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Calculate the length of the given curve

  • Thread starter bugatti79
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  • #1
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Homework Statement



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


The Attempt at a Solution



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

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

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

Answers and Replies

  • #2
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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?
 
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  • #3
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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

[itex]||r'(t)||= \sqrt{2cos(t)^2+5^2+(2sin(t)^2)}=\sqrt{29}[/itex] where sin^2(t)+cos^2(t)=1...?
 
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  • #4
I like Serena
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Good! :smile:

Do you have a formula for the length of a curve?
 
  • #5
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The length of the curve is the integral of this value with respect to t from -5 to 5...?

[itex]= |\sqrt{29}t|^{5}_{-5}[/itex].....?

If the above is correct, how do you justify removing the basis i,j and k suddenly to proceed with the calculation?
 
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  • #6
I like Serena
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That's it.

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

And if you really want to include i, j, and k, consider that [itex]\vec i \cdot \vec i = 1[/itex] and that [itex]\vec i \cdot \vec j= 0[/itex]
 
  • #7
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Homework Statement


[itex]|| d\vec r(t) ||= \sqrt{( (2cos(t) \vec i)^2+(5 \vec j)^2+ (2 sin(t) \vec k )^2)}[/itex]....?
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
 
  • #8
I like Serena
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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.
 
  • #9
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Ok, but it seems strange to suddenly drop the basis. Thanks anyhow!
 

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