Question

What is the instantaneous velocity of an object moving in accordance to `f(t)= (t-e^t,te^(2t))` at `t=3`?

Answer 1

Instantaneous velocity is `dotx=(1-e^3, 7e^6 )`,

or in vector form
`((1-e^3), (7e^6) )`,

or `(1-e^3)hati + 7e^6hatj` in cartesian vector form

Explanation:

I assume that f(t) is a position, Standard convention for a position in space would be

`x(t)=(t-e^t,te^(2t))`

Then we can get the velocity by differentiaiting (we will need to use the product rule and chain rules) as `dotx=dx/dt`,

Differentiating wrt `t` gives us:
`dotx(t) = (1-e^t, (t)(d/dte^(2t)) + (d/dtt)(e^(2t)) )`
`:. dotx(t) = (1-e^t, t(2)e^(2t) + 1(e^(2t)) )`
`:. dotx(t) = (1-e^t, 2te^(2t) + e^(2t) )`

When `t=3 =>dotx(3) = (1-e^3, 2(3)e^6 + e^6 )`
`:. dotx(3) = (1-e^3, 7e^6 )`

Hence the instantaneous velocity is `dotx=(1-e^3, 7e^6 )`,

or in vector form
`((1-e^3), (7e^6) )`,

or `(1-e^3)hati + 7e^6hatj` in cartesian vector form