# Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/230

206
DIFFERENTIAL CALCULUS

the parameter $\alpha$ being the same in both cases. The slope of (A) at any point is

 (C) $\frac{dy}{dx}=\frac{\psi'(\alpha)}{\phi'(\alpha)},$ (D), p. 80

and the slope of (B) at any point is

 (D) $\frac{dy}{dx}=- \frac{f'_x(x,\,y,\,\alpha)}{f'_y(x,\,y,\,\alpha)}.$ (57a), p. 199

Hence if the curves (A) and (B) are tangent, the slopes (C) and (D) will be equal (for the same value of $\alpha$), giving

$\frac{\psi'(\alpha)}{\phi'(\alpha)} = - \frac{f'_x(x,\,y,\,\alpha)}{f'_y(x,\,y,\,\alpha)},$ or

 (E) $f'_x(x,\,y,\,\alpha)\phi'(\alpha) + f'_y(x,\,y,\,\alpha)\psi'(\alpha)=0.$

By hypothesis (A) and (B) are tangent for every value of $\alpha$; hence for all values of $\alpha$ the point $(x,\, y)$ given by (A) must lie on a curve of the family (B). If we then substitute the values of $x$ and $y$ from (A) in (B), the result will hold true for all values of $\alpha$; that is,

 (F) $f[\phi(\alpha),\,\psi(\alpha),\,\alpha] = 0.$

The total derivative of (F) with respect to $\alpha$ must therefore vanish, and we get

 (G) $f'_x(x,\,y,\,\alpha)\phi'(\alpha) + f'_y(x,\,y,\,\alpha)\psi'(\alpha) + f'_\alpha(x,\,y,\,\alpha) = 0,$

where $x = \phi(\alpha),\,y = \psi(\alpha)$.

Comparing (E) and (G) gives

 (H) $f'_\alpha(x,\,y,\,\alpha) = 0.$

Therefore the equations of the envelope satisfy the two equations (B) and (H), namely,

 (I) $f(x,\,y,\,\alpha) = 0$ and $f'_\alpha(x,\,y,\,\alpha) = 0;$

that is, the parametric equations of the envelope may be found by solving the two equations (I) for $x$ and $y$ in terms of the parameter $\alpha$.

General directions for finding the envelope.

First Step. Differentiate with respect to the variable parameter, considering all other quantities involved in the given equation as constants.

Second Step. Solve the result and the given equation of the family of curves for $x$ and $y$ in terms of the parameter. These solutions will be the parametric equations of the envelope.