# Vector with relation to time

#### Kris

##### New member
A particle moves such that its position vector, as a function of time is
r(t) = (5*t^2)*i + (3*t^2)*j − (5*t)*k

Im trying to find the unit tangent to the trajectory as a function of time. However I can't seem to find any formula of how to do this.

#### Chris L T521

##### Well-known member
Staff member
A particle moves such that its position vector, as a function of time is
r(t) = (5*t^2)*i + (3*t^2)*j − (5*t)*k

Im trying to find the unit tangent to the trajectory as a function of time. However I can't seem to find any formula of how to do this.

The unit tangent vector is defined to be
$\mathbf{T}(t)=\frac{\mathbf{r}^{\prime}(t)} {|\mathbf{r}^{\prime}(t)|}.$
Since you're given $\mathbf{r}(t)$, it shouldn't be too difficult computing $\mathbf{r}^{\prime}(t)$ and it's norm $|\mathbf{r}^{\prime}(t)|$.

Can you take things from here?

#### Kris

##### New member
Thanks for the help

The answer I worked out to be was
(10*t*i+6*t*j-5*k)/(sqrt(136*t^2+25))

I at first got it wrong because I tried this answer
(10*t+6*t-5)/(sqrt(136*t^2+25))

But that didn't account for the i,j,k in the original vector hence the answer is
(10*t*i+6*t*j-5*k)/(sqrt(136*t^2+25))

Thanks for the help guys and I hope this can help someone else

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