Problem 42 - Calculus Early Transcendentals

Question

Consider the following functions \(f\) and points \(P .\) Sketch the \(x y\) -plane showing \(P\) and the level curve through \(P\). Indicate (as in Figure 70 ) the directions of maximum increase, maximum decrease, and no change for \(f\). $$f(x, y)=\tan (2 x+2 y) ; P(\pi / 16, \pi / 16)$$

Step-by-Step Solution

Answer

Question: Sketch the xy-plane showing point P and the level curve through P. Indicate the directions of maximum increase, maximum decrease, and no change for the function \(f(x, y) = \tan(2x + 2y)\) at point \(P(\pi/16, \pi/16)\). Answer: Since the function has no finite limit at point P, it does not lie on a regular level curve. The direction of maximum increase at point P is along the gradient vector \(\langle 2, 2 \rangle\), the direction of maximum decrease is opposite the gradient vector \(\langle -2, -2 \rangle\), and the directions of no change are perpendicular to the gradient vector, given by \(\langle 2, -2 \rangle\) and \(\langle -2, 2 \rangle\).

Step 1: Compute the gradient vector of f at point P

Let's compute the gradient vector of the function, denoted as \(\nabla f(x, y)\). To do this, we need to find the partial derivatives of \(f(x,y)\) with respect to \(x\) and \(y\): $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \tan(2x + 2y) = 2\sec^2(2x + 2y)$$ $$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \tan(2x + 2y) = 2\sec^2(2x + 2y)$$ Thus, the gradient vector is given by $$\nabla f(x, y) = \langle 2\sec^2(2x+2y), 2\sec^2(2x+2y) \rangle$$ Now, we evaluate the gradient vector \(\nabla f\) at point P \((\pi/16, \pi/16)\): $$\nabla f(\pi/16, \pi/16) = \langle 2\sec^2(2\frac{\pi}{16}+2\frac{\pi}{16}), 2\sec^2(2\frac{\pi}{16}+2\frac{\pi}{16}) \rangle = \langle 2\sec^2(\frac{\pi}{2}), 2\sec^2(\frac{\pi}{2})\rangle = \langle 2, 2 \rangle$$

Step 2: Determine Level Curve and Sketch

To find the points on the level curve, we find the points where the function is equal to the value at point P: $$f(\pi/16, \pi/16) = \tan(2(\pi/16) + 2(\pi/16)) = \tan(\pi/2) = \infty$$ Since this implies that the function does not have a finite limit as it approaches point P, it indicates this point doesn't lie on a regular level curve. Now, sketch the xy-plane, marking the point P \((\pi/16, \pi/16)\) and the direction of the gradient vector we calculated in Step 1. Since the gradient vector at P is \(\langle 2, 2 \rangle\), mark the directions of maximum increase along this vector, maximum decrease along the opposite vector (\(\langle -2, -2 \rangle\)), and no change along the vectors perpendicular to the gradient vector.

Step 3: Indicate Directions of Max Increase, Max Decrease, and No Change

To summarize, we have the follow directions for point P: 1. Direction of maximum increase: Along the gradient vector \(\langle 2, 2 \rangle\) 2. Direction of maximum decrease: Opposite the gradient vector \(\langle -2, -2 \rangle\) 3. Direction of no change: Perpendicular to gradient vector, which will be \(\langle 2, -2 \rangle\) and \(\langle -2, 2 \rangle\) In your sketch, indicate these directions by drawing arrows from point P in these directions.

Key Concepts

Level Curve

A level curve of a function of two variables, like our function \(f(x, y) = \tan(2x+2y)\), is a set of points in the \(xy\)-plane where the function takes on a constant value. In simpler terms, imagine it as a line or a curve on a map that connects points of equal altitude. Considering the specific function at point \(P(\pi/16, \pi/16)\), the exercise attempts to find where \(f\) equals its value at \(P\). However, since \(f(\pi/16, \pi/16) = \tan(\pi/2) = \infty\), this suggests that traditional level curves don't exist through \(P\) as they do not meet a finite value criterion here. Although we cannot draw a customary level curve crossing \(P\), it's still essential to visualize the point and conceptualize the tendencies of the function around it. This visualization helps in understanding other concepts like gradients and directional changes.

Partial Derivatives

Partial derivatives are a foundational concept in calculus, especially when dealing with functions of multiple variables. They represent how the function changes as each of its variables changes, while keeping other variables constant. For our function \(f(x, y) = \tan(2x + 2y)\), the partial derivatives with respect to \(x\) and \(y\) highlight how the function scales when \(x\) or \(y\) independently changes:

• The partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x} = 2\sec^2(2x + 2y)\).
• Likewise, \(\frac{\partial f}{\partial y} = 2\sec^2(2x + 2y)\) denotes how the function behaves with a change in \(y\).

These derivatives are crucial as they form the components of the gradient vector, which provides deeper insights into the function's behavior at specific points, like \(P(\pi/16, \pi/16)\).

Direction of Maximum Increase

The gradient vector, \(abla f(x, y)\), reflects the direction of the steepest ascent or maximum increase of a function at a given point. For our function, it is calculated as \(\langle 2\sec^2(2x+2y), 2\sec^2(2x+2y) \rangle\). At \(P(\pi/16, \pi/16)\), this simplifies to \(\langle 2, 2 \rangle\).

• The vector \(\langle 2, 2 \rangle\) indicates the direction in which \(f\) increases most rapidly from point \(P\).
• Conversely, the opposite vector, \(\langle -2, -2 \rangle\), shows the direction of maximum decrease.
• Vectors perpendicular to the gradient, \(\langle 2, -2 \rangle\) and \(\langle -2, 2 \rangle\), indicate directions of no change in function value.

Recognizing these directions is vital for optimizing functions and understanding behavior in multivariable calculus scenarios.

Calculus Visualization

Visualizing calculus concepts can significantly enhance comprehension, especially when dealing with multiple variables. In the exercise, we sketch the \(xy\)-plane to identify various directions of change:

• Marking point \(P(\pi/16, \pi/16)\) helps in locating where we need to understand the function behavior.
• The gradient vector \(\langle 2, 2 \rangle\), originating at \(P\), visually represents the direction of steepest ascent.
• Indicating the opposite and perpendicular directions further enriches understanding of the surface's geometry at \(P\).

Creating such a visual framework simplifies the complex interplays of calculus, making it more approachable. Whether tracing contours or modeling behavior dynamics, visualization paves the path to intuitive grasps of mathematical phenomena.