Find the work done by F over the curve in the direction of increasing t.
-$\mathbf { F } = 6 x \mathbf { i } - 6 \mathrm { x } ^ { 5 } \mathrm { y } ^ { 4 } \mathbf { j } + \left( - 3 \mathrm { z } - 7 \mathrm { y } ^ { 2 } \right) \mathbf { k }$; the path is $\mathrm { C } _ { 1 } \cup \mathrm { C } _ { 2 } \cup \mathrm { C } _ { 3 }$ where $\mathrm { C } _ { 1 }$ is the straight line from $( 0,0,0 )$ to $( 1,0,0 ) , \mathrm { C } _ { 2 }$ is the straight line from $( 1,0,0 )$ to $( 1,1,0 )$, and $C _ { 3 }$ is the straight line from $( 1,1,0 )$ to $( 1,1,1 )$
A) $W = - \frac { 67 } { 10 }$
B) $W = \frac { 9 } { 5 }$
C) $W = - \frac { 57 } { 10 }$
D) $W = 3$

Question

Quizplus · Accepted Answer

We need to find the work done by the vector field $\mathbf F$ over the curve $\mathrm C_1 \cup \mathrm C_2 \cup \mathrm C_3$ in the direction of increasing $t$. Using the linearity of work integrals, we can find the work done over each of the three segments of the curve and sum them up.
$\mathrm C_1$ is the straight line from $(0,0,0)$ to $(1,0,0)$, which can be parameterized as $\mathbf r_1(t) = (t,0,0)$ for $0\leq t\leq 1$. The velocity vector $\mathbf r_1'(t)$ is $\mathbf r_1'(t) = \mathbf i$, which means the dot product $\mathbf F(\mathbf r_1(t))\cdot \mathbf r_1'(t)$ is simply $6t$. Therefore, the work done by $\mathbf F$ over $\mathrm C_1$ is given by:
$$W_\mathrm{C_1} = \int_{\mathrm C_1} \mathbf F \cdot d\mathbf r = \int_0^1 \mathbf F(\mathbf r_1(t)) \cdot \mathbf r_1'(t) \ dt = \int_0^1 6t\ dt = 3.$$
$\mathrm C_2$ is the straight line from $(1,0,0)$ to $(1,1,0)$, which can be parameterized as $\mathbf r_2(t) = (1,t,0)$ for $0\leq t\leq 1$. The velocity vector $\mathbf r_2'(t)$ is $\mathbf r_2'(t) = \mathbf j$, which means the dot product $\mathbf F(\mathbf r_2(t))\cdot \mathbf r_2'(t)$ is simply $-6t^5$. Therefore, the work done by $\mathbf F$ over $\mathrm C_2$ is given by:
$$W_\mathrm{C_2} = \int_{\mathrm C_2} \mathbf F \cdot d\mathbf r = \int_0^1 \mathbf F(\mathbf r_2(t)) \cdot \mathbf r_2'(t) \ dt = \int_0^1 (-6t^5)\ dt = -\frac{6}{6} = -1.$$
$\mathrm C_3$ is the straight line from $(1,1,0)$ to $(1,1,1)$, which can be parameterized as $\mathbf r_3(t) = (1,1,t)$ for $0\leq t\leq 1$. The velocity vector $\mathbf r_3'(t)$ is $\mathbf r_3'(t) = \mathbf k$, which means the dot product $\mathbf F(\mathbf r_3(t))\cdot \mathbf r_3'(t)$ is simply $-3-7 = -10$. Therefore, the work done by $\mathbf F$ over $\mathrm C_3$ is given by:
$$W_\mathrm{C_3} = \int_{\mathrm C_3} \mathbf F \cdot d\mathbf r = \int_0^1 \mathbf F(\mathbf r_3(t)) \cdot \mathbf r_3'(t) \ dt = \int_0^1 (-10)\ dt = -10.$$
The total work done is the sum of the work done over each segment of the curve:
$$W = W_\mathrm{C_1} + W_\mathrm{C_2} + W_\mathrm{C_3} = 3 - 1 - 10 = -\frac{67}{10}.$$
Therefore, the answer is (A).

Find the Work Done by F Over the Curve in the Direction