Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the tangent plane to the given surface at the indicated point. \(z=x^{1 / 2}+y^{1 / 2} ;(1,4,3)\)

Step-by-Step Solution

Answer

The tangent plane is \( z = \frac{1}{2}x + \frac{1}{4}y + \frac{5}{2} \).

Step 1: Identifying Variables and Function

The function given is a surface represented by \[ z = x^{1/2} + y^{1/2} \] We are tasked with finding the equation of the tangent plane at the point (1, 4, 3), given as (x_0, y_0, z_0).

Step 2: Finding Partial Derivatives

To find the tangent plane, we first determine the partial derivatives of the function with respect to x and y.The partial derivative with respect to x is:\[ \frac{\partial z}{\partial x} = \frac{1}{2} x^{-1/2} \]The partial derivative with respect to y is:\[ \frac{\partial z}{\partial y} = \frac{1}{2} y^{-1/2} \]

Step 3: Evaluating Partial Derivatives at Given Point

Substitute the point (1, 4) into each partial derivative to find their values at the point:\[ \frac{\partial z}{\partial x} (1, 4) = \frac{1}{2} (1)^{-1/2} = \frac{1}{2} \]\[ \frac{\partial z}{\partial y} (1, 4) = \frac{1}{2} (4)^{-1/2} = \frac{1}{4} \]

Step 4: Using the Tangent Plane Formula

The formula for the tangent plane to a surface \( z = f(x, y) \) at point \((x_0, y_0, z_0)\) is:\[ z - z_0 = \left( \frac{\partial z}{\partial x} \right) (x - x_0) + \left( \frac{\partial z}{\partial y} \right) (y - y_0) \]Substitute the values:\[ z - 3 = \frac{1}{2} (x - 1) + \frac{1}{4} (y - 4) \]

Step 5: Simplifying the Equation

Simplify the equation for the tangent plane:\[ z - 3 = \frac{1}{2}x - \frac{1}{2} + \frac{1}{4}y - 1 \]Combine like terms:\[ z = \frac{1}{2}x + \frac{1}{4}y + \frac{5}{2} \]

Key Concepts

Partial Derivatives

In mathematics, a partial derivative measures how a function changes as its variables change independently of each other. This concept is crucial when dealing with functions of several variables, such as the surface equation given in the problem.

• The partial derivative of a function with respect to one variable considers all other variables as constants.
• For the surface z = x^1/2 + y^1/2, the derivative with respect to x, denoted as \( \frac{\partial z}{\partial x} \), captures how z changes when x changes while keeping y constant.
• Similarly, \( \frac{\partial z}{\partial y} \) represents the rate of change of z with change in y while x remains fixed.

Understanding partial derivatives is essential for computing tangent planes, which involves analyzing how a surface behaves in the vicinity of a point through these directional changes.

Surface Equation

The surface equation is a mathematical expression that defines a surface in three-dimensional space. In this exercise, we have the surface given by the equation:

• \( z = x^{1/2} + y^{1/2} \)

This surface is represented in terms of two variables, x and y. The "z" value is the height of the surface at any point (x, y).
The surface equation tells us, for every pair of x and y values, what the corresponding z value would be, which describes the shape of the surface. It's important to understand that a surface equation combines two changeable variables, offering insight into both the surface's topology and the principles governing its spatial change.

Tangent Plane Formula

The tangent plane to a surface at a certain point provides a planar approximation to the surface near that point. It is akin to a flat piece of paper just touching the surface. The formula for finding the tangent plane is derived from using partial derivatives and is given by:

• \[ z - z_0 = \left( \frac{\partial z}{\partial x} \right) (x - x_0) + \left( \frac{\partial z}{\partial y} \right) (y - y_0) \]

Here, \((x_0, y_0, z_0)\) is the specific point where the tangent plane touches the surface. Partial derivatives here act as the slopes of the plane in the x and y directions. These slopes tell us how steeply the tangent plane inclines away from the point in each direction. Applying the tangent plane formula allows us to derive a linear equation representing the tangent plane itself.

Evaluating Derivatives at a Point

When we evaluate derivatives at a specific point, we determine the exact change rates of the surface at that exact position. In our problem, we have:

• The partial derivative \( \frac{\partial z}{\partial x} \) becomes \( \frac{1}{2} \) when x is 1.
• Similarly, \( \frac{\partial z}{\partial y} \) becomes \( \frac{1}{4} \) when y is 4.

These values are critical because they provide the specific slopes of the tangent plane along the x and y axes. By substituting these evaluated derivatives back into the tangent plane equation, we paint a precise picture of how the surface behaves at the given point (1, 4, 3), enabling us to define the tangent plane's exact orientation and slope in space.