# Finding the Curvature K

• Oct 27th 2010, 11:59 PM
Paymemoney
Finding the Curvature K
Hi
Need help on the following question:

A curve has vector equation $\displaystyle r=(3t-t^3)i+(3t^2)j+(3t++t^3)k$ Find its curvature K

i know that $\displaystyle K = \frac{|v * a|}{|v^3|}$

This is what i have done.

$\displaystyle v= (3-3t^2)i+6tj+(3++3t^2)k$
$\displaystyle a= -6ti+6j+6tk$

$\displaystyle |v * a| = \begin {bmatrix} 3-t^2 & 6t & 3+3t^2 || -6t & 6 & 6t \end{bmatrix}$

$\displaystyle i(36t^2-18+18t^2)-j(18t-18^3+18t+18t^3)+k(18-18t^2+36t^2)$
$\displaystyle i(18t^2-18)-j(36t)+k(18-18t^2)$
$\displaystyle 18t^2-18-36t+18-18t^2$
$\displaystyle |v*a|=-36t$

$\displaystyle |v| = 3-3t^2+6t+3+3t^2$

This is where i am not sure what to do

how do you work out $\displaystyle |v|^3$???
P.S
• Oct 28th 2010, 12:10 AM
Sudharaka
Dear Paymemoney,

I hope the correct vector equation must be, $\displaystyle r=(3t-t^3)i+(3t^2)j+(3t+t^3)k$

If you want to find the curvature first find $\displaystyle \underline{T}=\frac{dr}{ds}=\frac{r'(t)}{\mid{r'(t )}\mid}$; where $\displaystyle \underline{T}$ is the unit tangent vector.

$\displaystyle K=\mid\frac{dT}{ds}\mid=\mid\frac{dT}{dt}\times\fr ac{dt}{ds}\mid=\mid\frac{1}{\mid{r'(t)}\mid}\frac{ dT}{dt}\mid$

Hope you could continue.
• Oct 28th 2010, 12:48 AM
Paymemoney
oh yeh, the first part i found the unit tangent so i get this:

$\displaystyle K =\frac{|-36t|}{|\sqrt{2}(1+t^2)|^3}$

what would i do next?
• Oct 28th 2010, 12:58 AM
Sudharaka
Dear Paymemoney,

That is the answer. Since the denominator is positive, $\displaystyle K=\frac{36\mid{t}\mid}{(\sqrt{2}(1+t^2))^3}$
• Oct 28th 2010, 04:06 AM
Paymemoney
the answer in the book is $\displaystyle K=\frac{1}{3(1+t^2)^2}$ unless this is incorrect
• Oct 28th 2010, 04:46 AM
Sudharaka
Quote:

Originally Posted by Paymemoney
the answer in the book is $\displaystyle K=\frac{1}{3(1+t^2)^2}$ unless this is incorrect

Dear Paymemoney,

I went all through this problem and found out $\displaystyle K=\frac{1}{3(1+t^2)^2}$ is the answer. Probably you have done something wrong in your calculations. Try to solve the problem according to the method I have given and if you have any problem please dont hesitate to ask.
• Oct 28th 2010, 04:31 PM
Paymemoney
in relation to this question the book had an example on these types of question and i didn't understand how they got

$\displaystyle r^{'}(t) * r^{''}(t) = \begin{matrix} i & j & k \\ 1 & 2t & 3t^2 \\ 0 & 2 & t \end{bmatrix}$ $\displaystyle = 6^2i-6tj+2k$

to this:

Hence $\displaystyle |r^{'}(t) * r^{''} (t)| = \sqrt{4+36^2+36t^4}=2\sqrt{1+9t^2+9t^4}$

why is the values inside 4+36&2+36^4?
• Oct 28th 2010, 05:02 PM
Sudharaka
Quote:

Originally Posted by Paymemoney
in relation to this question the book had an example on these types of question and i didn't understand how they got

$\displaystyle r^{'}(t) * r^{''}(t) = \begin{matrix} i & j & k \\ 1 & 2t & 3t^2 \\ 0 & 2 & t \end{bmatrix}$ $\displaystyle = 6^2i-6tj+2k$

to this:

Hence $\displaystyle |r^{'}(t) * r^{''} (t)| = \sqrt{4+36^2+36t^4}=2\sqrt{1+9t^2+9t^4}$

why is the values inside 4+36&2+36^4?

Dear Paymemoney,

Cross product of two vectors can be written in matrix notation. That is, if $\displaystyle A=a_{1}i+a_{2}j+a_{3}k~and~B=b_{1}i+b_{2}j+b_{3}k~ then~A\times{B}=\left(\begin{matrix} i & j & k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{matrix}\right)$

• Oct 28th 2010, 05:58 PM
Paymemoney
Quote:

Originally Posted by Sudharaka
Dear Paymemoney,

Cross product of two vectors can be written in matrix notation. That is, if $\displaystyle A=a_{1}i+a_{2}j+a_{3}k~and~B=b_{1}i+b_{2}j+b_{3}k~ then~A\times{B}=\left(\begin{matrix} i & j & k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{matrix}\right)$

not really, i understand that part, its the part where you square root i don't understand
• Oct 29th 2010, 02:48 AM
Sudharaka
Quote:

Originally Posted by Paymemoney
not really, i understand that part, its the part where you square root i don't understand

Dear Paymemoney,

The magnitude of a vector $\displaystyle \underline{r}$ is denoted by $\displaystyle \mid\underline{r}\mid$. Suppose you have a vector $\displaystyle \underline{r}=xi+yj+zk$. Then the magnitude of this vector is given by, $\displaystyle \mid{\underline{r}\mid=\sqrt{x^2+y^2+z^2}$.

Hope this helps.
• Oct 29th 2010, 01:56 PM
Paymemoney
yep i understand now, but as for the original question i still cannot get the answer would you show me your solutions to it?
• Oct 29th 2010, 08:17 PM
Sudharaka
Quote:

Originally Posted by Paymemoney
yep i understand now, but as for the original question i still cannot get the answer would you show me your solutions to it?

Dear Paymemoney,

Okay. What did you get for the unit tangent vector? It should be, $\displaystyle \underline{T}=\frac{(1-t^2)i+2tj+(1+t^2)k}{\sqrt{2}(1+t^2)}$

Now differentiate this with respect to 's' and you would end up with, $\displaystyle \frac{dT}{ds}=\frac{1}{3(1+t^2)}\left[\frac{-2t}{(1+t^2)^2}i+\frac{1-t^2}{(1+t^2)^2}j\right]$

Therefore, $\displaystyle K=\mid\frac{dT}{ds}\mid=\frac{1}{3(1+t^2)^3}\mid(-2t)i+(1-t^2)j\mid$

Hope you can continue.