How do you differentiate the following parametric equation: $x(t)=e^(t^2-t) , y(t)=t^2-t$ ?

Question

1658 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-05T04:50:53

Explanation is given below.

$x(t) =e^(t^2-t)$
$y(t) = t^2 -t$

To find the derivative $dy/dx$ we need to find $y'(t)$ and $x'(t)$

$dy/dx=(y'(t))/(x'(t))$

$x(t)= e^(t^2-t)$
Differentiating with respect to $t$

$x'(t) = e^(t^2-t)d/dt(t^2-t)$
$x'(t)=e^(t^2-t)(2t-1)$
$x'(t)=(2t-1)e^(t^2-t)$

$y(t)=t^2-t$
Differentiating with respect to $t$

$y'(t)=2t-1$

$dy/dx = (2t-1)/((2t-1)e^(t^2-t))$
$dy/dx = cancel(2t-1)/(cancel(2t-1)e^(t^2-t))$

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

How do you differentiate the following parametric equation: x(t)=e^(t^2-t) , y(t)=t^2-t x(t)=e^(t^2-t) , y(t)=t^2-t ?