Curvature

Measuring how sharply a path turns independently of its parameterization, and the radius of the best-fitting circle at each point.

Active Calculus Multivariable, Curvature

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7 exercises

Activity 10.6.2

In this activity we will use our previous work in finding unit speed parameterizations of lines and circles to make sense of the definition of curvature.

Recall that in this example we found that the unit speed parameterization of a line through the points $(1,2,3)$ and $(2,0,5)$ to be $\vr_1(s)= \langle 1,2,3\rangle + \frac{s}{3} \langle 1,-2,2 \rangle$ . Calculate $\vT(s)$ , $\frac{d\vT}{ds}$ , and $\vecmag{ \frac{d\vT}{ds} }$ for the unit speed parameterization of this line. Note that because we are using the unit speed parameter…

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Activity 10.6.4

In this activity we will look at the measurement of curvature and determine if curvature is a property of the driver or a property of the road. This is a continuation of this activity. As a reminder, a measurement is a property of the driver if the value(s) of that measurement can be different for different drivers (when measured at the same location on the racetrack). A measurement is a property…

2.5

Exercise 11

Let $y = f(x)$ define a curve in the plane. We can consider this curve as a curve in three-space with $z$ -coordinate 0.

Find a parameterization of the form $\vr(t) = \langle x(t), y(t), z(t) \rangle$ of the curve $y=f(x)$ in three-space.
Use the formula to show that

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Exercise 12

Consider the single variable function defined by $y = 4x^2 - x^3.$

Find a parameterization of the form $\vr(t) = \langle x(t), y(t) \rangle$ that traces the curve $y = 4x^2 - x^3$ on the interval from $x = -3$ to $x = 3$ .

Write a definite integral which, if evaluated, gives the exact length of the given curve from $x = -3$ to $x = 3$ . Why is the integral difficult to evaluate exactly?

Determine the curvature, $\kappa(t)$ , of the parameterized curve. (Exercise might be useful here.)
Use appropriate tech…

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Exercise 13

In this exercise we verify the curvature formula

Explain why

Use the fact that $\vT(t) = \frac{\vr'(t)}{\vecmag{\vr'(t)}}$ and $\vecmag{\vr'(t)} = \frac{ds}{dt}$ to explain why

The Product Rule shows that Explain why

In Key Idea on $\vT$ and $\vN$ we showed that $\vecmag{\vT(t)} = 1$ implies that $\vT(t)$ is orthogonal to $\vT'(t)$ for every value of $t$ . Explain what this tells us about $\vecmag{\vT(t) \times \vT'(t)}$ and conclude that
Finally, use the fact that $\kappa = \frac{\vecmag{\vT'(t)} }{\vecmag{\vr'(t)}}$ to verify that

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Exercise 15

In this activity, you verified for a specific line that the curvature is $0$ . The generalization of this is that the unit speed parameterization of a line through $(x_0,y_0,z_0)$ with direction vector $\langle a,b,c\rangle$ can be given by .

Compute the speed of this parameterization to verify that this has unit speed. This will also show that $\vr\, '(s)=\vv(s)=\vT(s)$ (which is not true in general).
Calculate $\vT(s)$ , $\frac{d\vT}{ds}$ , and $\vecmag{ \frac{d\vT}{ds} }$ for this unit speed parameterization…

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Exercise 16

If $\vu \times \vw = \vzero$ , what does this say about the direction of $\vu$ and $\vw$ ?
Remember that the vector $\vv$ has magnitude and direction, so the derivative $\frac{d\vv}{dt} = \va$ will measure the change in both the magnitude and direction of of $\vv$ . Write a few sentences to describe how if $\vv$ and $\va= \frac{d\vv}{dt}$ are in parallel directions, then the unit normal vector will be undefined at that point.

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Adapted from Active Calculus Multivariable © 2013–2026 Nicholas Long, Mitchel T. Keller, and Steven Schlicker. Licensed under CC BY-SA 4.0. Source: Active Calculus Multivariable. Imported at revision 692c03c.