Parametric to Cartesian - # 5

Oct 2012
1,314
21
USA
Find the Cartesian equation from the parametric one.

\(\displaystyle \[x = \cos (\pi - t), y = \sin (\pi - t), 0 \leq t \leq \pi\]\)

Since we have some stuff on the inside, rather than coefficients can we just say it's

\(\displaystyle x^{2} + y^{2} = 1\)
 

topsquark

Forum Staff
Jan 2006
11,583
3,455
Wellsville, NY
Find the Cartesian equation from the parametric one.

\(\displaystyle \[x = \cos (\pi - t), y = \sin (\pi - t), 0 \leq t \leq \pi\]\)

Since we have some stuff on the inside, rather than coefficients can we just say it's

\(\displaystyle x^{2} + y^{2} = 1\)
Yup!

See if you can figure out what the difference between the problems ( x = cos(t), y = sin(t) ) and ( x = cos(pi - t) , y = sin(pi - t) ) is. The y(x) equations are the same. So what's different?

-Dan
 
Feb 2016
38
8
Brisbane
Sorry, beats me.
Employ the following Trig Identities -
\(\displaystyle \cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B) \)
\(\displaystyle \sin(A + B) = \sin(A)\cos(B) + \sin(B)\cos(A) \)
eg -
\(\displaystyle \cos(\pi - t) = \cos(\pi)\cos(t) - \sin(\pi)\sin(t) \)
 
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Oct 2012
1,314
21
USA
Employ the following Trig Identities -
\(\displaystyle \cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B) \)
\(\displaystyle \sin(A + B) = \sin(A)\cos(B) + \sin(B)\cos(A) \)
eg -
\(\displaystyle \cos(\pi - t) = \cos(\pi)\cos(t) - \sin(\pi)\sin(t) \)
\(\displaystyle y = \cos(\pi - x) = \cos(\pi)\cos(x) - \sin(\pi)\sin(t) \) ??
 

Plato

MHF Helper
Aug 2006
22,493
8,654
Find the Cartesian equation from the parametric one.
\(\displaystyle \[x = \cos (\pi - t), y = \sin (\pi - t), 0 \leq t \leq \pi\]\)
Since we have some stuff on the inside, rather than coefficients can we just say it's
\(\displaystyle x^{2} + y^{2} = 1\)
Given this contour is a semicircle. To see that $(\pi-t)$ goes $(-1,0)\to(1,0)$ as $0\le t\le \pi$.
$\Large y=\sqrt{1-x^2}$
 
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Oct 2012
1,314
21
USA
Given this contour is a semicircle. To see that $(\pi-t)$ goes $(-1,0)\to(1,0)$ as $0\le t\le \pi$.
$\Large y=\sqrt{1-x^2}$
How can you tell it's a semi-circle?
 

Plato

MHF Helper
Aug 2006
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8,654
How can you tell it's a semi-circle?
It is is very simple: I know the basics.
I have observed that you are trying to skip the grunt work.
You need to go back and take a solid pre-calculus course. It cannot be done online.
 
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