Problem 32 - Calculus with Concepts in Calculus

Question

Find an equation of the plane tangent to the graph of the given function at the indicated point(s). $$ f(x, y)=\frac{x+2}{y+1} ;(2,3,1) $$

Step-by-Step Solution

Answer

The equation of the tangent plane is \( z = \frac{1}{4}x - \frac{1}{4}y + \frac{5}{4} \).

Step 1: Understand the problem

We need to find the equation of a tangent plane to the surface defined by the function \( f(x, y) = \frac{x+2}{y+1} \) at the given point \((2, 3, 1)\). This involves the use of partial derivatives to obtain the normal vector to the plane.

Step 2: Verify the point

Plug \(x = 2\) and \(y = 3\) into the given function to check that it gives the \(z\)-coordinate of the point.\[f(2, 3) = \frac{2+2}{3+1} = \frac{4}{4} = 1\]This matches the \(z\)-coordinate given as 1, thus confirming that \((2, 3, 1)\) is indeed a point on the surface.

Step 3: Find partial derivatives

Calculate the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\). These represent the rate of change of \(f\) in the direction of \(x\) and \(y\), respectively:\[\frac{\partial f}{\partial x} = \frac{1}{y+1}\]\[\frac{\partial f}{\partial y} = -\frac{x+2}{(y+1)^2}\]

Step 4: Evaluate partial derivatives at the point

Substitute \(x = 2\) and \(y = 3\) into the partial derivatives to find their specific values:\[\frac{\partial f}{\partial x}(2, 3) = \frac{1}{3+1} = \frac{1}{4}\]\[\frac{\partial f}{\partial y}(2, 3) = -\frac{2+2}{(3+1)^2} = -\frac{4}{16} = -\frac{1}{4}\]

Step 5: Write equation of tangent plane

The equation of the tangent plane at \((x_0, y_0, z_0)\) is given by:\[z = f(x_0, y_0) + \frac{\partial f}{\partial x}(x_0, y_0)(x - x_0) + \frac{\partial f}{\partial y}(x_0, y_0)(y - y_0)\]Substitute the known values \((2, 3, 1)\) and partial derivatives:\[z = 1 + \frac{1}{4}(x - 2) - \frac{1}{4}(y - 3)\]

Step 6: Simplify the tangent plane equation

Expand and simplify the equation:\[z = 1 + \frac{1}{4}x - \frac{1}{2} - \frac{1}{4}y + \frac{3}{4}\]Combining like terms results in:\[z = \frac{1}{4}x - \frac{1}{4}y + \frac{5}{4}\]

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus, which deals with functions of multiple variables. In the context of this problem, we are looking at a function of two variables, \(x\) and \(y\), so we focus on finding the rate at which the function \(f(x, y)\) changes as each variable changes, holding the other constant. This is precisely what partial derivatives help us determine.

• To find the partial derivative of \(f(x, y) = \frac{x+2}{y+1}\) with respect to \(x\), we treat \(y\) as a constant and differentiate with respect to \(x\): \(\frac{\partial f}{\partial x} = \frac{1}{y+1}\).
• Similarly, the partial derivative with respect to \(y\) treats \(x\) as a constant: \(-\frac{x+2}{(y+1)^2}\).

Evaluating these derivatives at a specific point gives us the slope of the tangent line to the function's graph at that point on the respective axis. This information is crucial in defining the tangent plane, as it dictates how the function is inclining or declining at precise sections of the graph.

Surface Normal Vector

The surface normal vector is central in determining the tangent plane of a surface at a given point. It is a vector that is perpendicular to the tangent plane. In practical terms, it provides the direction along which the surface is rising steeply at a given point, like a mountain's peak.
When dealing with functions of two variables, the components of the surface normal vector can be determined using the partial derivatives. For a function \(z = f(x, y)\), the normal vector is derived from the partial derivatives as follows:

• The partial derivative \(\frac{\partial f}{\partial x}\) provides the slope in the \(x\) direction.
• The partial derivative \(\frac{\partial f}{\partial y}\) offers the slope in the \(y\) direction.

Together, these components form the basis of the normal vector to the surface at that point. In this exercise, the partial derivatives evaluated at \((2, 3)\) become key components \(\langle \frac{1}{4}, -\frac{1}{4}, -1 \rangle\) of the normal vector. This normal vector, when substituted in the tangent plane equation, assists in illustrating the orientation and exact positioning of the tangent plane in three-dimensional space.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions with more than one variable. It becomes crucial when analyzing surfaces and curves in three-dimensional space. This particular field of study focuses on understanding how different variables interact and influence the behavior of functions.

• In multivariable calculus, we consider functions like \(f(x, y)\), where each variable can independently affect the output.
• This leads to exploring partial derivatives, gradients, and tangent planes as tools to understand how these functions behave.

The concept of partial derivatives enables us to understand the rate of change of a function in respects to one variable at a time, playing a key role in visualizations like tangent planes. The tangent plane is a linear approximation of the function at a particular point, representing the best-fit plane that touches the surface at just one point, providing an accurate local description of the surface. Through multivariable calculus, students explore the spatial relationships and curvature of surfaces, making problems in three-dimensional analytic geometry solvable and more intuitive.