# Calculate the length of the given curve part 2

## Homework Statement

$\vec r (t) = \vec u +t \vec v for -4 \le t \le 6$ where u and v are constant vectors and v not equal 0

Do i proceed with the usual calculation of differentiating r wrt t, find the magnitude and then integrate between limits? Looks too simple for that. Thanks

Related Calculus and Beyond Homework Help News on Phys.org
Simon Bridge
Homework Helper
http://www.ucl.ac.uk/Mathematics/geomath/level2/fvec/fv4.html

but in this case it indeed looks like:

$$\frac{d\vec r}{dt}=\vec v$$

... think what differentiation means. How $\vec r$ changes with time is it grows in the $\vec v$ direction and the rate of that growth is the magnitude.

The intergral will be the area between the path mapped out by $\vec r(t)$ and ... something.

Of course, I'm only guessing that you have been asked to differentiate and integrate that function ... you didn't actually say.

If we consider r to be a displacement, then $\vec u = \vec r(0)$ and $\vec v$ is, indeed, the velocity.
It is not clear what the area under the displacement-time graph represents.
It's easier to think about if r is a velocity, u is initial velocity, and v is the acceleration.

Last edited:
http://www.ucl.ac.uk/Mathematics/geomath/level2/fvec/fv4.html

but in this case it indeed looks like:

$$\frac{d\vec r}{dt}=\vec v$$

... think what differentiation means. How $\vec r$ changes with time is it grows in the $\vec v$ direction and the rate of that growth is the magnitude.

The intergral will be the area between the path mapped out by $\vec r(t)$ and ... something.

Of course, I'm only guessing that you have been asked to differentiate and integrate that function ... you didn't actually say.

If we consider r to be a displacement, then $\vec u = \vec r(0)$ and $\vec v$ is, indeed, the velocity.
It is not clear what the area under the displacement-time graph represents.
It's easier to think about if r is a velocity, u is initial velocity, and v is the acceleration.
It says I have to calculate the length of the curve for the given problem.

LCKurtz
Homework Helper
Gold Member

## Homework Statement

$\vec r (t) = \vec u +t \vec v for -4 \le t \le 6$ where u and v are constant vectors and v not equal 0

Do i proceed with the usual calculation of differentiating r wrt t, find the magnitude and then integrate between limits? Looks too simple for that. Thanks
It is very simple. Just do it.

It is very simple. Just do it.
I get $10 \vec v$ by differentiating r(t) wrt t , getting its magnitude which is v and then integrating between the limits wrt t...

LCKurtz
Homework Helper
Gold Member
I get $10 \vec v$ by differentiating r(t) wrt t , getting its magnitude which is v and then integrating between the limits wrt t...
That is confusing. You don't get $10\vec v$ as the derivative if that is what you mean. And if you mean that is the answer it isn't correct either because the answer must be a scalar.

Well yes I realise the answer is a scalar because I know the vector v is a constant as stated in the problem hence 10 times a constant?

r(t)= u + tv.
dr(t)= v.

Magnitude = (v^2)^(1/2) =v. Then integrate this between the limits from -4 to 6....

LCKurtz
Homework Helper
Gold Member
Well yes I realise the answer is a scalar because I know the vector v is a constant as stated in the problem hence 10 times a constant?

r(t)= u + tv.
dr(t)= v.

Magnitude = (v^2)^(1/2) =v. Then integrate this between the limits from -4 to 6....
You need to use different notation for a vector $\vec v$ and its magnitude $|\vec v| = v$. Otherwise it is difficult to tell whether you are using the vector or its magnitude. You are using v for both above.

You need to use different notation for a vector $\vec v$ and its magnitude $|\vec v| = v$. Otherwise it is difficult to tell whether you are using the vector or its magnitude. You are using v for both above.
$d \vec r (t)= \vec v$. Therefore the magnitude is $| \vec v| = v$. Then

$\int_{-4}^{6} v dt = 10v$...?

That is what I did . Correct?

LCKurtz
Homework Helper
Gold Member
$d \vec r (t)= \vec v$. Therefore the magnitude is $| \vec v| = v$. Then

$\int_{-4}^{6} v dt = 10v$...?

That is what I did . Correct?
Yes. That is much better and it is correct.

Simon Bridge
Homework Helper
Neato :)
I'm a big fan of understanding the problem.

If r(t) is a position vector, then the thingy's motion has speed |v| in the direction v/|v| and passes through position u at t=0. Since v is a constant vector, the path is a straight line. We want to know how far the thingy has traveled between t=-4 and t=6 units.

Since it's linear, that's just the final position subtracted from the initial position - or the speed multiplied by the elapsed time.

Of course, that's probably not mathematically rigorous.

Simon Bridge
Homework Helper
Neato :)
I'm a big fan of understanding the problem.

If r(t) is a position vector, then the thingy's motion has speed |v| in the direction v/|v| and passes through position u at t=0. Since v is a constant vector, the path is a straight line. We want to know how far the thingy has traveled between t=-4 and t=6 units.

Since it's linear, that's just the final position subtracted from the initial position - or the speed multiplied by the elapsed time.

Of course, that's probably not mathematically rigorous.

good insight! Thanks