# Work done by a force Field

Homework for Calc III includes a problem about computing the work done by a force field (defined by a specific vector equation) on a moving particle. I was attempting to compute this using the equation $$\int_a^b {\textbf{F}(\textbf{r}(t)) \bullet \textbf{r}'(t) \space \mathrm{d}}t$$ defined by the following vectors: $$\textbf{F} (x,y) = xy\textbf{i}+3y^2\textbf{j} \space \mathrm{and} \space \textbf{r}(t)=11t^4\textbf{i}+t^3\textbf{j} \space\mathrm{for}\space 0 \le t \le 1$$

Alas, I computed $\textbf{r}'(t)=44t^3\textbf{i}+3t^2\textbf{j}$, which is never going to produce the answer of 45 that the book gives, unless I am doing something very very wrong..

Would someone be so kind as to resolve this for me?

Thanks!

Edit: Wow, when integrating, I forgot to multiply by $1\over n$ as in $\int x^n \space \mathrm{d}$. Thanks for the help!

• what are the values for $a$ and $b$? Commented May 6, 2013 at 17:19
• And $f$ and $F$ are meant to be the same, right? Commented May 6, 2013 at 17:20
• Right, good points guys, lemme fix that.... a little trouble with latex made me slip my details.. Commented May 6, 2013 at 17:20
• You are doing something wrong in your computation. The answer is 45. Commented May 6, 2013 at 17:25
• Your approach is correct, it will lead to the right answer. Commented May 6, 2013 at 17:25

$r(t) = (11 t^4, t^3)$. $r'(t) = (44 t^3, 3 t^2)$. $f(r(t)) = (11 t^7, 3 t^6)$. $\langle f(r(t)), r'(t) \rangle = 484 t^{10} + 9 t^8$. Integrating the latter over $[0,1]$ gives $45$.