Vector Fields — Question 9

Question 9 – Solution

Strategy. Substitute the curve into the field, then use vector projection onto the unit tangent.

Step 1: Restrict the field to the curve $$\boxed{\mathbf F(\mathbf r(t))=\langle\sin t,t,\cos t\rangle}.$$ Also, $$\mathbf r'(t)=\langle-\sin t,\cos t,1\rangle, \qquad |\mathbf r'(t)|=\sqrt 2,$$ so $$\boxed{\mathbf T(t)=\frac 1{\sqrt 2} \langle-\sin t,\cos t,1\rangle}.$$

Step 2: Data at $t=0$ $$\mathbf F(\mathbf r(0))=\langle 0,0,1\rangle,\qquad \mathbf T(0)=\frac 1{\sqrt 2}\langle 0,1,1\rangle.$$ The scalar tangential component is $$\mathbf F\cdot\mathbf T=\frac 1{\sqrt 2}.$$

Step 3: Vector decomposition $$\mathbf F_{\parallel}=(\mathbf F\cdot\mathbf T)\mathbf T =\boxed{\left\langle 0,\frac 12,\frac 12\right\rangle},$$ $$\mathbf F_{\perp}=\mathbf F-\mathbf F_{\parallel} =\boxed{\left\langle 0,-\frac 12,\frac 12\right\rangle}.$$

Verification The two boxed vectors sum to $\langle 0,0,1\rangle$, and $\mathbf F_{\perp}\cdot\mathbf T(0)=0$.