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

Apr 8, 2018

$v = 3.5791$(Length unit/unit of time)
and it's direction $\theta = 253.7756$degrees

#### Explanation:

$t$ can't be negative as the time can't be negative but the following answer demonstrates the way to solve this kind of equations for $t = 1$ instead of $t = - 1$ and I will put the answer on double check.

This function is a displacement function for a particle moving in the Cartesian plane:
The function is divided into two parts $x , y$ which are both parametric equations
so You can differentiate each part alone to get the velocity-time function
$f ' \left(t\right) = \left(\left(1 - 2 t\right) {e}^{t - {t}^{2}} , 2 - {e}^{t} - t {e}^{t}\right)$

and by substituting with the value $t = - 1$ You get:

$f ' \left(1\right) = \left(\begin{matrix}- 1 {e}^{0} \\ 2 - {e}^{1} - 1 {e}^{1}\end{matrix}\right)$
and by simplification You get:
$f ' \left(- 1\right) = \left(- 1 , 2 \left(1 - e\right)\right)$
so the magnitude of the velocity of the particle at that moment will be:
$v$=$\sqrt{{\left({x}^{o}\right)}^{2} + {\left({y}^{0}\right)}^{2}}$

and it's direction will be $\tan \theta = \frac{{y}^{0}}{{x}^{0}}$

And by the substituting, You get $v = 3.5791$(Length unit/unit of time)
and $\theta = 253.775 \mathrm{de} g r e e s$