tangent plane to $\sqrt{x} + \sqrt{y} + \sqrt{z} = \sqrt{C}$

Question

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.

Answer 1

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 $P_1+Q_1+R_1$.

Answer 2

Your surface, $z=f(x,y)$ is implicitly defined by the equation $\sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{C}$. The tangent plane to the surface is given by:

$z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$

Calculate the partial derivatives using implicit differentiation and you are done.

Answer 3

There is a very easy way of getting this with some more conceptual reasoning (essentially, the chain rule). Namely, we define intermediate variables $u = \sqrt x$, $v = \sqrt y$, $w = \sqrt z$; then the given equation is $u + v + w = \sqrt C$ in $(u,v,w)$-space, which is a plane, so is its own tangent plane everywhere. Let's call this plane $P$. Furthermore, we have a map $$f \colon \mathbb{R}^3_{(x,y,z)} \to \mathbb{R}^3_{(u,v,w)}, \qquad f(x,y,z) = (\sqrt x, \sqrt y, \sqrt z) $$ by definition of $(u,v,w)$, and its Jacobian matrix at $(x_0, y_0, z_0)$ (otherwise known as its derivative) is a linear transformation sending the tangent plane, let's call it $T_{(x_0,y_0,z_0)}$, in $(x,y,z)$-space to that in $(u,v,w)$-space. In other words, it sends $T_{(x_0,y_0,z_0)}$ to $P$. Of course, the Jacobian is easy to compute for this map: $$J = Df_{(x_0,y_0,z_0)} = \begin{pmatrix} 1/2\sqrt{x_0} & 0 & 0 \\ 0 & 1/2\sqrt{y_0} & 0 \\ 0 & 0 & 1/2\sqrt{z_0} \end{pmatrix}$$ This means that we have $$du = \frac{1}{2\sqrt{x_0}} dx \qquad dv = \frac{1}{2\sqrt{y_0}} dy \qquad dw = \frac{1}{2\sqrt{z_0}} dz $$ so replacing $du = u - f(x_0,y_0,z_0) = u - \sqrt{z_0}$ and so on, and $dx = x - x_0$ and so on, we transform the equation $u + v + w = \sqrt{C}$ back to: $$\frac{x - x_0}{2\sqrt{x_0}} + \frac{y - y_0}{2 \sqrt{y_0}} + \frac{z - z_0}{2 \sqrt{z_0}} = \sqrt{C}.$$ That is the equation of your tangent plane, which I called $T_{(x_0,y_0,z_0)}$.