Parametric representation of a line

In summary, the conversation discusses the creation of a new parametric representation for a given curve using the arc length as the parametric variable. The arc length is calculated and the inverse function is suggested as a way to continue solving the problem.
  • #1
brunette15
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I am give the following curve r(t) = (t+1,0.5(1-t),0) where t ranges from -1 to 1. I am now trying to derive a new parametric representation of this line segment using the arc length as the parametric variable.

I have integrated r'(t) from -1 to 1 and found that the length of the segment ranges from 0 to 5^0.5 however am unsure where to go from here.

Can anyone please help me finish off this problem? :)
 
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  • #2
brunette15 said:
I am give the following curve r(t) = (t+1,0.5(1-t),0) where t ranges from -1 to 1. I am now trying to derive a new parametric representation of this line segment using the arc length as the parametric variable.

I have integrated r'(t) from -1 to 1 and found that the length of the segment ranges from 0 to 5^0.5 however am unsure where to go from here.

Can anyone please help me finish off this problem? :)

Hey brunette15! ;)

You have correctly calculated the full arc length, but let's get the arc length up to some arbitrary $t$.
That arc length is given by:
$$s(t) = \int_{-1}^t \|\mathbf r'(t)\| \,dt = \int_{-1}^t \|(1,-0.5,0)\| \,dt
= \int_{-1}^t \frac 12 \sqrt 5 \,dt =\frac 12 \sqrt 5(t+1)$$Now let $\mathbf{\tilde r}(s)$ be the new parametric representation of this line segment using the arc length as the parametric variable.
Then:
$$\mathbf{\tilde r}(s) = \mathbf{r}(t(s))$$
where $t(s)$ is the inverse function of the arc length $s(t)$.How about finding the inverse of $s(t)$ and substituting it in $\mathbf r(t)$? (Wondering)
 
  • #3
I like Serena said:
Hey brunette15! ;)

You have correctly calculated the full arc length, but let's get the arc length up to some arbitrary $t$.
That arc length is given by:
$$s(t) = \int_{-1}^t \|\mathbf r'(t)\| \,dt = \int_{-1}^t \|(1,-0.5,0)\| \,dt
= \int_{-1}^t \frac 12 \sqrt 5 \,dt =\frac 12 \sqrt 5(t+1)$$Now let $\mathbf{\tilde r}(s)$ be the new parametric representation of this line segment using the arc length as the parametric variable.
Then:
$$\mathbf{\tilde r}(s) = \mathbf{r}(t(s))$$
where $t(s)$ is the inverse function of the arc length $s(t)$.How about finding the inverse of $s(t)$ and substituting it in $\mathbf r(t)$? (Wondering)

Thankyou so much! I know how to figure it out from here :D
 

FAQ: Parametric representation of a line

What is a parametric representation of a line?

A parametric representation of a line is a way to describe a line in terms of one or more parameters. It involves expressing the coordinates of points on the line as functions of those parameters.

Why is parametric representation useful?

Parametric representation allows for a more flexible and concise way to describe a line compared to the standard slope-intercept form. It also makes it easier to perform calculations and transformations on the line.

How is a line represented parametrically?

A line can be represented parametrically by using the equations x = x0 + at and y = y0 + bt, where (x0, y0) is a point on the line and a and b are the parameters. This form is also known as the vector equation of a line.

What are the advantages of using two parameters in the parametric representation of a line?

Using two parameters allows for more flexibility in describing a line, as it can represent any direction and magnitude. It also makes it easier to find specific points on the line, such as the midpoint or a point a certain distance away from the initial point.

Can a line be represented parametrically in three-dimensional space?

Yes, a line can be represented parametrically in three-dimensional space using three parameters and the equations x = x0 + at, y = y0 + bt, and z = z0 + ct. This allows for even more flexibility in describing a line in three-dimensional space.

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