- $-\dfrac{1}{4}$
- $-\dfrac{1}{4\sqrt{2}}$
- $0$
- $\sqrt{2}$
- $3\sqrt{2}$

Solution :

Let $\text{F}$ be a constant force that is given by the vector $\left(\begin{array}{c}-1\\0\\1\end{array}\right)$. What is the work done by $\text{F}$ on a particle that moves along the path given by $\left(\begin{array}{c}t\\t^2\\t^3\end{array}\right)$ between time $t=0$ and $t=1$?

**Choice 3 is the answer.**

Recall that work can be calculated according to the line integral: \[W=\int_C{\text{F}\cdot d\text{r}}\] along a curve $C$ with force $\text{F}$ and position vector $\text{r}$. In this case, we have \[\text{r}=\left(\begin{array}{c}t\\t^2\\t^3\end{array}\right)\implies d\text{r}=\left(\begin{array}{c}1\\2t\\3t^2\end{array}\right)dt.\] The curve goes from 0 to 1, so our integral is: \[W=\int_0^1{\left(\begin{array}{c}-1\\0\\1\end{array}\right)\cdot \left(\begin{array}{c}1\\2t\\3t^2\end{array}\right)}dt=\int_0^1{(-1+0+3t^2)}dt=0\] which gives zero work as we wanted.

- $-\dfrac{1}{4}$
- $-\dfrac{1}{4\sqrt{2}}$
- $0$
- $\sqrt{2}$
- $3\sqrt{2}$

Solution :

Recall that work can be calculated according to the line integral: \[W=\int_C{\text{F}\cdot d\text{r}}\] along a curve $C$ with force $\text{F}$ and position vector $\text{r}$. In this case, we have \[\text{r}=\left(\begin{array}{c}t\\t^2\\t^3\end{array}\right)\implies d\text{r}=\left(\begin{array}{c}1\\2t\\3t^2\end{array}\right)dt.\] The curve goes from 0 to 1, so our integral is: \[W=\int_0^1{\left(\begin{array}{c}-1\\0\\1\end{array}\right)\cdot \left(\begin{array}{c}1\\2t\\3t^2\end{array}\right)}dt=\int_0^1{(-1+0+3t^2)}dt=0\] which gives zero work as we wanted.

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