Answer :
Answer:
Step-by-step explanation:To find the work done by the force \(\vec{F} = \langle P, Q, R \rangle\) when moving a particle along the curve \(\vec{r}(t) = \langle x(t), y(t), z(t) \rangle\) from \(t = 0\) to \(t = 2\), we use the formula:
\[ W = \int_{a}^{b} \vec{F} \cdot \vec{r}'(t) \, dt \]
where \(\vec{r}'(t)\) is the derivative of \(\vec{r}(t)\) with respect to \(t\).
Given:
- \(\vec{F} = \langle P, Q, R \rangle = \langle x, y, z \rangle\)
- \(\vec{r}(t) = \langle t, t^2, t^3 \rangle\)
- \(t\) ranges from \(0\) to \(2\)
First, find \(\vec{r}'(t)\):
\[ \vec{r}'(t) = \langle 1, 2t, 3t^2 \rangle \]
Now, compute \(\vec{F} \cdot \vec{r}'(t)\):
\[ \vec{F} \cdot \vec{r}'(t) = \langle t, t^2, t^3 \rangle \cdot \langle 1, 2t, 3t^2 \rangle \]
\[ \vec{F} \cdot \vec{r}'(t) = t \cdot 1 + t^2 \cdot 2t + t^3 \cdot 3t^2 \]
\[ \vec{F} \cdot \vec{r}'(t) = t + 2t^3 + 3t^5 \]
Now, integrate \(\vec{F} \cdot \vec{r}'(t)\) from \(t = 0\) to \(t = 2\):
\[ W = \int_{0}^{2} (t + 2t^3 + 3t^5) \, dt \]
Compute each term separately:
\[ \int_{0}^{2} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{2} = 2 \]
\[ \int_{0}^{2} 2t^3 \, dt = 2 \left[ \frac{t^4}{4} \right]_{0}^{2} = 2 \cdot 4 = 8 \]
\[ \int_{0}^{2} 3t^5 \, dt = 3 \left[ \frac{t^6}{6} \right]_{0}^{2} = 3 \cdot \frac{64}{6} = 32 \]
Add them together:
\[ W = 2 + 8 + 32 = 42 \]
Therefore, the work done by the force \(\vec{F} = \langle P, Q, R \rangle = \langle x, y, z \rangle\) when it moves the particle along the arc of the curve from \(t = 0\) to \(t = 2\) is \( \boxed{42} \).