# Math Help - Finding the Curvature K

1. ## Finding the Curvature K

Hi
Need help on the following question:

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

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

This is what i have done.

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

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

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

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

This is where i am not sure what to do

how do you work out $|v|^3$???
P.S

2. Dear Paymemoney,

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

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

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

Hope you could continue.

3. oh yeh, the first part i found the unit tangent so i get this:

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

what would i do next?

4. Dear Paymemoney,

That is the answer. Since the denominator is positive, $K=\frac{36\mid{t}\mid}{(\sqrt{2}(1+t^2))^3}$

5. the answer in the book is $K=\frac{1}{3(1+t^2)^2}$ unless this is incorrect

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

I went all through this problem and found out $K=\frac{1}{3(1+t^2)^2}$ 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.

7. in relation to this question the book had an example on these types of question and i didn't understand how they got

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

to this:

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

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

8. 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

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

to this:

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

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 $A=a_{1}i+a_{2}j+a_{3}k~and~B=b_{1}i+b_{2}j+b_{3}k~ then~A\times{B}=\left(\begin{matrix} i & j & k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{matrix}\right)$

9. Originally Posted by Sudharaka
Dear Paymemoney,

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

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

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

Hope this helps.

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

12. 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, $\underline{T}=\frac{(1-t^2)i+2tj+(1+t^2)k}{\sqrt{2}(1+t^2)}$

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

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

Hope you can continue.