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

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Answer 1 · 2017-02-22T13:43:51

Instantaneous velocity is given by the vector:

$vec(v) = << 1-e^2, 2e^4 >>$ , or, $(1-e^2) hat(i)+2e^4hat(j)$ , or, $( (1-e^2),(2e^4) )$

We have:

$f(t) = ( x(t), y(t) )$ where $x(t)=t-e^t$ , $y(t)=e^(2t)$

Where $f(t)$ represents the position at tme $t$ ; Then;

$dx/dt = 1-e^t = 1/(2sqrt(t+2))$
$dy/dt = 2e^(2t)$

So, when $t=2$ :

$dx/dt = 1-e^2 ~~ -6.389$
$dy/dt = 2e^4 \ \ \ \ \ ~~ 109.196$

And so the instantaneous velocity is given by the vector:

$vec(v) = << 1-e^2, 2e^4 >>$ , or, $(1-e^2) hat(i)+2e^4hat(j)$ , or, $( (1-e^2),(2e^4) )$

If we want the instantaneous speed , it is given by:

$v = || vec(v) ||$
$\ \ \= sqrt( (1-e^2)^2 + (2e^4)^2 )$
$\ \ \= sqrt( 11964.651 ... )$
$\ \ \~~ 109.383 ...$

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