Calculus with Vector Functions — Question 5

Question 5 – Solution

Strategy Differentiate the squared magnitude rather than the square root.

Proof Since $\vec r\cdot\vec r=25$, differentiation gives $2\vec r\cdot\vec r'=0$. Thus $\boxed{\vec r\cdot\vec r'=0}$, proving orthogonality wherever $\vec r'$ is nonzero.

Application Here $\vec r'(t)=\left\langle-3\sin t,3\cos t,0\right\rangle$ and $$\vec r\cdot\vec r'=-9\cos t\sin t+9\sin t\cos t=0.$$ Also $\|\vec r(t)\|=\sqrt{9+16}=5$, as required.