In the previous section, we learned how to find the slope of a surface in two specific directions—parallel to the x-axis and parallel to the y-axis. But what about approximating the surface itself? In Calculus I, the tangent line was the best linear approximation to a curve near a point. In Calculus III, we extend this idea to three dimensions with the tangent plane, which serves as the best linear approximation to a surface.
Recall that the equation of a tangent line to a curve $y=f(x)$ at a point $(a, f(a))$ is given by $y - f(a) = f'(a)(x-a)$. This line is the best linear approximation of the function near the point of tangency. This means that for values of $x$ close to $a$, the y-values on the tangent line are very close to the y-values on the curve.
For a surface $z=f(x,y)$, the equivalent concept to a tangent line is a tangent plane. This is a plane that "just touches" the surface at a point $(x_0, y_0, z_0)$. How is this plane defined? It is the unique plane that contains both of the tangent lines we studied in the last section:
The interactive graph below shows the surface $f(x,y) = 4 - x^2 - \frac{1}{2}y^2$ at the point $(1,1)$. You can see the two tangent lines and the single tangent plane that contains both of them, providing a "platform" that approximates the surface at that point.
We can derive the equation of the tangent plane directly from the normal vector. We know the plane passes through the point $P_0(x_0, y_0, z_0)$. A normal vector $\vec{n} = \langle a,b,c \rangle$ is perpendicular to any vector lying in the plane, such as $\vec{v} = \langle x-x_0, y-y_0, z-z_0 \rangle$. This gives the standard equation $\vec{n} \cdot \vec{v} = 0$, or $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.
If we rearrange this equation to look more like the function form $z = ...$, we get:
$$ c(z-z_0) = -a(x-x_0) - b(y-y_0) $$
$$ z-z_0 = -\frac{a}{c}(x-x_0) - \frac{b}{c}(y-y_0) $$
This equation must represent our tangent plane. The slope of the tangent line in the x-direction is $f_x(x_0,y_0)$, which must be the coefficient of the $(x-x_0)$ term. Similarly, the slope of the tangent line in the y-direction is $f_y(x_0,y_0)$, which must be the coefficient of the $(y-y_0)$ term. This gives us:
$$ f_x(x_0, y_0) = -\frac{a}{c} \quad \text{and} \quad f_y(x_0, y_0) = -\frac{b}{c} $$
Substituting these back into the equation gives us the formula we need.
The equation of the tangent plane to the surface $z=f(x,y)$ at the point $(x_0, y_0, z_0)$ is:
$$ z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) $$
Notice the beautiful analogy to the tangent line equation. The total change in $z$ is a combination of the change contributed by moving in the x-direction and the change contributed by moving in the y-direction.
The derivation above shows one way to connect the normal vector $\langle a,b,c \rangle$ to the partial derivatives. A more direct way is to construct two vectors in the plane and find their cross product.
The normal vector $\vec{n}$ must be perpendicular to both of these, so we can find it using the cross product:
$$ \vec{n} = \vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & f_x \\ 0 & 1 & f_y \end{vmatrix} = \langle -f_x, -f_y, 1 \rangle $$
Plugging this normal vector into the standard equation of a plane gives:
$$ (-f_x(x_0, y_0))(x-x_0) + (-f_y(x_0, y_0))(y-y_0) + (1)(z-z_0) = 0 $$
Rearranging this equation to solve for $z-z_0$ gives us our new formula exactly:
$$ z-z_0 = f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0) $$
This confirms that both methods yield the same plane. The benefit of our new formula is its directness; it allows us to build the plane's equation immediately from the partial derivatives without having to set up and compute a cross product. It's important to remember that this direct method is specifically for finding a tangent plane to a surface $z=f(x,y)$. For more general cases, such as finding a plane through three arbitrary points, we must use the vector methods learned in Section 12.5.
Find the equation of the tangent plane to the surface $f(x,y) = \sin(x) + \cos(y)$ at the point $(\pi/2, 0, 2)$.
First, we need the partial derivatives at the point $(\pi/2, 0)$.
$$ f_x(x,y) = \cos(x) \quad \implies \quad f_x(\pi/2, 0) = \cos(\pi/2) = 0 $$
$$ f_y(x,y) = -\sin(y) \quad \implies \quad f_y(\pi/2, 0) = -\sin(0) = 0 $$
Now we use the formula with $(x_0, y_0, z_0) = (\pi/2, 0, 2)$, $f_x=0$, and $f_y=0$.
$$ z - 2 = 0(x-\pi/2) + 0(y-0) $$
$$ z - 2 = 0 $$
$$ z = 2 $$
The tangent plane is the horizontal plane $z=2$. This makes sense because the point $(\pi/2, 0, 2)$ is a maximum point (a "peak") on the surface, where the slopes in both the x and y directions are zero.
Find the equation of the tangent plane to the surface $f(x,y) = xy^2 + x^2y$ at the point $(1,2,6)$.
The tangent plane is useful because it provides a simple, linear approximation of the function $f(x,y)$ for points $(x,y)$ that are near the point of tangency $(a,b)$. A key idea in calculus is that differentiable functions are "locally flat". This means that if you zoom in far enough on a smooth surface at a point, the surface becomes indistinguishable from its tangent plane.
Zoom in on the graph above to see how the surface and the tangent plane become nearly identical near the point of tangency.
The linear function whose graph is the tangent plane, $$ L(x,y) = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b) $$ is called the linearization of $f$ at $(a,b)$. The approximation $$ f(x,y) \approx L(x,y) $$ is the linear approximation or tangent plane approximation of $f$ at $(a,b)$.
Find the linearization of the function $f(x,y) = x e^{xy}$ at the point $(1,0)$ and use it to approximate the value of $f(1.1, -0.1)$.
First, we need the partial derivatives and function value at $(a,b)=(1,0)$.
The linearization is:
$$ L(x,y) = f(1,0) + f_x(1,0)(x-1) + f_y(1,0)(y-0) $$
$$ L(x,y) = 1 + 1(x-1) + 1(y-0) = 1 + x - 1 + y = x+y $$
Now we use this simple function to approximate $f(1.1, -0.1)$:
$$ f(1.1, -0.1) \approx L(1.1, -0.1) = 1.1 + (-0.1) = 1.0 $$
The actual value is $f(1.1, -0.1) = 1.1e^{(1.1)(-0.1)} = 1.1e^{-0.11} \approx 0.985$, so our approximation is quite good.
Use linear approximation to estimate the value of $\sqrt{2.9^2 + 4.1^2}$.
We need to choose a function and a "nice" point near $(2.9, 4.1)$.
Let $f(x,y) = \sqrt{x^2+y^2}$ and let our point of tangency be $(a,b) = (3,4)$.
The linearization at $(3,4)$ is:
$$ L(x,y) = 5 + \frac{3}{5}(x-3) + \frac{4}{5}(y-4) $$
Now approximate the value:
$$ f(2.9, 4.1) \approx L(2.9, 4.1) = 5 + \frac{3}{5}(2.9-3) + \frac{4}{5}(4.1-4) $$
$$ = 5 + \frac{3}{5}(-0.1) + \frac{4}{5}(0.1) = 5 - \frac{0.3}{5} + \frac{0.4}{5} = 5 + \frac{0.1}{5} = 5 + 0.02 = 5.02 $$
The actual value is approximately $5.02095$, so our estimate is very close.
Find the linearization $L(x,y)$ of the function $f(x,y) = \ln(x-3y)$ at the point $(7,2)$.
Find the equation of the tangent plane to the surface $f(x,y) = y \ln(x)$ at the point $(1, 4, 0)$.
Use linearization to approximate the value of $(1.01)^3 (1.98)^2$.
The equation of the tangent plane to the surface $z=f(x,y)$ at the point $(2, -1, 5)$ is given by $4x - 3y + z = 16$. Find the values of $f_x(2, -1)$ and $f_y(2, -1)$.
Today, we extended the concept of the tangent line to the tangent plane, providing a way to locally approximate a multivariable function with a much simpler linear function. This is a cornerstone of multivariable calculus, allowing us to analyze complex surfaces by examining their linear approximations at various points. This idea of local linearity is a recurring theme and a powerful tool in nearly every branch of applied mathematics.
After this lecture, you should be able to: