# Thread: The curvature for a vector

1. ## The curvature for a vector

Find the t-value where $\displaystyle \kappa(t)=0.0001m^{-1}$
for this vectorfunction:
$\displaystyle \overrightarrow{r}(t)=\binom{450t}{1200-30t^2-7t^3}, t\in[-5,5]$

I know that I must use this formula:

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

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

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

I know that I must use this formula:

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

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

3. Oh, I get it now!

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?

4. That's it !

5. Thanks!