Problem 47 - Calculus Early Transcendentals

Question

Find an equation of the plane tangent to the following surfaces at the given point. $$z=\tan ^{-1}(x y) ;(1,1, \pi / 4)$$

Step-by-Step Solution

Answer

Answer: The equation of the tangent plane at the given point is $$\frac{1}{2}x + \frac{1}{2}y - z = -\frac{\pi}{4}+\frac{3}{2}$$

Step 1: Find the partial derivatives

To determine the gradient of the surface at the given point, we need to find the partial derivatives of the function with respect to x and y. The function is given by $$z = \tan^{-1}(xy)$$. Let's find the partial derivatives: $$\frac{\partial z}{\partial x} = \frac{y}{1+(xy)^2}$$ $$\frac{\partial z}{\partial y} = \frac{x}{1+(xy)^2}$$

Step 2: Evaluate the partial derivatives at the given point

Now, let's evaluate these partial derivatives at the given point \((1,1,\pi / 4)\): $$\frac{\partial z}{\partial x} \Big|_{(1,1,\pi/4)} = \frac{1}{1+(1\cdot1)^2} = \frac{1}{2}$$ $$\frac{\partial z}{\partial y} \Big|_{(1,1,\pi/4)} = \frac{1}{1+(1\cdot1)^2} = \frac{1}{2}$$ The gradient at the point \((1,1,\pi / 4)\) is \(\left( \frac{1}{2}, \frac{1}{2} \right)\).

Step 3: Write the equation of the tangent plane

The equation of the tangent plane at the point \((1,1,\pi / 4)\) is given by: $$\frac{\partial z}{\partial x}(x-1) + \frac{\partial z}{\partial y}(y-1) = z - \frac{\pi}{4}$$ Plugging in the values we found in Step 2: $$\frac{1}{2}(x-1) + \frac{1}{2}(y-1) = z - \frac{\pi}{4}$$ Rearrange the equation to make it look like a standard plane equation (in the form of \(Ax + By + Cz = D\)): $$\frac{1}{2}x + \frac{1}{2}y - z = -\frac{\pi}{4}+\frac{3}{2}$$ This is the equation of the plane tangent to the surface $$z = \tan^{-1}(xy)$$ at the point \((1,1,\pi/4)\).

Key Concepts

Partial Derivatives

Partial derivatives are fundamental in understanding how a surface changes in different directions. When dealing with a function of two variables, like \( z = \tan^{-1}(xy) \), you take partial derivatives to see how changes in \(x\) or \(y\) affect \(z\) independently.
To find the partial derivatives, we hold one variable constant while differentiating with respect to the other. Here, the partial derivative with respect to \(x\), denoted \( \frac{\partial z}{\partial x} \), assumes \(y\) is constant. Similarly, \( \frac{\partial z}{\partial y} \) considers \(x\) constant. \(
\)For our surface, \( \frac{\partial z}{\partial x} = \frac{y}{1+(xy)^2} \) and \( \frac{\partial z}{\partial y} = \frac{x}{1+(xy)^2} \). These provide the rates of change in the surface’s height in the \(x\) and \(y\) directions, respectively.

Gradient of a Surface

The gradient of a surface at a point is a vector that combines the partial derivatives. It points in the direction of the steepest ascent on the surface and its magnitude represents how steep this ascent is.
For the function \( z = \tan^{-1}(xy) \), the gradient at a point \((x, y)\) is expressed as the vector \( abla z = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \right) \).
At the point \((1,1)\), both partial derivatives evaluated to \( \frac{1}{2} \). This gives the gradient vector \( \left( \frac{1}{2}, \frac{1}{2} \right) \) at the point \((1,1, \pi/4)\). This vector tells us that the surface rises equally in the \(x\) and \(y\) directions at this point.

Surface Equation

The surface equation describes a surface in three-dimensional space. In this context, the surface is given by \( z = \tan^{-1}(xy) \).
We are interested in finding the equation of the tangent plane to this surface at a specific point \((1,1, \pi/4)\). A tangent plane gives us the linear approximation of the surface near the point. It touches the surface at exactly one point without cutting through it.
To find this plane, we use the partial derivatives at the point, as these are crucial for constructing the equation. The general tangent plane formula is:

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

For our surface, substituting the values gives us:

• \( \frac{1}{2}(x-1) + \frac{1}{2}(y-1) = z - \frac{\pi}{4} \)

This simplifies to \( \frac{1}{2}x + \frac{1}{2}y - z = -\frac{\pi}{4} + \frac{3}{2} \), which is the precise equation of our tangent plane.