Differentiating for velocity

Question

The position $\vec r$ of a particle moving in an xy plane is given by $\vec r$ = $(1.00t^3 − 3.00t)i + (8.00 − 9.00t^4)j$, with $\vec r$ in meters and $t$ in seconds. In unit-vector notation, calculate the following for $t = 2.10 s$.

I found $\vec r$ by subbing in $2.1$ into the equation.

I differentiated the equation, then subbed $2.1$ to get velocity. The answer I have is $(3t^2 - 3)i + (0-9t^3)$, then $10.23 - 83.35$. It's not accepting that, though.

https://i.stack.imgur.com/VOCQB.jpg

"Velocity" is a vector not a number. Yes, the velocity is the derivative of the position so is $(3t^2- 3)i+ (-36t^3)j$ (you forgot the "j" and the coefficient of $t^3$ is -9(4)= -36). Yes, $3(2.1)^2- 3= 3(4.41)- 3= 10.23$ and $-36(2.1)^3= -36(9.261)= -333.396 so the velocity vector is 10.23i333.396j, NOT "10.23- 83..35". Apparently you forgot both the "i" and the "j" as well as forgetting the "4" in the j component. — user247327, Jul 11 '17 at 12:50

score 2 · Accepted Answer · answered Sep 23 '16 at 21:56

2

If $\vec r(t) = \hat i(t^3-3t)+\hat j(8-9t^4)$, then we find that

$$\vec r'(t) =\hat i (3t^2-3)+\hat j(-36t^3)$$

answered Sep 23 '16 at 21:56

Mark Viola

179,405

Doesn't 8 differentiate to 0? And leave -9t^4 to become -(3)9t^3? – user366783 Sep 23 '16 at 21:59
1

Yes, it does. So, $\frac{d}{dt}(8-9t^4)=(0-9\times 4\times t^3)=-36t^3$. – Mark Viola Sep 23 '16 at 22:00

score 1 · Answer 2 · answered Sep 23 '16 at 21:55

1

You forgot to multiply $9t^3$ by $4$ along the lines of the Power Rule.

answered Sep 23 '16 at 21:55

AlgorithmsX

4,560

Differentiating for velocity

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