Line Integrals of Vector Fields — Question 6

Question 6

For a constant $a$, let $$\mathbf F_a(x,y)=\langle ay,x\rangle.$$ Along the parabola $\mathbf r(t)=\langle t,t^2\rangle$, $0\le t\le 1$, the work done is $2$. Determine $a$.

Tasks

1. Express the work in terms of $a$.

2. Solve for $a$.

3. Substitute the result and verify the specified work.

Question 6 – Solution

Strategy. Restrict the one-parameter family of fields to the given curve; the work condition becomes a linear equation in $a$.

Step 1: Build the integrand $$\mathbf F_a(\mathbf r(t))=\langle at^2,t\rangle,\qquad \mathbf r'(t)=\langle 1,2t\rangle.$$ Thus $$\mathbf F_a(\mathbf r(t))\cdot\mathbf r'(t) =at^2+2t^2=(a+2)t^2.$$

Step 2: Impose the work condition $$2=\int_0^1(a+2)t^2\,dt=\frac{a+2}{3}.$$ Therefore $a+2=6$, and $$\boxed{a=4}.$$

Step 3: Verify With $a=4$, $$\int_0^1(4+2)t^2\,dt =6\left[\frac{t^3}{3}\right]_0^1=2,$$ which is the required work.

Uniqueness check The coefficient of $a$ in the work is $\int_0^1t^2dt=1/3\ne 0$, so no second value of $a$ can satisfy the condition.