Calculate the length of the given curve part 2

[bugatti79]

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

 

[Simon Bridge]

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.

 

[bugatti79]

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 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.

 

[bugatti79]

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]

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.

 

[bugatti79]

Well yes I realize 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]

Well yes I realize 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.

 

[bugatti79]

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]

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]

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]

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.

 

[bugatti79]

good insight! Thanks