Math 344: Calculus III

14.4 Tangent Planes

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14.4.1 Linear Approximations

One of the reasons calculus is so widely studied is that it helps us with studying functions locally . We know that for single-variable functions the derivative (if it exists) tells us the slope of the tangent line at a point along the function. If we choose a particular point and draw the tangent line to the curve, near the point the function is indistinguishable from the tangent line.

This function is relatively tame and so this property is quite apparent here, but it is important to note that is true for all differentiable functions . If you zoom in closely enough to the point where the tangent line is found, the function will appear to be a straight line. Near the point, we can take advantage of this by thinking of the tangent line as a linear approximation to the function. That is a line that behaves like the function in a neighborhood of the point where we computed the derivative. The primary usages of this in our previous calculus courses were tangent line approximations and differentials . Should you take other courses such as differential equations , you will find variations of this principle applied to compute numerical solutions.

Thinking about linear approximations in the context of multivariate functions, we find that this is the reason why the earth we stand on appears to be flat. We are so small relative to the radius of the planet that local regions without significant topographical features (such as Kansas) are effectively a plane.

In fact, the derivatives of multivariate functions generate tangent planes if there are two independent variables and hyperplanes for more independent variables.

Typically, we just use the phrase "tangent plane" regardless of how many dimensions there happen to be. We'll be focusing on functions of two variables for this section.

14.4.2 Tangent Planes

As in the previous section, we want to utilize partial derivatives for our analysis of the behaviors of these functions. However, we have some work to do to make sure that this would be appropriate. Let us begin by considering the above image of the sphere and the plane tangent. The lines that are present in that drawing represent the tangent lines to the curve in the $x$ and $y$ directions. So, if we were to intersect the sphere with planes parallel to the $yz$- and $xz$-planes, we would have the tangent lines corresponding to $\newcommand{\pzi}[2][]{\dfrac{\partial #1}{\partial #2}} \pzi{x}$ and $\pzi{y}$, respectively. There are several nice properties that come form having these lines arise from the partial derivatives. First, we know that these directions are both different and orthogonal . This means that we can use them to define a plane, since they are guaranteed to be intersecting (and the point on the surface), non-parallel lines. Second, this fact allows us to not have to worry about a potentially complicated set of things arising from limit paths. We have discussed previously how limits are allowed to approach from any possible direction or path in 2D space, including very complicated ones. If we are worried about the tangent plane, the tangent plane would always be the same, regardless of which path(s) are taken. This means that we can compute the plane by using the simplest method via partial derivatives.

To find the tangent plane, we appeal to the equation of a plane:

$$ A(x-x_0) + B(y - y_0) + C(z - z_0) = 0 $$
We assume $C\neq 0$ since we are interested in $z = f(x,y)$, and so after dividing by this value, the equation may be rearranged as

$$ z - z_0 = a(x - x_0) + b(y - y_0) $$
where $a = -A/C$ and $b = -B/C$. To find the values of these constants, we can set $y=y_0$ which is a plane parallel to the $xz$-plane. This reduces the equation to

$$ z - z_0 = a(x - x_0) $$
but we know that this corresponds to the $\pzi{x}$, so it must be the case that $a = f_x(x_0,y_0)$ since this is the equation of the line tangent to the curve defined by the intersection of the surface $z = f(x,y)$ and $y = y_0$. Hence,

$$ z - z_0 = f_x(x_0,y_0)(x - x_0) $$
The is a precisely the analogue to the tangent line equation for a single-variable function $y - y_0 = f'(x_0)(x - x_0)$. Likewise, we can set $x = x_0$ and get an intersecting plane parallel to the $yz$-plane corresponding to $\pzi{y}$ and see that $b$ is the partial with respect to $y$ of $f$ evaluated at $(x_0,y_0,z_0)$:

$$ z - z_0 = f_y(x_0,y_0)(y - y_0) $$
These taken together yield our result:

Equation of a Tangent Plane

The equation for a plane tangent to $z = f(x,y)$ at $(x_0,y_0,z_0)$ is given by

$$ z - z_0 = f_x(x_0,y_0)(x - x_0) + f_y(x_0,y_0)(y - y_0) $$

14.4.3 Finding Tangent Planes

Example 14.4.1

Find an equation of the line tangent to the surface

$$ z = 2x^2 + y^2 - 5y $$
at $(1,2,-4)$.

Solution

We begin by computing the first partials and evaluating them at the given point.

$$ \pzi[z]{x} = (2x^2 + y^2 - 5y)_x = 4x \qquad\qquad\qquad \pzi[z]{y} = (2x^2 + y^2 - 5y)_y = 2y - 5 $$
Evaluating at $(1,2,-4)$:

$$ \left.\pzi[z]{x}\right|_{\,(1,2,-4)} = 4(1) = 4 \qquad\qquad\qquad \left.\pzi[z]{y}\right|_{\,(1,2,-4)} = 2(2) - 5 = -1 $$

Writing $f_x(1,2)$ and $f_y(1,2)$ would have been fine here. I am just demonstrating other notation that sees use.

Our equation for the tangent plane is $z - z_0 = f_x(x_0,y_0)(x - x_0) + f_y(x_0,y_0)(y - y_0)$. We evaluate this at $(1,2,-4)$ to find

$$ z = 4(x - 1) - 1(y - 2) - 4 = 4x - y - 6 $$

The surface and tangent plane are shown below in Geogebra.

14.4.4 Linearization

A linearization of a function is its local reduction to a linear function. In the single-variable case, this is a line. In the multivariate case, this is a (hyper)plane. To find the linearization $L(x,y)$ of a function $f(x,y)$ at a point, we use the same formula as we did for computing the tangent plane with $z_0 = f(x_0,y_0)$. This new function $L$ allows to approximate the value of $f$ in a region near the point where the linearization was computed.

Example 14.4.2

Find the linearization of

$$ f(x,y) = e^{x-y} $$
at $(x,y) = (2,2)$ and use it to approximate $f(2.1,1.9)$.

Solution

The first partials are

$$ f_x = e^{x-y} \qquad\qquad\qquad f_y = -e^{x-y} $$
and evaluating them at the point $(2,2)$ yields

$$ f_x(2,2) = e^0 = 1 \qquad\qquad\qquad f_y(2,2) = -e^0 = -1 $$
Therefore, the linearization is given by

$$ L(x,y) = f(2,2) + f_x(2,2)(x - 2) + f_y(2,2)(y-2) = 1 + 1(x-2) - 1(y-2) = 1 + x - y $$
Under this linearization, the approximate value of $f(2.1,1.9) = 1 + 2.1 - 1.9 = 1.2 $.

14.4.5 Differentials

An application of linearization we are familiar with from Calculus 1 and 2 is differentials . We took the derivative expression

$$ f'(x) = \dfrac{dy}{dx} $$
and made the assumption that $dy/dx\approx \Delta y/\Delta x$ so that we could approximate $\Delta y$ using the formula

$$ dy = f'(x)\,dx \color{#307FE2}{\approx\Delta y} $$

We see here that there is approximate, but not exact, relationship between the differential $dy$ and the increment of the function $\Delta y$. For small changes in the inputs, labeled $dx = \Delta x$, we should see a small corresponding change in the outputs which can be approximated using the differential.

This analogue also carries over to the multivariate case. Our tangent line is now a tangent plane, and as depicted here

the increment $\Delta z$ here can be approximated by the differential $dz$, but instead depends on both the change in the $x$ and $y$ directions. The formula is

$$ dz = f_x(x,y)\,dx + f_y(x,y)\,dy = \pzi[z]{x}\,dx + \pzi[z]{y}\,dy \color{#307FE2}{\approx\Delta z} $$
where $dz$ is called the total differential .

Exercise 14.4.1

Suppose a cardboard box used in shipping is $7\text{ cm}\times 25\text{ cm}\times 15\text{ cm}$ and the cardboard is $5\text{ mm}$ thick. Use differentials to estimate the total volume of cardboard that constitutes the box.

Check Your Work

The volume of the box is given by $V = lwh$. It does not really matter for the purposes of volume which dimension corresponds to each measurement. If you're shipping something, you probably want the box to sit in a particular orientation, but that is not relevant to the problem at hand. We will, for simplicity, set $l = 25\text{ cm}$, $w = 15\text{ cm}$, and $h = 7\text{ cm}$ since this is the way most people would tend to orient the box. Computing the total differential is simple, and we find

$$ dV = wh\,dl + lh\,dw + lw\,dh $$
To find our total volume, we need to plug in our information from the problem, noting two things. First, we have to convert $5\text{ mm}$ to $\text{cm}$. Secondly, the differentials $dl$, $dw$, and $dh$ are reach represented twice on the box since there are two sides of the box whose normal is the $l$ direction, two sides whose normal is the $w$ direction, and two whose normal is the $h$ direction. Hence, we will need to double the thickness value to $1\text{ cm}$ when we plug in the numbers. Putting everything together, approximation to the total volume of cardboard comprising the box is

$$ dV = (15)(7)(1) + (25)(7)(1) + (25)(15) = 655\text{ cm}^3 $$

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