Problem 53 - Calculus Early Transcendentals

Question

Find the points at which the following surfaces have horizontal tangent planes. $$z=\cos 2 x \sin y \text { in the region }-\pi \leq x \leq \pi,-\pi \leq y \leq \pi$$

Step-by-Step Solution

Answer

Points with horizontal tangent planes are: $$ (-\pi, -\pi) , (-\pi,0), (-\pi, \pi), \left(-\frac{\pi}{2}, -\frac{\pi}{2}\right), \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), (0, -\pi), (0,0), (0, \pi), \left(\frac{\pi}{2}, -\frac{\pi}{2}\right), \left(\frac{\pi}{2}, \frac{\pi}{2}\right), (\pi, -\pi) , (\pi,0), (\pi, \pi) $$.

Step 1: Compute the partial derivatives of z with respect to x and y

The given function is \(z = \cos(2x)\sin(y)\). Let's compute the partial derivatives with respect to x and y: 1. Partial derivative with respect to x: $$\frac{\partial z}{\partial x} = -2\sin(2x)\sin(y)$$ 2. Partial derivative with respect to y: $$\frac{\partial z}{\partial y} = \cos(2x)\cos(y)$$

Step 2: Set up the system of equations

Horizontal tangent plane exists at points where both partial derivatives are equal to zero: $$ \begin{cases} -2\sin(2x)\sin(y) = 0 \\ \cos(2x)\cos(y) = 0 \end{cases} $$

Step 3: Solve the system of equations

We'll solve the system of equations by analyzing each equation separately and finding the points where the equations are simultaneously satisfied. 1. Analyze the first equation: $$-2\sin(2x)\sin(y) = 0$$ This equation is true when either \(\sin(2x) = 0\) or \(\sin(y)=0\). a) When \(\sin(2x) = 0\): $$2x = n\pi, \quad x = \frac{n\pi}{2}, \quad n \in \mathbb{Z}$$ b) When \(\sin(y) = 0\): $$y = m\pi, \quad m \in \mathbb{Z}$$ 2. Analyze the second equation: $$\cos(2x)\cos(y) = 0$$ This equation is true when either \(\cos(2x) = 0\) or \(\cos(y)=0\). a) When \(\cos(2x) = 0\): $$2x = (2n+1)\frac{\pi}{2}, \quad x = (2n+1)\frac{\pi}{4}, \quad n \in \mathbb{Z}$$ b) When \(\cos(y) = 0\): $$y = (2m+1)\frac{\pi}{2}, \quad m \in \mathbb{Z}$$ Now we need to find the points where both above conditions (1 and 2) are satisfied in the region \(-\pi \leq x \leq \pi,-\pi \leq y \leq \pi\).

Step 4: Find the points with horizontal tangent planes

We can see that for every point \((x,y)\) such that: $$ x = \frac{n\pi}{2}, y = m\pi $$ or $$ x = (2n+1)\frac{\pi}{4}, y = (2m+1)\frac{\pi}{2} $$ we have a horizontal tangent plane on the surface. Since we have a restricted region \(-\pi \leq x \leq \pi,-\pi \leq y\leq \pi\), we can find the points in this region: Points with horizontal tangent planes are: $$ (-\pi, -\pi) , (-\pi,0), (-\pi, \pi), \left(-\frac{\pi}{2}, -\frac{\pi}{2}\right), \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), (0, -\pi), (0,0), (0, \pi), \left(\frac{\pi}{2}, -\frac{\pi}{2}\right), \left(\frac{\pi}{2}, \frac{\pi}{2}\right), (\pi, -\pi) , (\pi,0), (\pi, \pi) $$