# The curvature for a vector

• May 2nd 2009, 10:00 AM
No Logic Sense
The curvature for a vector
Find the t-value where $\kappa(t)=0.0001m^{-1}$
for this vectorfunction:
$\overrightarrow{r}(t)=\binom{450t}{1200-30t^2-7t^3}, t\in[-5,5]$

I know that I must use this formula:

$\kappa(t)=\frac{x'(t)y''(t)-x''(t)y'(t)}{(x'(t)^2+y'(t)^2)^\frac{3}{2}}$

It's just that I don't know where to begin and what to do. :(
• May 2nd 2009, 10:26 AM
running-gag
Quote:

Originally Posted by No Logic Sense
Find the t-value where $\kappa(t)=0.0001m^{-1}$
for this vectorfunction:
$\overrightarrow{r}(t)=\binom{450t}{1200-30t^2-7t^3}, t\in[-5,5]$

I know that I must use this formula:

$\kappa(t)=\frac{x'(t)y''(t)-x''(t)y'(t)}{(x'(t)^2+y'(t)^2)^\frac{3}{2}}$

It's just that I don't know where to begin and what to do. :(

Hi

x(t) = 450t and y(t) = 1200-30t^2-7t^3

You can substitute in the formula but it is going to be a hard job (Wondering)
• May 2nd 2009, 10:36 AM
No Logic Sense
Oh, I get it now!(Evilgrin)

I can just insert x(t) and y(t) in the respective places and then solve the fraction with the solve button on my calculator, right?
• May 2nd 2009, 10:50 AM
running-gag
That's it !
• May 2nd 2009, 10:55 AM
No Logic Sense
Thanks! :D