# Calculate the length of the given curve

## 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
Mentor
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?

Last edited:
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

$||r'(t)||= \sqrt{2cos(t)^2+5^2+(2sin(t)^2)}=\sqrt{29}$ where sin^2(t)+cos^2(t)=1...?

Last edited by a moderator:
I like Serena
Homework Helper
Good!

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

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?

Last edited:
I like Serena
Homework Helper
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 $\vec i \cdot \vec j= 0$

## 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
Homework Helper
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.

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