Rate of change of a function

Question

Consider the surface $$ (S): z=f(x, y)=\dfrac{1}{x^2+y^2} $$ and the point $P_0\left(1, 1, \dfrac12 \right)$. Find the rate of change of $f$ at the point $P_0$ in the direction of the vector $\vec{u}=\hat{\imath}+\hat{\jmath}$.

I start solving in a straightforward proof, first we have $$ \|\vec{u}\|=\sqrt{1+1}=\sqrt{2} \implies \vec{v}=\dfrac{1}{\sqrt{2}}\hat{\imath}+\dfrac{1}{\sqrt{2}}\hat{\jmath} $$ and $$ \vec{\nabla f}=\left(\dfrac{-2x}{(x^2+y^2)^2}, \dfrac{-2y}{(x^2+y^2)^2}\right) \implies \vec{\nabla f}(P_0)=-\dfrac{1}{2}\hat{\imath}-\dfrac{1}{2}\hat{\jmath}. $$ Consequently, $$ D_{\vec{v}}f(P_0)=\vec{\nabla f}(P_0) \cdot \vec{v}=-\dfrac{\sqrt{2}}{2} $$

But my question is, if this answer is correct, why on introduce a point of 3 dimensions? when I find the directional derivative, do I need to introduce a new function $F(x,y,z)=f(x,y)-z$? Is my answer correct?

Next, how one can write the equation of the plane parallel to (S) at $P_0$?

Answer 1

I think the wording is "tangent to S".

Let's find the tangent plane of the surface $S: z=\frac{1}{x^2+y^2}$ at its point $P_0(1,1,\frac{1}{2})$.

A plane in $3D$ is determined by two things: a vector perpendicular to it which is called a normal vector of the plane and a point on it. What is the analytic geometry formula then? The plane passing through the point $P_0$ and with normal vector $\vec{n}$ is given by the quation $$\vec{n}\cdot \vec{P}(x,y,z)=\vec{n}\cdot\vec{P_0}(x_0,y_0,z_0).$$

The tangent plane is passing through $P_0$ so we need to find its normal vector: The gradient of a function $F(x,y,z)$ at $P_0(x_0,y_0,z_0)$ is a normal vector of the surface $F(x,y,z)=c$ for any constant $c$.

Our surface $S$ can be written as $F(x,y,z)=z-\frac{1}{x^2+y^2}=0.$ then $\vec\nabla F(x,y,z)=<\frac{2x}{(x^2+y^2)^2},\frac{2y}{(x^2+y^2)^2},1>$ and so $\vec{n}=\vec\nabla F(1,1,\frac{1}{2})=<\frac{1}{2},\frac{1}{2},1>$.

Hence, the tangent plane is $<\frac{1}{2},\frac{1}{2},1>\cdot<x,y,z>=<\frac{1}{2},\frac{1}{2},1>\cdot<1,1,\frac{1}{2}>$ which in the smpliest form is $x+y+2z=3$.

Hard question. I made many mistakes taking derivatives.

Answer 2

Strictly when you write $\vec{\nabla}f(P_{0})$ does not make sense since $f$ has the input $(x,y)\in \mathbf{R}^{2}$ but $P_{0}\in \mathbf{R}^{3}$. Instead, you can consider the surface of level as $F(x,y,z)=\frac{1}{x^{2}+y^{2}}-z=0$, then $\nabla F(P_{0})=(-1/2,-1/2,-1)$ and then $D_{u}f(P_{0})=\nabla F(P_{0})\cdot (1/\sqrt{2},1/\sqrt{2},0)= -\sqrt{2}/2$. On other hand, if the second question is find the plane tangent to surface $S$ at the point $P_{0}(x_{0},y_{0},z_{0})$, so since $\nabla F(P_{0})=(a,b,c)$ is normal to $S$ at $P_{0}$ by geometry of the gradient vector, then the plane tangent to $S$ at $P_{0}$ is given by $a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=0$.