Let $ S = \left\{ (z,y,z) \in R^3 : \sqrt{x} + \sqrt{y} + \sqrt{z} = \sqrt{C} \right\} $ be a surface.
a) Find the tangent plane to $S$ at $(x_{0}, y_{0}, z_{0})$.
b) Let $P_{0}, Q_{0}$ and $R_{0}$ be the points where the tangent plane to $S$ at $(x_{0}, y_{0}, z_{0})$ crosses the axes. If $P_{1}, Q_{1}$ and $R_{1}$ are the points where the tangent plane to $S$ at other point $(x_{1}, y_{1}, z_{1})$ crosses the axes , prove that $P_{0} + Q_{0} + R_{0} =P_{1} + Q_{1} + R_{1}$
I'm not sure how to even start. If I want to compute the tangent plane by using partial derivatives I need a function first, so I was thinking of:
$ \sqrt{x} + \sqrt{y} + \sqrt{z} = \sqrt{C} \iff (\sqrt{x} + \sqrt{y} + \sqrt{z})^2 = C \iff f(x,y,z) = C \iff f(x,y,z) = (\sqrt{x} + \sqrt{y} + \sqrt{z})^2 $
And now I would compute the tangent plane with $\nabla{f}(x_o,y_o,z_o) \cdot (x-x_o,y-y_o,z-z_o)$
Does this make any sense? Or can I just take partial derivatives of $\sqrt{x} + \sqrt{y} + \sqrt{z} = \sqrt{C}$?
For the second part, I'm even more lost. I guess I need to figure out how $P_{x}, Q_{x}$ and $R_{x}$ for any given tangent plane and prove what we asked.
Best Answer
Use the tangent plane formula you gave, plus the right side should equal to zero by geometric meaning of tangent, i.e., perpendicular to the normal $\nabla f(x_0,y_0,z_0)$: $$ \nabla{f}(x_0,y_0,z_0) \cdot (x-x_0,y-y_0,z-z_0) = 0, $$ directly use $f = \sqrt{x} + \sqrt{y} + \sqrt{z}$, we have: $$ (\frac{1}{2\sqrt{x_0}} ,\frac{1}{2\sqrt{y_0}} ,\frac{1}{2\sqrt{z_0}} )\cdot (x-x_0,y-y_0,z-z_0) = 0, $$ or $$ \frac{x-x_0}{\sqrt{x_0}}+\frac{y-y_0}{\sqrt{y_0}}+\frac{z-z_0}{\sqrt{z_0}} = 0.\tag{1} $$ This is the formula for tangent plane.
Now onto finding $P_0,Q_0,R_0$: for a point on $x$-axis, its $y$- and $z$-coordinates are zero in (1): $$ \frac{x-x_0}{\sqrt{x_0}}+\frac{-y_0}{\sqrt{y_0}}+\frac{-z_0}{\sqrt{z_0}} = 0, $$ solving for the $x$-coordinate for $P_0$ leads: $$ x = \sqrt{x_0}(\sqrt{x_0} + \sqrt{y_0}+\sqrt{z_0}), $$ similarly we have for $Q_0$'s $y$-coordinate, and $R_0$'s $z$-coordinate: $$ y = \sqrt{y_0}(\sqrt{x_0} + \sqrt{y_0}+\sqrt{z_0}), \quad\text{ and }\quad z = \sqrt{z_0}(\sqrt{x_0} + \sqrt{y_0}+\sqrt{z_0}). $$ by $P_0,Q_0,R_0$, I am assuming you are referring to the $x$-, $y$-, $z$-coordinates of the intercept: $$ P_0+Q_0+R_0 = (\sqrt{x_0} + \sqrt{y_0}+\sqrt{z_0})^2 = C, $$ which is a constant. This implies for any point on this surface, this sum for its tangent plane is a constant, hence the sum is the same for $P_1+Q_1+R_1$.