Unit tangent vector and curvature with arc length parameterization

  • #1
songoku
2,321
331
Homework Statement
Please see below
Relevant Equations
ds/dt = |r'(t)|

T(t) = r'(t) / |r'(t)|

K = |dT/ds|
1695568095854.png


(a)
$$\frac{ds}{dt}=|r'(t)|$$
$$=\sqrt{(x(t))^2+(y(t))^2+(z(t))^2}$$
$$=\frac{2}{9}+\frac{7}{6}t^4$$

$$s=\int_0^t |r'(a)|da=\frac{2}{9}t+\frac{7}{30}t^5$$

Then I think I need to rearrange the equation so ##t## is the subject, but how?

Thanks

Edit: wait, I realize my mistake. Let me redo
 
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  • #2
I think I can do (a). I got ##s=t^2##

For (b), I just want to ask about the correct approach. I think I need to use the chain rule to find r'(t) so ##r'(t)=\frac{dr}{ds}\times \frac{ds}{dt}##

Am I correct? Thanks
 

Related to Unit tangent vector and curvature with arc length parameterization

What is a unit tangent vector?

A unit tangent vector is a vector that is tangent to a curve at a given point and has a magnitude (length) of one. It indicates the direction in which the curve is proceeding at that point.

How do you find the unit tangent vector using arc length parameterization?

To find the unit tangent vector using arc length parameterization, you first need to parameterize the curve by its arc length, denoted as \( s \). The unit tangent vector \( \mathbf{T}(s) \) is then given by the derivative of the position vector \( \mathbf{r}(s) \) with respect to \( s \), normalized to have unit length: \( \mathbf{T}(s) = \frac{d\mathbf{r}(s)}{ds} \).

What is curvature in the context of a curve parameterized by arc length?

Curvature is a measure of how sharply a curve bends at a given point. For a curve parameterized by arc length \( s \), the curvature \( \kappa(s) \) is defined as the magnitude of the derivative of the unit tangent vector with respect to \( s \): \( \kappa(s) = \left| \frac{d\mathbf{T}(s)}{ds} \right| \).

How do you compute curvature using the arc length parameterization?

To compute the curvature using arc length parameterization, you take the derivative of the unit tangent vector \( \mathbf{T}(s) \) with respect to the arc length \( s \) and then find the magnitude of this derivative. Mathematically, it is expressed as \( \kappa(s) = \left| \frac{d\mathbf{T}(s)}{ds} \right| \).

Why is arc length parameterization useful for studying curves?

Arc length parameterization is useful for studying curves because it provides a natural and intrinsic way to describe the curve that is independent of the specific coordinate system. It simplifies the computation of geometric properties like the unit tangent vector and curvature, since the arc length \( s \) directly measures distance along the curve.

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