Curvature of r(t) = (3 sin t) i + (3 cos t) j + 4t k

In summary: Your work for r' x r'' is wrong. Your determinant for the cross product assumes that r'(t) is 3i + 3j + 4k, and that...
  • #1
chetzread
801
1

Homework Statement


[/B]
find the curvature of the vector valued function r(t) = 3sint i + 3cost j +4t k

Homework Equations

The Attempt at a Solution


For the unit tangent vector , i got T(t) = (3cost i -3sint j +4k) / sqrt (9 ((sint)^2 ) + 9 ((cost)^2 ) + 4^2 ) =
(3cost i -3sint j +4k) / 5

For the principal unit vector , N(t) , i got -3sint i -3cost j ) / sqrt (9 ((sint)^2 ) + 9 ((cost)^2 )
= -sint i -cost j

We can conclude that |r'(t) | = 5 , |T'(t) | = 3 , so , curvature = |T'(t) | / |r'(t) | = 3/5 , but the ans is 3/25 , which part of my working is wrong ?
 
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  • #2
chetzread said:

Homework Statement


[/B]
find the curvature of the vector valued function r(t) = 3sint i + 3cost j +4t k

Homework Equations

The Attempt at a Solution


For the unit tangent vector , i got T(t) = (3cost i -3sint j +4k) / sqrt (9 ((sint)^2 ) + 9 ((cost)^2 ) + 4^2 ) =
(3cost i -3sint j +4k) / 5

For the principal unit vector , N(t) , i got -3sint i -3cost j ) / sqrt (9 ((sint)^2 ) + 9 ((cost)^2 )
= -sint i -cost j

We can conclude that |r'(t) | = 5 , |T'(t) | = 3 , so , curvature = |T'(t) | / |r'(t) | = 3/5 , but the ans is 3/25 , which part of my working is wrong ?

You need ##|\hat{T}'|## not ##|T'|##.
 
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  • #3
PeroK said:
You need ##|\hat{T}'|## not ##|T'|##.
what's the difference between ##|\hat{T}'|## and ##|T'|## ?
 
  • #4
chetzread said:
what's the difference between ##|\hat{T}'|## and ##|T'|## ?

The first is the time derivative of the unit tangent vector. The second is the time derivative of the tangent vector.
 
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  • #5
PeroK said:
The first is the time derivative of the unit tangent vector. The second is the time derivative of the tangent vector.
Unit tangent vector and tangent vector are not the same ?
 
  • #6
chetzread said:
Unit tangent vector and tangent vector are not the same ?

No, why would they be? The tangent vector can have any length.
 
  • #7
PeroK said:
The first is the time derivative of the unit tangent vector. The second is the time derivative of the tangent vector.
so , how to get time derivative of the tangent vector. ?
 
  • #8
chetzread said:
so , how to get time derivative of the tangent vector. ?

It's the time derivative of the unit tangent vector you want. You just calculate the unit tangent vector, then differentiate it.
 
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  • #9
PeroK said:
It's the time derivative of the unit tangent vector you want. You just calculate the unit tangent vector, then differentiate it.
do you mean it's |T'(t) | ? if so , then my ans of |T'(t) | is 3 , since k = |T'(t) | / |r'(t) | , so i gt k = 3/5 , but not 3/25
 
  • #10
chetzread said:
do you mean it's |T'(t) | ? if so , then my ans of |T'(t) | is 3 , since k = |T'(t) | / |r'(t) | , so i gt k = 3/5 , but not 3/25

That's just what you did the first time! You need to differentiate ##\hat{T}##.
 
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  • #11
removed
 
  • #12
PeroK said:
It's the time derivative of the unit tangent vector you want. You just calculate the unit tangent vector, then differentiate it.
do you mean differentiate |T| to get |T'(t) | ?

If so , then my |T'(t) | = 3
 
  • #13
chetzread said:
do you mean differentiate |T| to get |T'(t) | ?

If so , then my |T'(t) | = 3

What happened to the ##1/5##?
 
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  • #14
PeroK said:
What happened to the ##1/5##?
1/5 ?
i don't have 1/5 in my working
 
  • #15
chetzread said:
1/5 ?
i don't have 1/5 in my working

Yes you do!

chetzread said:
For the unit tangent vector , i got T(t) = (3cost i -3sint j +4k) / sqrt (9 ((sint)^2 ) + 9 ((cost)^2 ) + 4^2 ) =
(3cost i -3sint j +4k) / 5
 
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  • #16
PeroK said:
Yes you do!
OK , but I didn't get 3 /25 . . . .
 
  • #17
chetzread said:
OK , but I didn't get 3 /25 . . . .

That's because you dropped the 1/5!
 
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  • #18
ok , i have tried in another way , but i didnt get the ans though ...
here's my ans , i gt k = 0.1357
instead of 4/25
 

Attachments

  • IMG_20161031_142025.jpg
    IMG_20161031_142025.jpg
    21.6 KB · Views: 692
  • #19
PeroK said:
That's because you dropped the 1/5!
for the previous question , i gt the question already , thanks!
 
  • #20
chetzread said:
ok , i have tried in another way , but i didnt get the ans though ...
here's my ans , i gt k = 0.1357
instead of 4/25
bump , what's wrong with my working ?
 

Attachments

  • IMG_20161031_142025.jpg
    IMG_20161031_142025.jpg
    21.8 KB · Views: 604
  • #21
chetzread said:
bump , what's wrong with my working ?
Your work for r' x r'' is wrong. Your determinant for the cross product assumes that r'(t) is 3i + 3j + 4k, and that r''(t) is -3i - 3j + 0k. Neither of these is correct.
 
  • #22
Mark44 said:
Your work for r' x r'' is wrong. Your determinant for the cross product assumes that r'(t) is 3i + 3j + 4k, and that r''(t) is -3i - 3j + 0k. Neither of these is correct.
after correction , my ans is 12i -12j -18k , so magnitude = 6sqrt (17) ,
 
  • #23
chetzread said:
after correction , my ans is 12i -12j -18k
No If this is for r'(t) X r''(t), it's wrong. In the attachment in post #20, you have the calculations for r'(t) and r''(t) correct, but the determinant leaves out all of the sin(t) and cos(t) factors.
The determinant for calculating r'(t) x r''(t) should look like this:
##\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3\cos(t) & -3\sin(t) & 4 \\ -3\sin(t) & -3\cos(t) & 0 \end{vmatrix}##
chetzread said:
, so magnitude = 6sqrt (17) ,
No
 
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Related to Curvature of r(t) = (3 sin t) i + (3 cos t) j + 4t k

What is the equation for the given curvature?

The equation for the given curvature is r(t) = (3 sin t) i + (3 cos t) j + 4t k.

What is the parametric equation for the given curvature?

The parametric equation for the given curvature is x = 3 sin t, y = 3 cos t, and z = 4t.

What is the unit tangent vector for the given curvature?

The unit tangent vector for the given curvature is T(t) = (3 cos t) i - (3 sin t) j + 4k.

What is the principal normal vector for the given curvature?

The principal normal vector for the given curvature is N(t) = -3 sin t i - 3 cos t j.

What is the binormal vector for the given curvature?

The binormal vector for the given curvature is B(t) = -4 cos t i + 4 sin t j + 3k.

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