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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?


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

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what are the values for $a$ and $b$? – Shuhao Cao May 6 '13 at 17:19
And $f$ and $F$ are meant to be the same, right? – Milind May 6 '13 at 17:20
Right, good points guys, lemme fix that.... a little trouble with latex made me slip my details.. – user1833028 May 6 '13 at 17:20
You are doing something wrong in your computation. The answer is 45. – copper.hat May 6 '13 at 17:25
Your approach is correct, it will lead to the right answer. – Milind May 6 '13 at 17:25
up vote 1 down vote accepted

Your approach is correct, it will lead to the right answer.

\begin{align} \textbf{F}(x,y) &= (xy,3y^2)\\ \textbf{r}(t) &= (11t^4, t^3)\\ \textbf{r}'(t) &= (44t^3, 3t^2)\\ \end{align}

\begin{align} \int_0^1 \textbf{F}(\textbf{r}(t))\cdot\textbf{r}'(t)dt &= \int_0^1 (11t^4* t^3,3t^6)\cdot (44t^3, 3t^2)dt \\ &= \int_0^1 484t^{10} + 9t^8dt \\ &=\left.44t^{11} + t^9\right |_0^1 \\ &=45 \end{align}

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

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