Problem 23 - Calculus Early Transcendentals

Question

Find an equation of the plane tangent to the following surfaces at the given points. $$z=(x-y) /\left(x^{2}+y^{2}\right) ;\left(1,2,-\frac{1}{5}\right) \text { and }\left(2,-1, \frac{3}{5}\right)$$

Step-by-Step Solution

Answer

Question: Write the equations of the tangent planes to the surface $$z = \frac{x - y}{x^2 + y^2}$$ at the points (1,2,-1/5) and (2,-1,3/5). Answer: The tangent planes at the given points are $$81z + 16 = -2x + 5y$$ and $$25z - 15 = 12x + 7y$$.

Step 1: Find the gradient of the surface

To find the gradient of the surface $$z = \frac{x - y}{x^2 + y^2}$$, we must first differentiate the function with respect to x and y. Partial derivative with respect to x: $$\frac{\partial z}{\partial x} = \frac{(x^2+y^2)(1)-(x-y)(2x)}{(x^2+y^2)^2}$$ Partial derivative with respect to y: $$\frac{\partial z}{\partial y} = \frac{(x^2+y^2)(-1)-(x-y)(2y)}{(x^2+y^2)^2}$$

Step 2: Evaluate the gradient at the given points

Now we will evaluate the gradient of the surface at the given points (1,2,-1/5) and (2,-1,3/5). For point (1,2,-1/5): $$\frac{\partial z}{\partial x} = \frac{(1^2+2^2)-(-1)(2\cdot1)}{(1^2+2^2)^2} = \frac{5-2}{81}$$ $$\frac{\partial z}{\partial y} = \frac{(1^2+2^2)(1+2(2)}{(1^2+2^2)^2} = \frac{5+8}{81}$$ For point (2,-1,3/5): $$\frac{\partial z}{\partial x} = \frac{(2^2+(-1)^2)-(-3)(2\cdot2)}{(2^2+(-1)^2)^2} = \frac{5-12}{25}$$ $$\frac{\partial z}{\partial y} = \frac{(2^2+(-1)^2)-(-3)(2\cdot(-1))}{(2^2+(-1)^2)^2} = \frac{5+6}{25}$$

Step 3: Write the planar equation for both tangent planes

Now we will use the point-normal form of the plane equation $$z - z_0 = a(x - x_0) + b(y - y_0)$$ with the points and gradients we found in the previous steps. For point (1,2,-1/5): $$z + \frac{1}{5} = \frac{5-2}{81}(x-1) + \frac{5+8}{81}(y-2)$$ Simplifying this equation, we get: $$81z + 16 = -2x + 5y$$ For point (2,-1,3/5): $$z - \frac{3}{5} = \frac{5-12}{25}(x-2) + \frac{5+6}{25}(y+1)$$ Simplifying this equation, we get: $$25z - 15 = 12x + 7y$$ So the tangent planes are: $$81z + 16 = -2x + 5y$$ and $$25z - 15 = 12x + 7y$$

Key Concepts

Partial Derivatives

Partial derivatives are a key tool in calculus, especially when dealing with functions of multiple variables. They allow us to understand how a function changes when one variable is varied, while keeping others constant.
In the context of a surface like \(z = \frac{x-y}{x^2+y^2}\), the partial derivative with respect to \(x\), denoted as \(\frac{\partial z}{\partial x}\), measures how the value of \(z\) changes as \(x\) is incremented, while \(y\) is held fixed. Similarly, \(\frac{\partial z}{\partial y}\) looks at changes in \(z\) as \(y\) changes, with \(x\) constant.
These derivatives form the core components in the calculation of the gradient, which is central to determining the tangent planes, as we shall see in the next sections.

Gradient

The gradient of a function of two variables is a vector that indicates the direction of the greatest rate of increase of the function.
For a surface defined by \(z = f(x, y)\), the gradient \(abla f(x, y)\) is given by the vector \([\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}]\).
In our exercise, these are calculated as follows:

• \(\frac{\partial z}{\partial x} = \frac{(x^2+y^2)(1)-(x-y)(2x)}{(x^2+y^2)^2}\)
• \(\frac{\partial z}{\partial y} = \frac{(x^2+y^2)(-1)-(x-y)(2y)}{(x^2+y^2)^2}\)

Evaluating these at the given points gives us the specific gradients used to define the tangent planes. The gradient's components are essentially the coefficients in the tangent plane equation.

Point-Normal Form

When finding the equation for a tangent plane, we often use the point-normal form. This form is particularly useful as it employs the gradient vector of the function, which is perpendicular to the tangent plane.
The point-normal form of a plane can be represented as \(z - z_0 = a(x - x_0) + b(y - y_0)\), where \((x_0, y_0, z_0)\) is a point on the plane, and \(a\) and \(b\) are the components of the gradient vector at that point.
By substituting the calculated gradients and given points into this equation, we can derive the equations of the tangent planes. This form is powerful because it simplifies to the familiar Cartesian plane equation, making it easy to manipulate and understand. Thus, each tangent plane is described by the gradient's contribution at specific points, adding significant geometrical insight to how the surface behaves locally.