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enter image description here

The question was written with dark-blue pen. And I tried to solve this question. I obtained $x$ as it is below. But I cannot obtain $y$ Please show me how to do this.

By the way, $\gamma (t)$ may not be clearly readable. So, I wrote again.

$$\gamma (t)=( \cos ^2 (t)-1/2, \sin(t)\cos (t), \sin (t))$$

Thanks for helping.

-sorry for not writing with MathJax. -

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  • $\begingroup$ What does it mean to substruct an equation? $\endgroup$
    – dfeuer
    Oct 18 '13 at 21:23
  • $\begingroup$ That's, $x^2+y^2-[(x+1/2)^2+y^2+z^2]=1/4-1$ @dfeuer $\endgroup$
    – 1190
    Oct 18 '13 at 21:25
  • $\begingroup$ Ah, subtract. The structure of the parameterization suggests to me that the double angle formulas for sine and cosine may help simplify things. $\endgroup$
    – dfeuer
    Oct 18 '13 at 21:28
  • $\begingroup$ I didnt see, I wrote Wrong. Sorry:( @dfeuer $\endgroup$
    – 1190
    Oct 18 '13 at 21:29
  • $\begingroup$ Hmm, but I dont understand enough what you said. @dfeuer $\endgroup$
    – 1190
    Oct 18 '13 at 21:29
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From your work, $$x^2+y^2=\frac{1}{4} \Rightarrow y^2=\frac{1}{4}-x^2,$$ and $$x=\cos^2 t -\frac{1}{2}.$$

Substituting the latter into the former produces \begin{align*} y^2 &=\frac{1}{4}-x^2 \\ &=\frac{1}{4}-\left( \cos^2 t - \frac{1}{2} \right)^2 \\ &=\frac{1}{4}-\left( \cos^4 t - \cos^2 t + \frac{1}{4} \right) \\ &=\cos^2 t - \cos^4 t \\ &=\cos^2 t\left( 1-\cos^2 t \right) \\ &=\cos^2 t \sin^2 t \\ \Rightarrow y &= \sin t \cos t, \end{align*} as was your intention.

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  • $\begingroup$ Ahhh It is so easy, but I didnt notice this. :( thank you so much:) $\endgroup$
    – 1190
    Oct 18 '13 at 22:07
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    $\begingroup$ Forest through the trees. :) $\endgroup$
    – J. W. Perry
    Oct 18 '13 at 22:12
  • $\begingroup$ Nope. Everything is okay;)) $\endgroup$
    – 1190
    Oct 18 '13 at 22:12
  • $\begingroup$ Ahaha you're so clever:)) $\endgroup$
    – 1190
    Oct 18 '13 at 22:14
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Hint: To get a clearer picture, multiply $\gamma$ by $2$. Then use the double angle formulas.

For reference, the double angle identities are as follows:

  1. $\sin (2x)=2\sin x \cos x$

  2. $\cos(2x)=\cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$.

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  • $\begingroup$ Doesnt there exist a way like my first solution? I dont know the way you said. $\endgroup$
    – 1190
    Oct 18 '13 at 21:48
  • $\begingroup$ @B11b, do you know the double angle identities? They're pretty much the key to finishing your solution, as far as I can tell. $\endgroup$
    – dfeuer
    Oct 18 '13 at 21:55
  • $\begingroup$ No, I never have heart it. My diffrential geometry instructor didnt state this. $\endgroup$
    – 1190
    Oct 18 '13 at 21:56
  • $\begingroup$ @B11b, I wrote them up in my answer. $\endgroup$
    – dfeuer
    Oct 18 '13 at 22:00
  • $\begingroup$ Yeup I know this. Sorry, I dont know its name in English, so I didnt understand. Well, how to use these to get $y$? $\endgroup$
    – 1190
    Oct 18 '13 at 22:03

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