Proposition 4.3.1.
Let \(f(x,y)\) be a function \(\RR^2 \rightarrow
\RR\text{.}\) The equation of the tangent plane to \(f\) at \((a,b,f(a,b))\) is
\begin{equation*}
z - f(a,b) = f_x(a,b) (x-a) + f_y(a,b) (y-b)
\end{equation*}
Section 4.3 Tangent Planes
Subsection 4.3.1 Generalizing Tangents
In single variable calculus, derivatives calculated the slopes and equations of tangent lines to the graph of a function. I want to extend this idea. For functions of two variables, graphs are surfaces in \(\RR^3\) instead of curves in \(\RR^2\text{.}\) These surfaces have tangent planes instead of tangent lines.
Consider a function \(f(x,y)\) and look at a particular point \((a,b,f(a,b))\) on the graph of the function. I can calculate the partial derivatives \(f_x(a,b)\) and \(f_y(a,b)\text{.}\) At the point \((a,b,f(a,b))\text{,}\) these two partial derivatives calculate the rate of change in \(x\) and in \(y\text{.}\) That’s the slope of a tangent line in the \(x\) direction and a tangent line in the \(y\) direction. I’d rather have direction vectors than slopes, but I can construct these. For the \(x\) direciton, the \(y\) coordinate is \(0\) since there is no change in \(y\text{.}\) That gives the vector \((1, 0, f_x(a,b))\text{.}\) Likewise in the \(y\) direction, I can construct the vector \((0,1,f_y(a,b))\text{.}\) These are two local tangent direction vectors.
Given two vectors on a plane, the normal of the plane is found by the cross product. So I calculate the normal to the tangent plane.
\begin{equation*}
(1,0, f_x(a,b)) \times (0,1, f_y(a,b)) = (-f_x(a,b),
-f_y(a,b),1)\text{.}
\end{equation*}
Then I can use this normal to construct the equation of a plane in the usual way.
Proof.
I just calculated the normal at any point \((a,b,f(a,b))\) on the graph of the function. That vector was \((-f_x(a,b),
-f_y(a,b),1)\text{.}\) The equation of the plane is given by the dot product of the variables with the normal. Here is this dot product, with an unknown value \(c\text{.}\)
\begin{equation*}
- f_x(a,b) x - f_y(a,b) y + z = c
\end{equation*}
I already know a point on the plane: \((a,b,f(a,b))\text{.}\) By substitution, I can solve for \(c\text{.}\)
\begin{equation*}
c = f(a,b) - f_x(a,b) a - f_y(a,b) b
\end{equation*}
Putting this \(c\) in gives the equation of the plane.
\begin{equation*}
- f_x(a,b) x - f_y(a,b) y + z = f(a,b) - f_x(a,b)a -
f_y(a,b)b
\end{equation*}
From here, is it just a re-arrangement to get the form in the proposition.
Subsection 4.3.2 Tangent Plane Examples
Example 4.3.2.
Consider \(f(x,y) = \frac{1}{1 + x^2 + y^2}\text{.}\)
\begin{align*}
\frac{\del f}{\del x} \amp = \frac{-2x}{(1+x^2+y^2)^2}\\
\frac{\del f}{\del y} \amp = \frac{-2y}{(1+x^2+y^2)^2}
\end{align*}
At the point \((x,y) = (1,1)\text{,}\) the partial derivative values are \(f_x(1,1) = \frac{-2}{9}\) and \(f_y(1,1) =
\frac{-2}{9}\text{.}\) The normal is \(\left( \frac{2}{9},
\frac{2}{9}, 1 \right)\) and the point is \(\left(1,1,
\frac{1}{3} \right)\text{.}\) The tangent plane is
\begin{gather*}
\frac{-2}{9} (x-1) + \frac{-2}{9} (y-1) = z - \frac{1}{3} \\
\frac{2}{9} x + \frac{2}{9} y + z = \frac{7}{9}\text{.}
\end{gather*}
At the point \((x,y) = (0,0)\text{,}\) the partial derivative values are \(f_x(0,0) = 0\) and \(f_y(0,0) =
0\text{.}\) The normal is \((0,0,1)\) and the point is \((0,0,1)\text{.}\) The tangent plane is
\begin{equation*}
z=1\text{.}
\end{equation*}
At the point \((x,y) = (-2,2)\text{,}\) the partial derivative values are \(f_x(-2,2) = \frac{4}{81}\) and \(f_y(-2,2) = \frac{-4}{81}\text{.}\) The normal is \(\left(
\frac{4}{81}, \frac{-4}{81}, 1 \right)\) and the point is \(\left(-2,2, \frac{1}{9} \right)\text{.}\) The tangent plane is
\begin{equation*}
\frac{-4}{81} x + \frac{4}{81} y + z = \frac{1}{9}\text{.}
\end{equation*}
Subsection 4.3.3 Higher Dimensions
The definition of tangent planes for \(f: \RR^2 \rightarrow
\RR\) extends to many dimensions. A function \(f: \RR^3 \rightarrow \RR\) has a graph in \(\RR^4\text{.}\) Its tangent spaces are 3-spaces in \(\RR^4\text{.}\) I can understand those 3-spaces in a very similar method. I calcluate the three local tangent directions.
\begin{equation*}
v_1 = \left( 1, 0, 0, \frac{\del f}{\del x} \right)
v_2 = \left( 0, 1, 0, \frac{\del f}{\del y}
\right) v_3 = \left( 0, 0, 1, \frac{\del f}{\del
z} \right)
\end{equation*}
There isn’t a cross-product in \(\RR^4\text{,}\) but I can genearlize the pattern in this case to get the normal to the tangent 3-space.
\begin{equation*}
\left( - \frac{\del f}{\del x}, - \frac{\del f}{\del y}, -
\frac{\del f}{\del z}, 1 \right)
\end{equation*}
The equation of the tangent 3-space at \((a,b,c,f(a,b,c))\) is
\begin{equation*}
w - f(a,b,c) = \frac{\del f}{\del x}(a,b,c) (x-a) + \frac{\del
f}{\del y} (a,b,c) (y-b) + \frac{\del f}{\del z} (a,b,c)
(z-c)
\end{equation*}
And I could extend this to \(f: \RR^n \rightarrow \RR\text{,}\) which has a tangent hyperplane in \(\RR^{n+1}\text{.}\)
\begin{equation*}
v_1 = \left( 1, 0, \ldots, \frac{\del f}{\del x_1} \right)
v_2 = \left( 0, 1, 0, \ldots, \frac{\del f}{\del
x_2} \right) \ldots v_n = \left( 0,
\ldots, 0, 1, \frac{\del f}{\del x_n} \right)
\end{equation*}
The normal to the tangent hyperplane is
\begin{equation*}
\left( - \frac{\del f}{\del x_1}, - \frac{\del f}{\del x_2},
\ldots, - \frac{\del f}{\del x_n}, 1 \right)\text{.}
\end{equation*}
The equation of the tangent hyperplane at \((a_1, a_2, a_3,
\ldots, a_n, f(a_1, a_2, \ldots, a_n))\) is
\begin{align*}
x_0 - f(a_1, a_2, \ldots a_n) \amp = \frac{\del f}{\del
x_1}(a_1, a_2, \ldots a_n ) (x_1-a_2) + \\
\amp \frac{\del f}{\del x_2} (a_1, a_2, \ldots, a_n)
(x_2-a_2) + \ldots\\
\amp \ldots + \frac{\del f}{\del x_n} (a_1,
a_2, \ldots, a_n) (x_n - a_n) \text{.}
\end{align*}
To connect tangent (hyper)planes to tangents to parametric curves and derivatives along those curves, consider the following result.
Proposition 4.3.3.
Let \(\gamma(t)\) be a parametric curve in \(\RR^{n+1}\) and \(f: \RR^n \rightarrow \RR\) a differentiable function. Then if \(\gamma(t)\) lies on the graph of \(f\text{,}\) the tangents to \(\gamma(t)\) must lie on the tangent planes to the graph of \(f\text{.}\) (All these tangent vectors are local direction vectors).
Example 4.3.4.
Consider the same function as Example 4.3.2: \(f(x,y) =
\frac{1}{1+x^2+y^2}\text{.}\) Then consider the parametric curves \(\gamma_1(t) = \left(t,1, \frac{1}{2+t^2}
\right)\) and \(\gamma_2(t) = \left(1, t, \frac{1}{2+t^2}
\right)\text{.}\) It is easy to check that both curves lie on the graph of \(f\) and both pass through the point \(\left(1,1,\frac{1}{3} \right)\) at \(t=1\text{.}\) Then I can calculate the tangents to the curves at that point and the plane they span.
\begin{align*}
\gamma_1^\prime(t) \amp = \left( 1, 0
\frac{-2t}{(2+t^2)^2} \right) \\
\gamma_2^\prime(t) \amp = \left( 0, 1
\frac{-2t}{(2+t^2)^2} \right) \\
\gamma_1^\prime(1) \amp = \left( 1, 0, \frac{-2}{9}
\right) \\
\gamma_2^\prime(1) \amp = \left( 0, 1, \frac{-2}{9}
\right)\\
\gamma_1^\prime(1) \times \gamma_2^\prime(1) \amp =
\left( \frac{2}{9}, \frac{2}{9}, 1 \right)
\end{align*}
This gives exactly the same normal at the same point \(\left( 1,1,\frac{1}{3} \right)\text{,}\) so the same plane.
There is a key idea here, relating tangent planes for scalar fields to the calculus of parameteric curves from Section 2.2. A curve can lie on the graph of a function as long as the components of the curve satisfy the function: \(z(t) = f(x(t),y(t))\text{.}\) The above calculation was an example of the tangent of these curves. Their tangent vectors ended up being local direction vectors that lived in the tangent plane of the scalar fields. This can be a conceptual definition of the tangent plane: the tanget plane is the environment which is home to tangent to curves which lie on the graph of the function.
