Problem 214 - Calculus Volume 3

Question

Find the equation of the tangent plane to the surface \(z=f(x, y)=\sin \left(x+y^{2}\right)\) at point \(\left(\frac{\pi}{4}, 0, \frac{\sqrt{2}}{2}\right),\) and graph the surface and the tangent plane.

Step-by-Step Solution

Answer

The equation of the tangent plane is \(z = \frac{\sqrt{2}}{2}x + \frac{\sqrt{2}}{2} - \frac{\sqrt{2}\pi}{8}\).

Step 1: Understand the Problem

We need to find the equation of a tangent plane to the surface \( z = \sin(x+y^2) \) at the given point \(\left(\frac{\pi}{4}, 0, \frac{\sqrt{2}}{2}\right)\). A tangent plane is a plane that just "touches" a surface at a point and is "parallel" to the surface at that point.

Step 2: Express the General Formula of Tangent Plane

The equation for the tangent plane to a surface \( z = f(x, y) \) at the point \((x_0, y_0)\) is given by: \( z = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \), where \( f_x \) and \( f_y \) are the partial derivatives of \( f \) with respect to \( x \) and \( y \), respectively.

Step 3: Calculate Partial Derivatives

Calculate the partial derivative \( f_x(x,y) \): \[ f_x(x, y) = \cos(x + y^2) \times 1 = \cos(x + y^2) \]. Calculate \( f_y(x,y) \): \[ f_y(x, y) = \cos(x + y^2) \times 2y = 2y \cos(x + y^2) \].

Step 4: Evaluate Partial Derivatives at the Point

Now substitute \( x_0 = \frac{\pi}{4} \) and \( y_0 = 0 \) into the partial derivatives:- \( f_x\left(\frac{\pi}{4}, 0\right) = \cos\left(\frac{\pi}{4} + 0^2\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \).- \( f_y\left(\frac{\pi}{4}, 0\right) = 2 \times 0 \times \cos\left(\frac{\pi}{4} + 0^2\right) = 0 \).

Step 5: Write the Tangent Plane Equation

Using the formula of the tangent plane and the evaluated derivatives, the equation can be formed:\[ z = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) + 0(y - 0) \]Simplify it to:\[ z = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}x - \frac{\sqrt{2}}{2} \times \frac{\pi}{4} \]So, the equation is:\[ z = \frac{\sqrt{2}}{2}x + \frac{\sqrt{2}}{2} - \frac{\sqrt{2}\pi}{8} \].

Step 6: Graph the Surface and Tangent Plane

Use a graphing tool or software to plot both the surface \( z = \sin(x+y^2) \) and the tangent plane \( z = \frac{\sqrt{2}}{2}x + \frac{\sqrt{2}}{2} - \frac{\sqrt{2}\pi}{8} \). They should intersect exactly at the point \(\left(\frac{\pi}{4}, 0, \frac{\sqrt{2}}{2}\right)\). The plane will be a flat surface touching the sine wave-like surface at this point.

Key Concepts

Partial Derivative

A partial derivative represents how a function changes as one of the variables changes while others are kept constant. In multivariable calculus, when you have a function like \( z = f(x, y) \), the partial derivative with respect to \( x \), denoted as \( f_x(x, y) \), tells us how \( z \) changes as \( x \) changes, keeping \( y \) fixed.

To calculate the partial derivative \( f_x(x, y) \) of our function \( f(x, y) = \sin(x + y^2) \), we treat \( y^2 \) as a constant and differentiate \( \sin(x + y^2) \) with respect to \( x \). This process gives us \( f_x(x, y) = \cos(x + y^2) \).

Similarly, the partial derivative concerning \( y \), \( f_y(x, y) \), involves differentiating the function while treating \( x \) as a constant. For \( f(x, y) = \sin(x + y^2) \), it results in \( f_y(x, y) = 2y \cos(x + y^2) \). Calculating these derivatives helps understand how the function behaves when making small changes in \( x \) or \( y \).

Graphing Surfaces

To visualize a function of two variables, such as \( z = \sin(x + y^2) \), we graph it as a surface in 3D space. Here, \( x \) and \( y \) serve as the horizontal axes, while \( z \) is the vertical dimension.

Graphing helps us understand the structure of the function better, as it reveals features like peaks, valleys, and slopes. In our specific example, the surface resembles a wavy pattern due to the sine function, with variations dependent on both \( x \) and \( y^2 \).

Once we incorporate the tangent plane \( z = \frac{\sqrt{2}}{2}x + \frac{\sqrt{2}}{2} - \frac{\sqrt{2}\pi}{8} \), we can observe where the plane touches the surface at a single point. This intersection gives insight into the local linear approximation of the surface at that point and helps verify the correctness of our calculations.

Equation of a Plane

The equation of a plane in 3D space typically takes the form \( ax + by + cz = d \), where \( a \), \( b \), \( c \), and \( d \) are constants. For tangent planes to a surface, we use a specific form derived from the partial derivatives:

Substitute calculated partial derivatives and the point of tangency into the formula for a tangent plane:
\[z = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0).\]

In the problem, this formula gave us the equation \( z = \frac{\sqrt{2}}{2}x + \frac{\sqrt{2}}{2} - \frac{\sqrt{2}\pi}{8} \).

This equation represents the plane touching the surface \( z = \sin(x + y^2) \) at the point \( (\frac{\pi}{4}, 0, \frac{\sqrt{2}}{2}) \). The tangent plane can be regarded as a flat approximation of the surface at a given point and is particularly useful in optimization and linear approximation problems.