1) Describe the curve traced out by a point 2 feet from the top of an 8-foot ladder as the bottom of the ladder moves away from a vertical wall.

Thank you

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- Nov 25th 2010, 07:17 AMRaeQuestions on conic sections
1) Describe the curve traced out by a point 2 feet from the top of an 8-foot ladder as the bottom of the ladder moves away from a vertical wall.

Thank you - Nov 26th 2010, 06:44 AMOpalg
Take the origin to be at the base of the wall. When the ladder makes an angle $\displaystyle \theta$ with the wall, the coordinates (x,y) of the point 2 feet from the top of the ladder are $\displaystyle (x,y) = (2\sin\theta, 6\cos\theta)$. (Draw a diagram to make sure that you understand why that is.) Thus $\displaystyle \sin\theta = \frac12x$ and $\displaystyle \cos\theta = \frac16y$. Now use the fact that $\displaystyle \sin^2\theta + \cos^2\theta = 1$ to get an equation connecting x and y.

- Nov 28th 2010, 02:10 PMRae
Could you explain how you get (x,y) = (2sin theta, 6cos theta) ?

- Nov 29th 2010, 12:26 AMOpalg
Did you draw a diagram, as I suggested???

$\displaystyle \setlength{\unitlength}{2mm}

\begin{picture}(40,40)

\put(5.5,23.5){$\bullet$}

\put(0.5,28){$ \theta$}

\put(4.5,26.5){$2$}

\put(15,13){$ 6$}

\put(7,24){$(x,y)$}

\put(6,0){\line(0,1){24}}

\put(0,24){\line(1,0){6}}

\thicklines

\put(0,0){\line(0,1){40}}

\put(0,0){\line(1,0){40}}

\put(24,0){\line(-3,4){24}}

\end{picture}$ - Dec 2nd 2010, 04:15 PMRae
I did draw a diagram, but I guess I was thinking about the problem wrong because I didn't see it that way. Thanks for your help.