# Question on Arc Length parameterization.

1. Apr 8, 2013

### yungman

This is an example in book by Howard Anton:

Vector form of line is $\vec r=\vec r_0+t\vec v$ where $\vec v$ is parallel with the line. So both $\vec r$ and $\vec r_0$ are POSITION VECTORS.

To change parameters,
1)Let u=t $\Rightarrow\; \vec r=\vec r_0+u\vec v$.

2) $\frac {d\vec r}{du}=\vec v\;\Rightarrow\;|\frac {d\vec r}{du}|=|\vec v|$

$s=\int_0^t |\frac {d\vec r}{du}|du=t|\vec v|\;\Rightarrow\; t=\frac{s}{|\vec v|}$

3)$\Rightarrow\; \vec r=\vec r_0+\frac{s}{|\vec v|}\vec v$

My question is in #2 above. In order for $\frac {d\vec r}{du}=\vec v$ which is the tangent vector of the curve traced by $\vec r$, $\vec r$ has to be a VECTOR VALUE FUNCTION, NOT JUST A POSITION VECTOR. This means $\vec r =\vec r(w)$ where w is the independent variable that make the tip of $\vec r$ tracing out the line when w increases or decreases.( of cause it can be a vector value function of many variables also).

As you see, my problem is there are TWO parameters, t and w. The book only change parameter of t, which has nothing to do with the vector value function $\vec r(w)$. t only tell the line is multiple of $\vec v$. In another word, this example totally miss the point in changing parameter. The parameter needed to be change is w, not t.

Last edited: Apr 9, 2013
2. Apr 9, 2013

### yungman

Further more, the next example is to find arc length parametrization of the line where $x=2t+1,\;y=3t-2$ with $\vec r_0=<1,-2>$ and parallel to vector $\vec v=\hat x 2+\hat y 3$

$\Rightarrow \vec r=\vec r_0+s\frac{\vec v}{|\vec v|}=(\hat x-\hat y 2)+s\frac{\hat x 2 +\hat y 3}{\sqrt{13}}$

Thereby after change of parameter:

$x=\frac {2}{\sqrt{13}}s+1,\;y=\frac {3}{\sqrt{13}}s-2$

This example use the same variable t as the parameter for x and y. Can you even do that? As I explained in the last post, there should be two independent variables w and t, not just t alone. Please comment on this also. I am confused.

3. Apr 9, 2013

### SteamKing

Staff Emeritus
The components of a vector can be expressed in terms of some parameter t. In post 2 above, think of the x and y as being the number of unit vectors i and j combined, according to values for the third parameter t.

If the vector R(t) = [x(t), y(t)], then as t runs from 0 to 2, R(0) = [1,-2] thru R(2) = [5,4] and all values in between, consistent with the definition of x and y in terms of t.

Essentially, R could be defined as R(t) = [(2t+1),(3t-2)] instead.

4. Apr 9, 2013

### yungman

Thanks for the response, I also work it out while I was waiting. Thanks.