Line Integrals of Vector Fields — Question 7

Question 7

The field $\mathbf F(x,y)=\langle-y,x\rangle$ acts between $A=(0,0)$ and $B=(1,1)$. Compare the work along $$C_1:\mathbf r(t)=\langle t,t\rangle,\quad 0\le t\le 1,$$ and the broken path $C_2$ that goes from $(0,0)$ to $(1,0)$ to $(1,1)$.

Tasks

1. Compute the work along $C_1$.

2. Compute the work along both pieces of $C_2$.

3. Explain what the comparison demonstrates.

Question 7 – Solution

Strategy. Evaluate the same vector-field line integral on two explicitly parametrized paths with common endpoints.

Step 1: Diagonal path On $C_1$, $$\mathbf F(\mathbf r(t))=\langle-t,t\rangle,\qquad \mathbf r'(t)=\langle 1,1\rangle.$$ The dot product is $-t+t=0$, so $W_1=\boxed{0}$.

Step 2: Broken path On the horizontal piece $\langle t,0\rangle$, the dot product $\langle 0,t\rangle\cdot\langle 1,0\rangle$ is zero. On the vertical piece $\langle 1,s\rangle$, $$\langle-s,1\rangle\cdot\langle 0,1\rangle=1,\qquad 0\le s\le 1.$$ Thus $$W_2=0+\int_0^1 1\,ds=\boxed{1}.$$

Step 3: Compare Since $W_1\ne W_2$ despite identical endpoints, the work for this field depends on the path taken.

Verification Along the diagonal the field is perpendicular to the motion. Along the final vertical piece of $C_2$, its upward component is $1$, explaining the extra unit of work.