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You CAN Ace Calculus

17calculus > vector functions > curvature

### Calculus Main Topics

Single Variable Calculus

Multi-Variable Calculus

### Tools

math tools

general learning tools

Curvature Using Vector Functions

For the following discussion, we will consider a parameterized curve defined by the vector function $$\displaystyle{ \vec{r}(t) = \langle x(t), y(t), z(t) \rangle }$$ which is traversed once on the continuous interval $$a \leq t \leq b$$.

 Some books use the Greek letter κ for curvature.

The curvature of a smooth curve is a measure of how 'tight' or 'sharp' the curve is. If we have a smooth curve $$\vec{r}$$ and we have a function s which is the arc length function, the curvature is defined to be
$$\displaystyle{ K(s) = \left\| \frac{d\vec{T}}{ds} \right\|}$$.

This equation for the curvature is not particularly useful for calculations. So we have several ways to write the equation of the curvature. But first notice, that the curvature is a scalar function, not a vector function. And since it is the norm of a vector, the curvature will always be positive.

Curvature Formula #1
For our first equation to use when calculating the curvature, we will use the chain rule to write $$\displaystyle{ \frac{d\vec{T}}{dt} = \frac{d\vec{T}}{ds} \cdot \frac{ds}{dt} }$$. We can solve for $$d\vec{T}/ds$$ to get

$$\displaystyle{ K = \left\| \frac{d\vec{T}}{ds} \right\| = \frac{\| d\vec{T}/dt \|}{ \| ds/dt \| } = \frac{1}{\| \vec{v} \|} \left\| \frac{d\vec{T}}{dt} \right\| }$$

Notice in the previous equation, we used $$ds/dt = \|\vec{v}\|$$ to simplify the equation somewhat.
Now we can write the first curvature formula in a form that we can use for calculations.

$$\displaystyle{ K(t) = \frac{1}{\|\vec{v}\|} \left\| \frac{d\vec{T}}{dt} \right\| = \frac{\| \vec{T}'(t) \|}{\|\vec{r}'(t)\|} }$$

Curvature Formula #2
If we define a vector $$\vec{a} = d\vec{v}/dt$$ as an acceleration vector, a second curvature formula is
$$\displaystyle{ K=\frac{\|\vec{v} \times \vec{a} \|}{\|\vec{v}\|^3} }$$

Before working some practice problems, here is a quick video clip for you that should help you understand the curvature a bit better.

 MIP4U - Determining Curvature of a Curve Defined by a Vector Valued Function [3min-9secs]

Here is a longer video explaining the derivation of some of the equations.

 Louis Saumier - Curvature: Definition, Derivation [25min-30secs]

### Search 17Calculus

Practice Problems

Instructions - - Unless otherwise instructed, find the curvature of these vector functions. Give your answers in exact, completely factored form.

 Level A - Basic

Practice A01

$$\vec{r}(t)=3t\vhat{i}+4\sin t\vhat{j}+4\cos t\vhat{k}$$

solution

Practice A02

$$\vec{r}(t)=\langle 2\cos(t),2\sin(t)\rangle$$

solution

Practice A03

Find the curvature at $$t=1$$ for $$\vec{r}(t)=\langle t,t^2/2,t^3/3\rangle$$.

solution

Practice A04

Find the curvature of $$\vec{r}(t)=\langle -5t,2t^3,3t^4\rangle$$ at $$t=2$$.

Find the curvature of $$\vec{r}(t)=\langle 2\cos(3t),2\sin(3t),4t\rangle$$ at $$t=0$$.