
11
Original post by Zacken
No problem. :tongue:


I have another question,

So i have X = Sec T + Tan T
y = Sec T - Tan T

I need to differentiate this, so when i differentiate both and divide i get

SecT Tan T ( multiplied) - Sec^2t divided by Sec T Tan t + sec^2t.

How do i rearrange this to get Sin T - 1 / Sin T + 1
Reply 5
Avatar for Zacken
Zacken
22
Original post by SunDun111
I have another question,

So i have X = Sec T + Tan T
y = Sec T - Tan T

I need to differentiate this, so when i differentiate both and divide i get

SecT Tan T ( multiplied) - Sec^2t divided by Sec T Tan t + sec^2t.

How do i rearrange this to get Sin T - 1 / Sin T + 1


Factorise:

sec⁡t(tan⁡t−sec⁡t)sec⁡t(tan⁡t+sec⁡t)=tan⁡t−sec⁡ttan⁡t+cos⁡t=sin⁡tcos⁡t−1cos⁡t......+......=⋯=sin⁡t−1cos⁡tsin⁡t+1cos⁡t\displaystyle \frac{\sec t(\tan t - \sec t)}{\sec t(\tan t + \sec t)} = \frac{\tan t - \sec t}{\tan t + \cos t} = \frac{\frac{\sin t}{\cos t} - \frac{1}{\cos t}}{\frac{...}{...} + \frac{...}{...}} = \cdots = \frac{\frac{\sin t - 1}{\cos t}}{\frac{\sin t + 1}{\cos t}}

By writing tangent and secant as sine and cosine and recognising they have the same denominator of costcos t.

Then cancel the cosines t.
(edited 7 years ago)

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