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Chapter 13: Problem 50

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

Short Answer

Expert verified
Question: Determine the points in the specified region where the tangent plane is horizontal for the surface defined by \(z = \sin(x-y)\). Answer: There are infinitely many points in the given region where the tangent plane is horizontal. A few examples of those points include \((x,y) = (\pi, \frac{\pi}{2}), (\pi, -\frac{\pi}{2}), (2\pi, \frac{3\pi}{2}), (2\pi, -\frac{\pi}{2}), \dots\).

Step by step solution

01

Compute the partial derivatives

Find the partial derivative of the function \(z = \sin(x-y)\) with respect to x and y, which are denoted as \(f_x\) and \(f_y\). Apply the chain rule for differentiation: $$f_x = \frac{\partial z}{\partial x} = \cos(x-y)$$ $$f_y = \frac{\partial z}{\partial y} = -\cos(x-y)$$
02

Find when the gradient vector is orthogonal to the xy-plane normal vector

The gradient vector is orthogonal to the normal vector of the xy-plane (\((0, 0, 1)\)) when their dot product is zero. The gradient vector is \((f_x, f_y)\). Compute the dot product: $$(f_x, f_y) \cdot (0, 0, 1) = f_x \cdot 0 + f_y \cdot 0 + 0 \cdot 1 = 0$$ Since the dot product is always zero, the gradient vector is orthogonal to the normal vector at all points on the surface.
03

Solve for x and y

Now we have to find the non-trivial solutions \((x, y)\) from the partial derivatives. Here, we can't just use the criterion \(f_x = 0\) and \(f_y = 0\). We have to check when the gradient vector becomes the null vector: $$f_x = \cos(x-y) = 0$$ $$f_y = -\cos(x-y) = 0$$
04

Determine the points in the specified region

Since the cosine function is zero at odd multiples of \(\frac{\pi}{2}\), \(x - y = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \dots\) Now we need to find the points \((x, y)\) such that their difference corresponds to these values and they lie in the specified region. For example, when \(x - y = \frac{\pi}{2}\), we can take \((x,y) = (\pi, \frac{\pi}{2}), (\pi, -\frac{\pi}{2}), (2\pi, \frac{3\pi}{2}), (2\pi, -\frac{\pi}{2}), \dots\) Similarly, we can find points for other values of \(x - y\). Therefore, there are infinitely many points in the given region where the tangent plane is horizontal.

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