Problem 43 - Calculus with Concepts in Calculus

Question

Find an equation of the plane tangent to the given surface at the given point. $$ \sin (x y)=2-z^{2} ;\left(\pi, \frac{1}{2},-1\right) $$

Step-by-Step Solution

Answer

The tangent plane's equation is \( z = -1 \).

Step 1: Identify the Surface Function F(x, y, z)

The function from which the surface is derived is given by rearranging the equation \( F(x, y, z) = \sin(xy) - 2 + z^2 = 0 \). This function represents the surface in implicite form.

Step 2: Calculate the Gradient ∇F

The gradient of \( F \), \( abla F = (\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}) \), gives the normal vector to the surface. Calculate each partial derivative:- \( \frac{\partial F}{\partial x} = y \cos(xy) \)- \( \frac{\partial F}{\partial y} = x \cos(xy) \)- \( \frac{\partial F}{\partial z} = 2z \).Substitute \( (x, y, z) = (\pi, \frac{1}{2}, -1) \) into each derivative to find the gradient at that point:- \( \frac{\partial F}{\partial x} = \frac{1}{2} \cos(\pi \cdot \frac{1}{2}) = 0 \)- \( \frac{\partial F}{\partial y} = \pi \cos(\pi \cdot \frac{1}{2}) = 0 \)- \( \frac{\partial F}{\partial z} = 2(-1) = -2 \).Thus, \( abla F(\pi, \frac{1}{2}, -1) = (0, 0, -2) \).

Step 3: Write the Equation of the Tangent Plane

The equation of the tangent plane to the surface at a point \((x_0, y_0, z_0)\) with normal vector \((A, B, C)\) is given by:\[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \].Substituting the normal vector \((0, 0, -2)\) and the point \((\pi, \frac{1}{2}, -1)\):\[ 0(x - \pi) + 0(y - \frac{1}{2}) - 2(z + 1) = 0 \].Simplify this to:\[ z + 1 = 0 \].

Step 4: Finalize the Tangent Plane Equation

From the previous step, the equation of the tangent plane simplifies to \( z = -1 \). This plane is parallel to the xy-plane and passes through all points where \( z = -1 \).

Key Concepts

Surface Function

A surface function is a mathematical expression that represents a surface in a three-dimensional space. In the given exercise, the surface is presented in an implicit form through the function:

\[ F(x, y, z) = \sin(xy) - 2 + z^2 = 0 \]
Implicit forms of surface functions are valuable because they incorporate all three variables: x, y, and z. This allows us to understand a broader array of surfaces and apply calculus operations, such as finding tangent planes. By manipulating this function, we observe how changes in each of the variables affect the surface.

• The surface is defined by the zero set of the function.
• It is crucial for extracting information about the overall shape and orientation of the surface.

Gradient

The gradient of a function is a vector that indicates the direction of the steepest ascent. In our exercise, the surface function \( F(x, y, z) \) is used to find its gradient \( abla F \). This is calculated as follows:
\[ abla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) \]
The gradient is essential as it serves as the normal vector to the tangent plane at any point on the surface. Since it provides the most rapid rate of change, it helps us determine the tangent's orientation on the surface. In our example, the gradient at the specific point \( (\pi, \frac{1}{2}, -1) \) results in a vector:

• The gradient at the point is \( (0, 0, -2) \).
• It tells us that the surface rises steeply in the z-direction compared to the x and y directions, which have no rise.

Partial Derivatives

Partial derivatives measure how a function changes as each variable changes, while keeping other variables constant. For the function \( F(x, y, z) \), partial derivatives with respect to each variable are computed:

- \( \frac{\partial F}{\partial x} = y \cos(xy) \)- \( \frac{\partial F}{\partial y} = x \cos(xy) \)- \( \frac{\partial F}{\partial z} = 2z \)

Partial derivatives are instrumental in forming the gradient, revealing how each independent variable influences the function's value. By substituting the point \( (\pi, \frac{1}{2}, -1) \), students observe how each derivative particularly contributes:

• At the point, both \( \frac{\partial F}{\partial x} \) and \( \frac{\partial F}{\partial y} \) equal zero, implying no change along x and y directions.
• \( \frac{\partial F}{\partial z} = -2 \) indicates a decrease in the direction of the z-axis.

Normal Vector

The normal vector is a vector that is perpendicular to the tangent plane at a given point on a surface. In our task, calculating the normal vector involves finding the gradient of the function \( F(x, y, z) \) at the specified point.

This vector, \( (0, 0, -2) \), is perpendicular to the surface at \( (\pi, \frac{1}{2}, -1) \). Understanding the normal vector is critical because:

• It defines the exact orientation of the surface at a given point.
• It is used to formulate the equation of the tangent plane.
• The normal vector's components directly inform the coefficients in the tangent plane equation.

With the equation of the plane expressed as \( z = -1 \), the normal vector ensures the plane's alignment is consistent in three-dimensional space.