Cylindrical Coordinates — Question 5

Question 5 – Solution

Strategy. The cylinder fixes $r$; let $\theta$ be the curve parameter and use the plane for $z$.

Step 1: Constraints The cylinder gives $r=3$. The plane becomes $$z=r\cos\theta+r\sin\theta=3(\cos\theta+\sin\theta).$$ Thus $$\boxed{\mathbf r(\theta)=\left\langle 3\cos\theta,3\sin\theta,3(\cos\theta+\sin\theta)\right\rangle},\quad 0\le\theta<2\pi.$$

Step 2: Extrema Use $\cos\theta+\sin\theta=\sqrt 2\cos(\theta-\pi/4)$. Hence $z_{\max}=3\sqrt 2$ at $\theta=\pi/4$, and $z_{\min}=-3\sqrt 2$ at $\theta=5\pi/4$.

Step 3: Points $$\boxed{P_{\max}=(3/\sqrt 2,3/\sqrt 2,3\sqrt 2)},\qquad \boxed{P_{\min}=(-3/\sqrt 2,-3/\sqrt 2,-3\sqrt 2)}.$$ Both satisfy the original cylinder and plane equations.