Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/217

 Similarly, if we pass the plane BCD through P parallel to the YOZ-plane, its equation is $$x = a,$$

and for the section $$DPI, \tfrac{\partial z}{\partial y}$$ means the same as $$\tfrac{dz}{dy}$$. Hence


 * $$\frac{\partial z}{\partial y} = \frac{dz}{dy} = -\tan MT'P = $$ slope of section DI at P.

Given the ellipsoid $$\frac{x^2}{24} + \frac{y^2}{12} + \frac{z^2}{6} = 1$$; find the slope of the section of the ellipsoid made (a) by the plane $$y = 1$$ at the point where $$x = 4$$ and $$z$$ is positive; (b) by the plane $$x = 2$$ at the point where $$y = 3$$ and $$z$$ is positive.




 * Solution.
 * colspan="2"|Considering $$y$$ as constant,
 * style="text-align: right;"|$$\tfrac{2x}{24} + \tfrac{2z}{6} \tfrac{\partial z}{\partial x}$$
 * = 0, or $$\tfrac{\partial z}{\partial x} = -\tfrac{x}{4z}.$$
 * When $$x$$ is a constant
 * style="text-align: right;"|$$\tfrac{2y}{12} + \tfrac{2z}{6} \tfrac{\partial z}{\partial y}$$
 * = 0, or $$\tfrac{\partial z}{\partial y} = -\tfrac{y}{2z}.$$
 * }
 * style="text-align: right;"|$$\tfrac{2y}{12} + \tfrac{2z}{6} \tfrac{\partial z}{\partial y}$$
 * = 0, or $$\tfrac{\partial z}{\partial y} = -\tfrac{y}{2z}.$$
 * }


 * (a) When $$y = 1$$ and $$x = 4, z = \sqrt{\tfrac{3}{2}}$$. &there4; $$\tfrac{\partial z}{\partial x} = -\sqrt{\tfrac{2}{3}}.$$ Ans.


 * (b) When $$x = 2$$ and $$y = 3, z = \tfrac{1}{\sqrt{2}}$$. &there4; $$\tfrac{\partial z}{\partial y} = -\tfrac{3}{2} \sqrt{2}.$$ Ans.