Labware - MA35 Multivariable Calculus - Single Variable Calculus
 MA35 Labs 1 » Single Variable Calculus Contents1.1 Functions of One Variable 1.1.1 Demo Tutorial 1.1.2 Linear Functions 1.1.4 Slices 1.1.6 Continuity 1.1.7 Normal Vectors for Linear Function Graphs 1.4 Integration Search

Normal Vectors for Linear Function Graphs

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A normal vector to a linear function graph is any vector which is perpendicular to that function graph.

One way to find such a vector is to use the fact that two nonzero vectors with a dot product of zero must be perpendicular. If we choose a vector parallel to the linear function graph, we can find a normal vector by finding a vector whose dot product with the first is zero.

For example, for the graph of the function

f(x) = px + k,

we can choose the parallel vector (1, p). For the normal vector N, it must be true that

N(1, p) = 0.

One vector which satisfies this equation and is thus a normal vector is (-p, 1).

We can use the same method for the graph of the implicit function

ax + by = c.

One vector which is parallel to the graph is (b, -a).

(a, b) is a normal vector because its dot product with (b, -a) is 0.

Demos

 Normal Vectors for Linear Function Graphs This demo finds a normal vector, shown here in white, to the graph of f(x) = px + k given any p and k. The tangent vector to which this normal vector is perpendicular is shown in cyan. Note that the components of the normal vector are just the components of the tangent vector rotated 90 degrees counterclockwise.

 Normal Vectors for Linear Function Graphs This demo finds a normal vector, again shown in white, to the implicit curve for the equation ax + by = c given any a, b, and c.

Exercises

• 1. Find normal vectors for the graphs of the following explicit functions:
• f(x) = 0
• f(x) = 2x
• f(x) = 2x + 5
• f(x) = 3x + 4
• 2. Find normal vectors for the graphs of the following implicit functions:
• x + y = 1
• 3x + 4y = 12
• 4x + 3y = 12
• 4x + 3y = 1
• y = 5
• x = 3
• 3. How does any general expression for a normal vector to a line depend on k for f(x) = px + k and c for ax + by = c. Why is this?