Question

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

Short answer

In summary, the points with horizontal tangent planes on the surface \(z = \cos(2x) \cdot \sin(y)\) have the form: $$(x, y, z) = \left(x, \pm \pi, 0\right)$$ $$(x, y, z) = \left(x, 0, 0\right)$$ $$(x, y, z) = \left(-\pi+\frac{2k\pi}{n}, y, 0\right)$$ Where \(x \in \{-\pi, -\dfrac{\pi}{2}, 0, \dfrac{\pi}{2}, \pi\}\), \(y \in \{-\pi, 0, \pi \}\), and \(n, k \in \mathbb{Z}\).

Step-by-step solution

Find the gradient vector of the given surface

To find the gradient vector, we need to compute the partial derivatives of \(z(x, y)\) with respect to \(x\) and \(y\), as follows: $$\frac{\partial z}{\partial x} = \frac{d(\cos(2x) \cdot \sin(y))}{dx} = -2\sin(2x) \cdot \sin(y)$$ $$\frac{\partial z}{\partial y} = \frac{d(\cos(2x) \cdot \sin(y))}{dy} = \cos(2x) \cdot \cos(y)$$ So, the gradient vector is given by: $$\nabla z = \langle -2\sin(2x) \cdot \sin(y), \cos(2x) \cdot \cos(y) \rangle$$

Determine the conditions for horizontal tangent planes

For a horizontal tangent plane, the z-coordinate of the gradient vector equals zero. Thus: $$-2\sin(2x) \cdot \sin(y) = 0$$ This equation is true in three cases: 1. \(\sin(2x) = 0\) 2. \(\sin(y) = 0\) 3. Both \(\sin(2x)\) and \(\sin(y)\) are zero. We will now consider the given range for \(x\) and \(y\): \(-\pi \leq x \leq \pi\) and \(-\pi \leq y \leq \pi\)

Find the points satisfying the conditions

For each case, we find the specific points on the surface which satisfy the conditions: 1. \(\sin(2x) = 0\) This condition is met when \(2x = n\pi\), where \(n\) is an integer. However, \(x\) is in the range \([-\pi, \pi]\). In this range, we have: \(x = -\pi,-\dfrac{\pi}{2},0,\dfrac{\pi}{2},\pi\) Thus, \(x\) values are : \(-\pi,-\dfrac{\pi}{2}, 0, \dfrac{\pi}{2}, \pi\) At these values of \(x\), \(z\) equals $$z = \cos(2x) \cdot \sin(y) = 0$$ irrespective of the value of \(y\). 2. \(\sin(y) = 0\) This condition is met when \(y = m\pi\), where \(m\) is an integer. However, \(y\) is in the range \([-\pi, \pi]\). In this range, we have: \(y = -\pi,0,\pi\) Thus, \(y\) values are : \(-\pi, 0, \pi\) At these values of \(y\), \(z\) equals $$z = \cos(2x) \cdot \sin(y) = 0$$ irrespective of the value of \(x\).

Combine the values found in steps 3

To find the points with horizontal tangent planes on the surface \(z = \cos(2x) \cdot \sin(y)\), we combine the values of \(x\) and \(y\) from Step 3 (due to the conditions being satisfied for either \(\sin(2x) = 0\) or \(\sin(y) = 0\)). So, the points with horizontal tangent planes are of the form: $$(x, y, z) = \left(x, \pm \pi, 0\right)$$ $$(x, y, z) = \left(x, 0, 0\right)$$ $$(x, y, z) = \left(-\pi+\frac{2k\pi}{n}, y, 0\right)$$ Where \(x \in \{-\pi, -\dfrac{\pi}{2}, 0, \dfrac{\pi}{2}, \pi\}\), \(y \in \{-\pi, 0, \pi \}\), and \(n, k \in \mathbb{Z}\).

Key concepts

Partial Derivatives

Partial derivatives are key when dealing with functions of multiple variables. They help us understand how a function changes with respect to one variable, keeping others constant. In simple terms, it's similar to regular derivatives but in a multi-dimensional setting.

For the function \(z = \cos(2x) \cdot \sin(y)\), we need to find the partial derivatives with respect to both \(x\) and \(y\). This involves differentiating the trigonometric functions separately while treating one variable as a constant:

• The partial derivative with respect to \(x\) yields \(\frac{\partial z}{\partial x} = -2\sin(2x) \cdot \sin(y)\).
• The partial derivative with respect to \(y\) results in \(\frac{\partial z}{\partial y} = \cos(2x) \cdot \cos(y)\).

By analyzing these partial derivatives, you can determine how changes in \(x\) or \(y\) affect the surface.

Horizontal Tangent Planes

Horizontal tangent planes are a geometric concept where the tangent to a surface at a point is parallel to the \(xy\)-plane. This occurs when the gradient of the function at that point is zero.

The gradient vector \(abla z\) of our surface \(z = \cos(2x) \cdot \sin(y)\) provides information on the slope of the surface in all directions. For the tangent plane to be horizontal, the components of this gradient that point vertically must equal zero. This gives the conditions:

• \(-2\sin(2x) \cdot \sin(y) = 0\),which implies either \(\sin(2x) = 0\) or \(\sin(y) = 0\).

By solving these conditions, we can find precise points \( (x, y) \) where the tangent plane to the surface is horizontal.

Trigonometric Functions

Trigonometric functions like sine and cosine play a key role in this exercise. They are periodic functions which means they repeat their values in a regular pattern over intervals. This property is useful when solving equations in trigonometry.

For example, the function \(\sin(2x)\) becomes zero whenever \(2x\) is a multiple of \(\pi\). Similarly, \(\sin(y)\) is zero when \(y\) itself is an integer multiple of \(\pi\).

• These functions determine the points where the derivatives equate to zero.
• Because of periodicity, only specific values of \(x\) and \(y\) need to be considered within the interval \(-\pi \leq x, y \leq \pi\).

Understanding these properties helps in finding the points on the surface where the tangent plane is horizontal efficiently.