Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/231

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 In case the rectangular equation of the envelope is required we may either eliminate the parameter from the parametric equations of the envelope, or else eliminate the parameter from the given equation (B) of the family and the partial derivative (H).

Find the envelope of the family of straight lines $$x\,\cos\,\alpha + y\,\sin\,\alpha = p$$, $$\alpha$$ being the variable parameter.
 * Solution.


 * {| style="width:100%"


 * style="width:3%"|
 * align="center" |$$x\,\cos\, \alpha + y\, \sin\, \alpha = p.$$
 * }
 * First step. Differentiating (A) with respect to $$\alpha$$,


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 * style="width:3%"|
 * align="center" |$$-x \,\sin\, \alpha + y\, \cos\, \alpha = 0.$$
 * }


 * Second step. Multiplying (A) by $$\cos\,\alpha$$ and (B) by $$\sin\,\alpha$$ and subtracting, we get


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 * style="width:3%"|
 * align="center" |$$x = p\,\cos\,\alpha.$$
 * }
 * Similarly, eliminating $$x$$ between (A) and (B), we get


 * {| style="width:100%"


 * style="width:3%"|
 * align="center" |$$y = p\,\sin\,\alpha.$$
 * }




 * The parametric equations of the envelope are therefore


 * {| style="width:100%"


 * style="width:3%"|
 * align="center" |$$\begin{cases}x=p\,\cos\,\alpha,\\y=p\,\sin\,\alpha \end{cases} $$
 * }
 * $$\alpha$$ being the parameter. Squaring equations (C) and adding, we get

$x^2 + y^2 = p^2$


 * the rectangular equation of the envelope, which is a circle.

Find the envelope of a line of constant length $$a$$, whose extremities move along two fixed rectangular axes.
 * Solution. Let $$AB = a$$ in length, and let


 * {| style="width:100%"


 * style="width:3%"|
 * align="center" |$$x\, \cos\,\alpha + y\, \sin\, \alpha - p = 0 $$
 * }


 * be its equation. Now as $$AB$$ moves always touching the two axes, both $$\alpha$$ and $$p$$ will vary. But $$p$$ may be found in terms of $$\alpha$$. For $$AO = AB\, \cos\, \alpha = a\, \cos\, \alpha$$, and $$p = AO\, \sin\, \alpha = a\, \sin\, \alpha\, \cos\, \alpha$$. Substituting in (A),


 * {| style="width:100%"


 * style="width:3%"|
 * align="center" |$$x\, \cos\, \alpha + y\, \sin\, \alpha - a\, \sin\, \alpha\, \cos\, \alpha = 0,$$
 * }


 * where $$\alpha$$ is the variable parameter. Differentiating (B) with respect to $$\alpha$$,


 * {| style="width:100%"


 * style="width:3%"|
 * align="center" |$$- x\, \sin\, \alpha + y\, \cos\, \alpha + a\, \sin^2\, \alpha - a\, \cos^2\, \alpha = 0.$$
 * }


 * Solving (B) and (C) for $$x$$ and $$y$$ in terms of $$\alpha$$, we get


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 * style="width:3%"|
 * align="center" |$$\begin{cases}x = a\, \sin^3\, \alpha\\y = a\, \cos^3\, \alpha\end{cases} $$
 * }


 * the parametric equations of the envelope, a hypocycloid.




 * The corresponding rectangular equation is found from equations (D) by eliminating $$\alpha$$ as follows:


 * {| style="width:100%"

x^{\frac{2}{3}} = a^{\frac{2}{3}}\, \sin^2\, \alpha.\\ y^{\frac{2}{3}} = a^{\frac{2}{3}}\, \cos^2\, \alpha. \end{align}$$
 * align="center" |$$\begin{align}
 * }


 * {| style="width:100%"


 * style="width:3%"|Adding,
 * align="center" |$$x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}},$$
 * }


 * the rectangular equation of the hypocycloid.