Guichard 16.3.9

Zoë

Problem: Let $\vec F=\langle yz, xz, xy \rangle$. Find the work done by this force on an object that moves from (1,0,2) to (1,2,3).

I know that $W=\vec F \cdot \vec d$ and I'm able to find $\vec d$ by using the following method: $$\vec d = \langle 1,2,3 \rangle - \langle 1,0,2 \rangle = \langle 0,2,1 \rangle$$

When I go to take the dot product of $\vec F$ and $\vec d$, I get that $W=2(xz)+xy$. The answer is suppose to be $W=6$.

What am I doing wrong?

Zoë

In order to find the correct work done by the given force, try instead using the following equation: $$W=\int\limits_a^b \vec F(\vec r(t)) \cdot \vec r^{\prime}(t) \mathrm{d}t$$
The way you found $\vec d$ was correct but should then be plugged into the following to find $\vec r (t)$: $$\vec r (t) = \vec d * t + \langle 1,0,2 \rangle$$ $$\vec r (t) = \langle 0,2,1\rangle t + \langle 1,0,2 \rangle = \langle 1,2t,t+2 \rangle$$
Next, take the derivative of $\vec r (t)$.
$$\vec r^{\prime} (t) = \langle 0,2,1 \rangle$$
After finding $\vec r^{\prime} (t)$, solve for $\vec F (\vec r(t))$ by plugging the values found for $\vec r (t)$ in $\vec F $.
$$\vec F (\vec r(t)) = \langle (2t)(t+2),(1)(t+2),(1)(2t)\rangle=\langle 2t^2+4t,t+2,2t\rangle$$
Now you have found all the needed values to plug into the following equation and integrate from 0 to 1.
$$
\begin{aligned}
W&=\int\limits_0^1 \vec F(\vec r(t)) \cdot \vec r^{\prime}(t) \mathrm{d}t \newline
&=\int\limits_0^1(\langle 2t^2+4t,t+2,2t\rangle \cdot \langle 0,2,1 \rangle )\mathrm{d}t \newline
&=\int\limits_0^1 (0(2t^2+4t)+2(t+2)+1(2t))\mathrm{d}t \newline
&=\int\limits_0^1 (2t+4+2t)\mathrm{d}t \newline
&=\int\limits_0^1 (4t+4)\mathrm{d}t \newline
&=\frac{4(1)^2}{2}+4(1) \newline
W &=6
\end{aligned}
$$
Therefore, in order to obtain the correct answer, you need to use line integration.

mark

I gave your answer a like because I think it's correct. But I think you can do this a bit more easily. The vector field $\vec F$ is conservative. In fact, it arises as the gradient of $f(x,y,z)=xyz$. Thus, the work should be
$$f(1,2,3) - f(1,0,2).$$
Also, you really used the fact that the field is conservative in your answer because you assumed that the result is independent of path. That's why you can choose any old parametrization that begins and ends at the correct points.