Question

What is the derivative of `f(t) = (t^3-te^(1-t) , t^2-6t+e^t )`?

Answer 1

`f'(t) = (x'(t), y'(t)) = ([3t^2 + te^(1-t) - e^(1-t)], 2t - 6 + e^t)`

Explanation:

Recall that to compue the derivative of the function `f(t)`, we just need to compute the derivatives of each parametric equation.

If we have `f(x,y) = (x, y)`, then `f(t) = (x(t), y(t))`, where `x(t)` and `y(t)` are our two parametric equations. These equations describe `x` and `y` as functions of the single parameter `t`.

For our particular example:
`x(t) = t^3 - te^(1-t)`
`y(t) = t^2 - 6t + e^t`

So, we just need to take the derivative of each equation:

`x'(t) = 3t^2 - (-te^(1-t) + e^(1-t))`
`y'(t) = 2t - 6 + e^t`

Our final solution is:

`f'(t) = (x'(t), y'(t)) = ([3t^2 + te^(1-t) - e^(1-t)], 2t - 6 + e^t)`

Hope that helped :)