Line Integrals of Vector Fields — Question 9

Question 9 – Solution

Strategy. Carry the parameter derivative through both computations; the slower start must be included in $d\mathbf r$.

Step 1: First parametrization Here $$\mathbf F(\mathbf r_1(t))=\langle-t,3t\rangle,\qquad \mathbf r_1'(t)=\langle 1,2\rangle.$$ Thus $$\int_0^1(-t+6t)\,dt=\int_0^1 5t\,dt=\boxed{\frac 52}.$$

Step 2: Nonuniform parametrization Now $$\mathbf F(\mathbf r_2(u))=\langle-u^2,3u^2\rangle,\qquad \mathbf r_2'(u)=\langle 2u,4u\rangle.$$ Therefore $$\int_0^1(-2u^3+12u^3)\,du =\int_0^1 10u^3\,du=\boxed{\frac 52}.$$

Step 3: Interpret Although $\mathbf r_2'(0)=\mathbf 0$, the curve pauses only at one endpoint and then traces the segment once with the same orientation. The derivative factor compensates for its varying speed.

Verification Both maps begin at $(0,0)$, end at $(1,2)$, and move monotonically along the same segment, so parameter independence requires the matching values obtained above.