Professor Leonard – Calculus 3
13.4 – Finding Differentials of Multivariable Functions
13.7: Finding Tangent Planes and Normal Lines to Surfaces
Thomas – 14.6 Tangent Planes and Differentials
Review: Derivatives, Tangent Lines, and Linearization
One begins their Calculus journey learning about the derivative. We find out that the derivative defines the slope of a tangent line to a curve. That tangent line provides the best linear approximation to the curve near a chosen point. This process of moving from derivative to tangent to describing the curve with a function at a given point is known as linearization. Many of the original ideas in multivariate calculus are extensions of these foundational concepts.
Extending from Curves to Surfaces
Whereas a function with one independent variable produces a curve, functions with two independent variables yield a surface. Instead of approximating a curve with a tangent line, we approximate a surface with a tangent plane. Just as a tangent line captures the local behavior of a curve, a tangent plane zeroes in on the local behavior or a surface near a point.
Naturally, understanding the behavior of the surface is incredibly important. Our starting point for doing so is to first find a vector that is perpendicular to the tangent plane.
Level Curves – Looking at a Surface from Above
A surface described by z=f(x,y) can be sliced horizontally. Each slice produces a level curve represented by f(x,y)=C. That tells us that every point on a given level curve has the same function value. Level curves, like contours on a topographic map, are a useful means of examining three-dimensional surfaces. We can represent a level curve using a vector function \mathbf r(t)=\langle x(t),y(t)\rangle.
The Gradient and Why It Matters
We return to the gradient. As we touched on in the previous entry, differentiating f(x(t),y(t))=C shows us that the gradient is orthogonal to the tangent vector of a level curve. This then tells us that:
- The gradient is normal (perpendicular) to every level curve,
- The gradient points in the direction of the greatest increase
- The gradient provides the normal vector needed to construct the tangent planes
Returning to the Hill Analogy
Imagine a hill surrounded by contour lines representing constant elevation. Those contour lines are our level curves. If you walk along a contour, your elevation remains constant. Turning perpendicular to the contour line will point you in the quickest direction up to the succeeding contour line. Since the gradient is perpendicular to every contour line, it also serves as the normal vector to the tangent plane.
Tangent Planes
A plane is determined by a point and a normal vector. Since the gradient provides us with the normal vector, it allows us to write the equation of the tangent plane. The tangent plane can be defined as follows:
z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)
Or equivalently
z-f(x_0,y_0)=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)
Example:
Find a tangent plane to the surface z=x^2+y^2 at the point (1,2,5)
Step 1: Find the partial derivatives
f_x=2x,\qquad f_y=2y
Step 2: Evaluate them at the point
f_x(1,2)=2,\qquad f_y(1,2)=4
Step 3: Use the tangent plane formula
z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)
Substitute the values:
z=5+2(x-1)+4(y-2)
Step 4: Simplify (optional)
z=2x+4y-5
Normal Lines:
Once we know the normal vector, constructing the normal lines is straightforward.
We can define the normal line in symmetric and parametric form:
Parametric form: x=x_0+at,\qquad y=y_0+bt,\qquad z=z_0+ct
Where the normal vector is
\mathbf{n}=\langle a,b,c\rangle
Symmetric form: \frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}
Linearization of Functions
The tangent plane is more than just a geometric object. It’s a way of finding the best linear approximation to a surface near a chosen point.
The linearization formula is as follows:
L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)
Example
Find the linearization of
f(x,y)=x^2+y^2
at the point
(1,2)
Step 1: Evaluate the Function
f(1,2)=1^2+2^2=5
Step 2: Find the Partial Derivatives
f_x=2x,\qquad f_y=2y
Step 3: Evaluate the Partial Derivatives
f_x(1,2)=2,\qquad f_y(1,2)=4
Step 4: Apply the Linearization Formula
L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)
Substituting the values gives
L(x,y)=5+2(x-1)+4(y-2)
Simplifying,
L(x,y)=2x+4y-5
Differentials
Linearization naturally leads to differentials. Rather than computing the exact change in a function, differentials let us estimate the change using the tangent plane.
The differential dz estimates the change in height as we move from the point P(x_0,y_0) to a nearby point on the tangent plane. If we move horizontally by \Delta x and \Delta y, then for sufficiently small changes we take
dx=\Delta x,\qquad dy=\Delta y
and estimate the corresponding change in height by
dz=f_x,dx+f_y,dy.
In other words, the tangent plane predicts how much the function’s value changes for a small change in x and y.