MHB Parametric representation of a line

brunette15
Messages
58
Reaction score
0
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? :)
 
Physics news on Phys.org
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)
 
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
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top