# 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...

## Homework Statement

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

## 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 ?

Last edited by a moderator:

## Homework Statement

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

## 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'|##.

Last edited by a moderator:
PeroK said:
You need ##|\hat{T}'|## not ##|T'|##.
what's the difference between ##|\hat{T}'|## and ##|T'|## ?

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.

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 ?

Unit tangent vector and tangent vector are not the same ?

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

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. ?

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.

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

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}##.

removed

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

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

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

What happened to the ##1/5##?

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

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

PeroK said:
Yes you do!
OK , but I didn't get 3 /25 . . . .

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

That's because you dropped the 1/5!

ok , i have tried in another way , but i didnt get the ans though ...
here's my ans , i gt k = 0.1357

#### Attachments

• IMG_20161031_142025.jpg
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PeroK said:
That's because you dropped the 1/5!
for the previous question , i gt the question already , thanks!

ok , i have tried in another way , but i didnt get the ans though ...
here's my ans , i gt k = 0.1357
bump , what's wrong with my working ?

#### Attachments

• IMG_20161031_142025.jpg
21.8 KB · Views: 604
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.

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) ,

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

## 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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