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

[chetzread]

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 ?

 

[PeroK]

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 $|T'|$.

 

[chetzread]

You need $|\hat{T}'|$ not $|T'|$.

what's the difference between $|\hat{T}'|$ and $|T'|$ ?

 

[PeroK]

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.

 

[chetzread]

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 ?

 

[PeroK]

Unit tangent vector and tangent vector are not the same ?

No, why would they be? The tangent vector can have any length.

 

[chetzread]

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. ?

 

[PeroK]

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.

 

[chetzread]

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

 

[PeroK]

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 $\hat{T}$.

 

[chetzread]

removed

 

[chetzread]

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

 

[PeroK]

do you mean differentiate |T| to get |T'(t) | ?

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

What happened to the $1/5$?

 

[chetzread]

What happened to the $1/5$?

1/5 ?
i don't have 1/5 in my working

 

[PeroK]

1/5 ?
i don't have 1/5 in my working

Yes you do!

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

 

[chetzread]

Yes you do!

OK , but I didn't get 3 /25 . . . .

 

[PeroK]

OK , but I didn't get 3 /25 . . . .

That's because you dropped the 1/5!

 

[chetzread]

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

 

[chetzread]

That's because you dropped the 1/5!

for the previous question , i gt the question already , thanks!

 

[chetzread]

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 ?

 

[Mark44]

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.

 

[chetzread]

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) ,

 

[Mark44]

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:
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3\cos(t) & -3\sin(t) & 4 \\ -3\sin(t) & -3\cos(t) & 0 \end{vmatrix}$

, so magnitude = 6sqrt (17) ,

No