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Section 5.1 Tangent Planes

Recall that the graph associated with the function \(z=f(x,y)\) is a surface in \(\mathbb{R}^3\) (that passes the vertical line test). Wherever this surface does not have any discontinuities or cusp-like points it will have a tangent plane. Like the tangent line to the graph of a function of one variable, the tangent plane to the function \(z=f(x,y)\) at the point \((x_0,y_0)\) is the plane that “just touches” the surface at the point \((x_0,y_0,z_0)\text{.}\)

For the function \(f(x,y) = 5-\dfrac{x^2+y^2}{2}\) the graph below shows the graph of the function and it's tangent plane at the point \((x,y) = (2,1)\text{.}\)

Figure 5.1.2.

To find the equation of the plane tangent to the function \(z=f(x,y)\) at the point \((x_0,y_0,z_0)\text{,}\) firstly recall that the equation of a plane in Cartesian form is given by

\begin{equation*} ax+by+cz = k, \quad a,b,c,k \in \mathbb{R}, \end{equation*}

or in normal form, as illustrated in Figure 5.1.3, by

\begin{equation} \mathbf{n} \cdot (\mathbf{r}-\mathbf{r}_0) = 0\label{Eqn-Tangent_Plane_Vector_Form}\tag{5.1.1} \end{equation}

where \(\mathbf{n} = \langle a,b,c \rangle\) is a normal vector to the plane, \(\mathbf{r}_0\) is the position vector of a point on the plane and \(\mathbf{r} = \langle x,y,z \rangle\text{.}\)

Figure 5.1.3.

Find the equation of the tangent plane to the function \(f(x,y) = 5-\dfrac{x^2+y^2}{2}\) at \((x,y) = (2,1)\text{.}\)

Answer.

\(2x+y+z=\dfrac{15}{2}\)

Solution.

Now, we know that the partial derivative \(f_x(2,1)\) gives the slope of the tangent at \(x=2\text{,}\) to the curve of intersection of the surface associated with \(f\) and the plane \(y=1\text{.}\)

Figure 5.1.5.

Since this tangent line lies in the plane tangent to \(f\) at \((x,y) = (2,1)\) the vector

\begin{equation*} \langle 1,0,f_x(2,1) \rangle \end{equation*}

will be a vector that is parallel to the tangent plane, (or lies in the tangent plane if we place it’s tail at the point \((2,1,\frac{5}{2})\)). Similarly, the vector

\begin{equation*} \langle 0,1,f_y(2,1) \rangle \end{equation*}

will be another vector parallel to the tangent plane. Since these two non-parallel vectors are parallel to the tangent plane, their vector product will give a vector normal to the tangent plane, i.e.

\begin{align*} \mathbf{n} \amp = \langle 1,0,f_x(2,1) \rangle \times \langle 0,1,f_y(2,1) \rangle\\ \amp = \langle -f_x(2,1), -f_y(2,1), 1 \rangle\\ \amp = \langle 2,1,1 \rangle \end{align*}

Thus, using equation (5.1.1), the equation of the plane tangent to \(f(x,y) = 5 - \dfrac{x^2+y^2}{2}\) at \((x,y) = (2,1)\) is

\begin{equation*} \langle 2,1,1 \rangle \cdot \left( \langle x,y,z \rangle - \langle 2,1,\frac{5}{2} \rangle \right) = 0 \end{equation*}

which simplifies to

\begin{equation*} 2x+y+z=\frac{15}{2}\text{.} \end{equation*}

In general, to find the equation of the plane tangent to the function \(f(x,y)\) at the point \((x_0,y_0)\text{,}\) note that the vectors

\begin{equation*} \langle 1, 0, f_x(x_0,y_0) \rangle \: \text{and} \: \langle 0,1,f_y(x_0,y_0) \rangle \end{equation*}

lie in the tangent plane and hence a normal to the plane is

\begin{align*} \mathbf{n} \amp = \langle 1, 0, f_x(x_0,y_0) \rangle \times \langle0,1,f_y(x_0,y_0) \rangle\\ \amp = \langle -f_x(x_0,y_0), -f_y(x_0,y_0),1 \rangle\text{.} \end{align*}

Since the point \((x_0,y_0,z_0)\) lies on the plane, using equation (5.1.1), the equation of the tangent plane is

\begin{equation*} \langle -f_x(x_0,y_0), -f_y(x_0,y_0),1 \rangle \cdot \bigg( \langle x,y,z \rangle - \langle x_0, y_0, z_0 \rangle \bigg) = 0. \end{equation*}

On expanding and rearranging this we get the following result.

Find the equation of the plane tangent to \(z = x^2+2y^2\) at the point \((x,y)=(1,2)\text{.}\)

Figure 5.1.8.
Answer.

\(2x+8y-z-9=0\)

Solution.

Firstly, note that when \((x,y)=(1,2)\text{,}\) \(z=9\text{.}\) Now

\begin{equation*} z_x = 2x \: \text{ and so } \: z_x(1,2) = 2 \end{equation*}

and

\begin{equation*} z_y = 4y \: \text{ and so } \: z_y(1,2) = 8\text{.} \end{equation*}

Thus the equation of the tangent plane is

\begin{equation*} z = 9+2(x-1)+8(y-2) \end{equation*}

which simplifies to

\begin{equation*} 2x+8y-z-9=0\text{.} \end{equation*}

Find the equation of the line normal to the graph of the function \(z=x^2+2y^2\) at the point \((x,y) = (1,2)\text{.}\)

Answer.

\(\mathbf{r} = \langle 1, 2, 9 \rangle + t \langle -2, -8, 1 \rangle\)

Solution.

Recall that the vector equation of a line in \(\mathbb{R}^3\) is

\begin{equation*} \mathbf{r} = \mathbf{r}_0 + t \mathbf{d}, \: t \in \mathbb{R} \end{equation*}

where \(\mathbf{r} = \langle x, y, z \rangle\) is the position vector of a general point, \(\mathbf{r}_0\) is the position vector of a point that lies on the line and \(d\) is a direction vector for the line (i.e. a vector that is parallel to the line).

We know that a vector normal to the surface \(z=f(x,y)\) at the point \((x_0,y_0,z_0)\) is given by

\begin{equation*} \mathbf{n} = \langle -f_x(x_0,y_0), -f_y(x_0,y_0),1 \rangle\text{.} \end{equation*}

For the function \(z=x^2+2y^2\text{,}\) \(z_x=2x\) and \(z_y=4y\text{.}\) Thus a direction vector for the line normal to \(z=x^2+2y^2\) at the point \((x,y)=(1,2)\) will be

\begin{equation*} \mathbf{d} = \langle -2, -8, 1 \rangle\text{.} \end{equation*}

Since the normal line passes through the point \((1,2,9)\) its equation is

\begin{equation*} \mathbf{r} = \langle 1, 2, 9 \rangle + t \langle -2, -8, 1 \rangle\text{.} \end{equation*}

Exercises Example Tasks

1.

Find the equation of the tangent plane to \(z=x^2-5xy+2y^2\) at the point \((x,y)=(-1,-2)\text{.}\)

2.

Find the equation of the tangent plane and normal line to \(z = \sqrt{x+y} \sin (xy)\) at the point \((x,y)=(0,1)\text{.}\)

3.

Show that every line that is normal to the sphere \(x^2+y^2+z^2 = 1\) passes through the origin.