Calculus 8th Eighth Edition 10.3 #22

[calculushelp]

Homework Statement

Find the equations of the tangent lines at the point where the curves crosses itself.

x=2-piCos(t), y=2t-piSin(t)

Homework Equations

The Attempt at a Solution

 

[jhicks]

What have you tried so far?

 

[calculushelp]

well, all i need are the points and after that its easy. But the PROBLEM is the points that's is where i am stuck

 

[jhicks]

Points? The problem only mentions one. Anyway, there wasn't any trick to it rather than realizing that the textbook wouldn't make a calculus problem very difficult in precalculus terms. I'd observe where the curve intercepts y=0. Any time you see such a strange formulation of the x and y coordinates, you know something is bound to simplify dramatically, so start plugging in things!

 

[HallsofIvy]

At the point where the graph crosses itself, you must have the point $2- \pi cos(t)= 2- \pi sin(s)$ for some s and t and $2t- \pi sin(t)= 2s-\pi sin(s)$. Solve those equations for s and t.

 

[calculushelp]

sorry, but still confusion.. can you take it step by step please

 

[calculushelp]

$2- \pi cos(t)= 2- \pi sin(s)$ and is this right? or a typo because I thought its was suppose to be $2- \pi sin(t)= 2- \pi sin(s)$

 

[HallsofIvy]

$2- \pi cos(t)= 2- \pi sin(s)$ and is this right? or a typo because I thought its was suppose to be $2- \pi sin(t)= 2- \pi sin(s)$

Sorry that was a typo. It should be $2- \pi cos(t)= 2- \pi cos(s)$ which says that the x values for s and t are the same. Remember that $x= 2- \pi cos(t)$.
The other equation is [itex]2t- \pi sin(t)= 2s- \pi sin(s)[itex]which says that the y values for s and t are the same- from $y= 2t- \pi cos(t)$.<br /> <br /> It should be obvious that $2- \pi cos(t)= 2- \pi cos(x)$ gives cos(t)= cos(s) which does <b>not</b> mean t= s but rather that $t= 2\pi- s$.[/itex][/itex]