Expert: Paul Klarreich - 8/6/2009

Question
Hi, thanks for helping me out.
I'm studying calculus ABC, and the section would be functions & logarithms & parametric equations.

Find a cartesian equation for the curve that contains the parametricized curve.
1. x=sint y=cos2t -∞<t<∞

I tried to use the x^2 + y^2=1, and tried to square the terms by using cos^2 + sin^2=1, but I can't fit it into those equations, and I don't know how to get a y= equation.
I know the answer is y=-2x^2+1, but I have no idea how to get there.
Solve for x.

2. e^x + e^-x=3
I know I need to eliminate by using ln, and somehow eliminating the e., but other than that I don't understand how to get x by itself.  

Answer
Questioner: Jaye
Country: United States
Category: Calculus
Private: No
Subject: Parametric & Logarithm Help.
Question: Hi, thanks for helping me out.
I'm studying calculus ABC, and the section would be functions & logarithms & parametric equations.

Find a cartesian equation for the curve that contains the parametricized curve.
1. x = sin t
  y = cos 2t

I tried to use the x^2 + y^2=1, and tried to square the terms by using cos^2 + sin^2 = 1, but I can't fit it into those equations, and I don't know how to get a y= equation.
I know the answer is y = - 2x^2 + 1, but I have no idea how to get there.
Solve for x.
<< ------------------

Aha:  you know the answer is y = - 2x^2 + 1, which is the same as:

I know the answer is y = 1 - 2x^2

That might jog your memory:  Isn't  cos(2t) = 1 - 2 sin^2(t) ?? Didn't your high school teacher make you memorize that?

Now you will be inspired to write:

  y = cos 2t
  y = 1 - 2 sin^2(t)
  y = 1 - 2x^2

and you are done.

Isn't inspiration great?  Now you know why your mean old high school teacher made you memorize all those formulas.
........................................

2. e^x + e^-x = 3
I know I need to eliminate by using ln, and somehow eliminating the e., but other than that I don't understand how to get x by itself.
>> -----------------
I'm not sure what your question is -- are you asked to solve for x?  Try this:

Let  y = e^x
then e^-x = 1/e^x = 1/y

and you have:

e^x + e^-x = 3
y + 1/y = 3

y^2 + 1 = 3y

y^2 - 3y + 1 = 0,

********  y^2 - 3y + 2 would be nicer, but.....

for which you will solve using the quadratic formula, and then write:

x = ln(whatever you got)