Calculate the length of the given curve

[bugatti79]

Homework Statement

2sin(t) \vec i +5t \vec j -2cos(t) \vec k for -5 <= t<=5

The Attempt at a Solution

let \vec r (t)= 2sin(t) \vec i +5t \vec j -2cos(t) \vec k

Then d \vec r(t) = 2cos(t) \vec i +5 \vec j + 2sin(t) \vec k

|| d\vec r(t) ||= \sqrt( (2cos(t) \vec i)^2+(5 \vec j)^2+ (2 sin(t) \vec k )^2)...?

 

[Mark44]

All those i's, j's, and k's are getting in your way.
Let r(t) = <2sin(t), 5t, -2cos(t)>
dr/dt = <2cos(t), 5, 2sin(t)>
How do you calculate the magnitude of a vector?

 

[bugatti79]

All those i's, j's, and k's are getting in your way.
Let r(t) = <2sin(t), 5t, -2cos(t)>
dr/dt = <2cos(t), 5, 2sin(t)>
How do you calculate the magnitude of a vector?

It will be the square root of the sum of the squares

||r&#039;(t)||= \sqrt{2cos(t)^2+5^2+(2sin(t)^2)}=\sqrt{29} where sin^2(t)+cos^2(t)=1...?

 

[I like Serena]

Good!

Do you have a formula for the length of a curve?

 

[bugatti79]

The length of the curve is the integral of this value with respect to t from -5 to 5...?

= |\sqrt{29}t|^{5}_{-5}...?

If the above is correct, how do you justify removing the basis i,j and k suddenly to proceed with the calculation?

 

[I like Serena]

That's it.

The formula for the length of a vector is ||x \vec i + y \vec j + z \vec k|| = \sqrt{x^2 + y^2 + z^2}

And if you really want to include i, j, and k, consider that \vec i \cdot \vec i = 1 and that \vec i \cdot \vec j= 0

 

[bugatti79]

Homework Statement

|| d\vec r(t) ||= \sqrt{( (2cos(t) \vec i)^2+(5 \vec j)^2+ (2 sin(t) \vec k )^2)}...?

So based on your last post this is corect since the dot product of i by itself, j by itself and k by itself is one...

Thanks

 

[I like Serena]

So based on your last post this is corect since the dot product of i by itself, j by itself and k by itself is one...

Thanks

Yes, it is correct, but it is very unusual to write it like this.

 

[bugatti79]

Ok, but it seems strange to suddenly drop the basis. Thanks anyhow!