# What is the instantaneous velocity of an object moving in accordance to  f(t)= (e^(t^2),2t-te^t)  at  t=-1 ?

Mar 15, 2016

$f ' \left(- 1\right) = \left(- 2 e , 2\right)$.

#### Explanation:

The law $f \left(t\right)$ is the position of the object at the time $t$.

To find the instantaleous velocity we have to find the derivative of the previous law.

$f ' \left(t\right) = \left(2 t {e}^{{t}^{2}} , 2 - {e}^{t} - t {e}^{t}\right)$

and so:

$f ' \left(- 1\right) = \left(- 2 e , 2 - \frac{1}{e} + \frac{1}{e}\right) = \left(- 2 e , 2\right)$.