Need help with Parametric Represntation

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Kristally

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Nov 12, 2006
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Can someone please help me with this problem. I'm not sure how to do it.

The problem:
Consider the curve x^2+xy+y^2=3
1. Show that a parametic representation of the above curve is:
------{x=cost - (3)^(1/2)sint
------{y=cost +(3)^(1/2)sint
2. Use this parametric representation to find the slope of the tangent to the curve at (1,1) (t=0).

So far I have this work:
For x --> x^2+xsint+sint^2=3
==> xsint+sint^2=3 - x^2
==> (sint)(x+sint)=3 - x^2
==> ?
For y ---> cost^2+ycost+y^2=3
==> ycost+cost^2=3 - y^2
==> (cost)(x+cost)=3 - y^2
==> ?

I don't know where to go from there and what to do. Can someone please help me?
 
Hello, Kristally!

Consider the curve: \(\displaystyle \,x^2\,+xy\,+\,y^2\:=\:3\)

(1) Show that a parametic representation of the above curve is:

. . \(\displaystyle \begin{array}{cc}x\:=\:\cos\theta\,-\,\sqrt{3}\sin\theta \\ y\:=\:\cos\theta\,+\,\sqrt{3}\sin\theta\end{array}\)

(2) Use this parametric representation to find the slope
. . .of the tangent to the curve at (1,1) (t = 0).
Click to expand...

(1) This is just a plug-in problem.
Substitute the parametric expressions into the equation ... and see if it equals 3.


So \(\displaystyle \,x^2\,+\,xy\,+\,y^2\) becomes:
. . \(\displaystyle \left(\cos\theta\,-\,\sqrt{3}\sin\theta\right)^2\,+\,\left(\cos\theta\,-\,\sqrt{3}\sin\theta\right)\left(\cos\theta\,+\,\sqrt{3}\sin\theta\right)\,+\,\left(\cos\theta\,+\,\sqrt{3}\sin\theta\right)^2\)

\(\displaystyle =\;\left(\cos^2\theta\,-\,2\sqrt{3}\sin\theta\cos\theta\,+\,3\sin^2\theta\right)\,+\,\left(\cos^2\theta\,-\,3\sin^2\theta\right) \,+\,\left(\cos^2\theta\,+\,2\sqrt{3}\sin\theta\cos\theta\,+\,3\sin^2\theta\right)\)

\(\displaystyle =\;3\cos^2\theta\,+\,3\sin^2\theta \;=\;3\left(\cos^2\theta\,+\,\sin^2\theta\right) \;= \; 3(1) \;=\;3\;\;\) . . . Yes!


(2) You are expected to know this parametric formula:
. . . \(\displaystyle \L\frac{dy}{dx} \:=\:\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\)

We have: \(\displaystyle \,\begin{array}{cc}\frac{dy}{d\theta}\:=\:-\sin\theta\,+\,\sqrt{3}\cos\theta \\ \frac{dx}{d\theta}\:=\:-\sin\theta\,-\,\sqrt{3}\cos\theta\end{array}\;\;\) Hence: \(\displaystyle \,\frac{dy}{dx} \:=\:\frac{-\sin\theta\,+\,\sqrt{3}\cos\theta}{-\sin\theta\,-\,\sqrt{3}\cos\theta}\)

At \(\displaystyle t\,=\,0:\;\;\frac{dy}{dx}\:=\:\frac{-\sin0\,+\,\sqrt{3}\cos0}{-\sin0\,-\,\sqrt{3}\cos0} \:=\:\frac{-0\,+\,\sqrt{3}\cdot1}{-0\,-\,\sqrt{3}\cdot1} \:=\:\frac{\sqrt{3}}{-\sqrt{3}} \:=\:\fbox{-1}\)

 
Thank you so much for helping.

But aren't I trying to prove that they equal to:
------{x=cost - (3)^(1/2)sint
------{y=cost +(3)^(1/2)sint
 
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