Calculus III - Tangent Planes and Linear Approximations

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Section 14.1 : Tangent Planes and Linear Approximations

Earlier we saw how the two partial derivatives ${f_x}$ and ${f_y}$ can be thought of as the slopes of traces. We want to extend this idea out a little in this section. The graph of a function $z = f\left( {x,y} \right)$ is a surface in ${\mathbb{R}^3}$ (three dimensional space) and so we can now start thinking of the plane that is “tangent” to the surface as a point.

Let’s start out with a point $\left( {{x_0},{y_0}} \right)$ and let’s let ${C_1}$ represent the trace to $f\left( {x,y} \right)$ for the plane $y = {y_0}$ (i.e. allowing $x$ to vary with $y$ held fixed) and we’ll let ${C_2}$ represent the trace to $f\left( {x,y} \right)$ for the plane $x = {x_0}$ (i.e. allowing $y$ to vary with $x$ held fixed). Now, we know that ${f_x}\left( {{x_0},{y_0}} \right)$ is the slope of the tangent line to the trace ${C_1}$ and ${f_y}\left( {{x_0},{y_0}} \right)$ is the slope of the tangent line to the trace ${C_2}$ . So, let ${L_1}$ be the tangent line to the trace ${C_1}$ and let ${L_2}$ be the tangent line to the trace ${C_2}$ .

The tangent plane will then be the plane that contains the two lines ${L_1}$ and ${L_2}$ . Geometrically this plane will serve the same purpose that a tangent line did in Calculus I. A tangent line to a curve was a line that just touched the curve at that point and was “parallel” to the curve at the point in question. Well tangent planes to a surface are planes that just touch the surface at the point and are “parallel” to the surface at the point. Note that this gives us a point that is on the plane. Since the tangent plane and the surface touch at $\left( {{x_0},{y_0}} \right)$ the following point will be on both the surface and the plane.

$\left( {{x_0},{y_0},{z_0}} \right) = \left( {{x_0},{y_0},f\left( {{x_0},{y_0}} \right)} \right)$

What we need to do now is determine the equation of the tangent plane. We know that the general equation of a plane is given by,

$a\left( {x - {x_0}} \right) + b\left( {y - {y_0}} \right) + c\left( {z - {z_0}} \right) = 0$

where $\left( {{x_0},{y_0},{z_0}} \right)$ is a point that is on the plane, which we have. Let’s rewrite this a little. We’ll move the $x$ terms and $y$ terms to the other side and divide both sides by $c$ . Doing this gives,

$z - {z_0} = - \frac{a}{c}\left( {x - {x_0}} \right) - \frac{b}{c}\left( {y - {y_0}} \right)$

Now, let’s rename the constants to simplify up the notation a little. Let’s rename them as follows,

$A = - \frac{a}{c}\hspace{0.25in}B = - \frac{b}{c}$

With this renaming the equation of the tangent plane becomes,

$z - {z_0} = A\left( {x - {x_0}} \right) + B\left( {y - {y_0}} \right)$

and we need to determine values for $A$ and $B$ .

Let’s first think about what happens if we hold $y$ fixed, i.e. if we assume that $y = {y_0}$ . In this case the equation of the tangent plane becomes,

$z - {z_0} = A\left( {x - {x_0}} \right)$

This is the equation of a line and this line must be tangent to the surface at $\left( {{x_0},{y_0}} \right)$ (since it’s part of the tangent plane). In addition, this line assumes that $y = {y_0}$ (i.e. fixed) and $A$ is the slope of this line. But if we think about it this is exactly what the tangent to ${C_1}$ is, a line tangent to the surface at $\left( {{x_0},{y_0}} \right)$ assuming that $y = {y_0}$ . In other words,

$z - {z_0} = A\left( {x - {x_0}} \right)$

is the equation for ${L_1}$ and we know that the slope of ${L_1}$ is given by ${f_x}\left( {{x_0},{y_0}} \right)$ . Therefore, we have the following,

$A = {f_x}\left( {{x_0},{y_0}} \right)$

If we hold $x$ fixed at $x = {x_0}$ the equation of the tangent plane becomes,

$z - {z_0} = B\left( {y - {y_0}} \right)$

However, by a similar argument to the one above we can see that this is nothing more than the equation for ${L_2}$ and that it’s slope is $B$ or ${f_y}\left( {{x_0},{y_0}} \right)$ . So,

$B = {f_y}\left( {{x_0},{y_0}} \right)$

The equation of the tangent plane to the surface given by $z = f\left( {x,y} \right)$ at $\left( {{x_0},{y_0}} \right)$ is then,

$z - {z_0} = {f_x}\left( {{x_0},{y_0}} \right)\left( {x - {x_0}} \right) + {f_y}\left( {{x_0},{y_0}} \right)\left( {y - {y_0}} \right)$

Also, if we use the fact that ${z_0} = f\left( {{x_0},{y_0}} \right)$ we can rewrite the equation of the tangent plane as,

$\begin{align*}z - f\left( {{x_0},{y_0}} \right) & = {f_x}\left( {{x_0},{y_0}} \right)\left( {x - {x_0}} \right) + {f_y}\left( {{x_0},{y_0}} \right)\left( {y - {y_0}} \right)\\ z & = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)\left( {x - {x_0}} \right) + {f_y}\left( {{x_0},{y_0}} \right)\left( {y - {y_0}} \right)\end{align*}$

We will see an easier derivation of this formula (actually a more general formula) in the next section so if you didn’t quite follow this argument hold off until then to see a better derivation.

Example 1 Find the equation of the tangent plane to

$z = \ln \left( {2x + y} \right)$ at

$\left( { - 1,3} \right)$ .

Show Solution

There really isn’t too much to do here other than taking a couple of derivatives and doing some quick evaluations.

$\begin{align*}f\left( {x,y} \right) & = \ln \left( {2x + y} \right)\hspace{0.25in}& {z_0}& = f\left( { - 1,3} \right) = \ln \left( 1 \right) = 0\\ {f_x}\left( {x,y} \right) & = \frac{2}{{2x + y}}\hspace{0.25in} &{f_x}\left( { - 1,3} \right) & = 2\\ {f_y}\left( {x,y} \right) & = \frac{1}{{2x + y}}\hspace{0.25in} & {f_y}\left( { - 1,3} \right) & = 1\end{align*}$

The equation of the plane is then,

$\begin{align*}z - 0 & = 2\left( {x + 1} \right) + \left( 1 \right)\left( {y - 3} \right)\\ z & = 2x + y - 1\end{align*}$

One nice use of tangent planes is they give us a way to approximate a surface near a point. As long as we are near to the point $\left( {{x_0},{y_0}} \right)$ then the tangent plane should nearly approximate the function at that point. Because of this we define the linear approximation to be,

$L\left( {x,y} \right) = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)\left( {x - {x_0}} \right) + {f_y}\left( {{x_0},{y_0}} \right)\left( {y - {y_0}} \right)$

and as long as we are “near” $\left( {{x_0},{y_0}} \right)$ then we should have that,

$f\left( {x,y} \right) \approx L\left( {x,y} \right) = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)\left( {x - {x_0}} \right) + {f_y}\left( {{x_0},{y_0}} \right)\left( {y - {y_0}} \right)$

Example 2 Find the linear approximation to

$z = 3 + \frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9}$ at

$\left( { - 4,3} \right)$ .

Show Solution

So, we’re really asking for the tangent plane so let’s find that.

$\begin{align*}f\left( {x,y} \right) & = 3 + \frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9}\hspace{0.25in} & f\left( { - 4,3} \right) & = 3 + 1 + 1 = 5\\ {f_x}\left( {x,y} \right) & = \frac{x}{8}\hspace{0.25in} & {f_x}\left( { - 4,3} \right) & = - \frac{1}{2}\\ {f_y}\left( {x,y} \right) & = \frac{{2y}}{9}\hspace{0.25in} & {f_y}\left( { - 4,3} \right) & = \frac{2}{3}\end{align*}$

The tangent plane, or linear approximation, is then,

$L\left( {x,y} \right) = 5 - \frac{1}{2}\left( {x + 4} \right) + \frac{2}{3}\left( {y - 3} \right)$

For reference purposes here is a sketch of the surface and the tangent plane/linear approximation.

This is a graph with the standard 3D coordinate system. The positive z-axis is straight up, the positive x-axis moves off to the left and slightly downward and positive y-axis move off the right and slightly downward. This is a cup shaped object that is centered on the z-axis, starts at z=3 and opens upwards. At the point (-4, 3, 5) there is a plane that is tangent to the cup shaped object.