Problem 41 - 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 \(f\). $$f(x, y)=x^{2}+x y+y^{2}+7 ; P(-3,3)$$

Step-by-Step Solution

Answer

Answer: At the point P(-3, 3), the direction of maximum increase is (-3, 3), the direction of maximum decrease is (3, -3), and the direction of no change is (1, 1).

Step 1: Calculate the value of the function at point P

To find the level curve that passes through P(-3, 3), we first need to find the value of the function at this point: $$ f(-3, 3) = (-3)^2 + (-3)(3) + (3)^2 + 7 = 9 -9 + 9 + 7 = 16 $$ The value of the function at point P is 16.

Step 2: Find the gradient of the function

Next, we need to find the gradient of the function \(f(x, y)\), which will give us information about the directions of maximum increase and maximum decrease. The gradient is given by: $$ \nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) = (2x + y, x + 2y) $$ Now we need to find the gradient at point P(-3, 3): $$ \nabla f(-3, 3) = (2(-3) + 3, -3 + 2(3)) = (-3, 3) $$

Step 3: Determine the directions

From the gradient at point P, we can find the directions we need: - Maximum increase: The direction of the maximum increase is the same as the gradient. Thus, the direction of the maximum increase is \((-3, 3)\). - Maximum decrease: The direction of the maximum decrease is the opposite of the gradient. Therefore, the direction of the maximum decrease is \((3, -3)\). - No change: The perpendicular direction to the gradient vector has no change. Using the dot product, we can find the vector that is perpendicular to the gradient vector, which results in no change in the function value. The dot product between the gradient vector and the perpendicular vector \((a, b)\) should equal 0. We have: $$ (-3)a + (3)b = 0 $$ We can choose any scalar multiple of a solution to this equation as our vector for no change. One such solution is \((a, b) = (1, 1)\).

Step 4: Sketch the xy-plane

Now we can sketch the xy-plane, showing the level curve, point P(-3, 3), and the directions of maximum increase, maximum decrease, and no change. The level curve is given by the equation \(f(x, y) = 16\), which is $$ x^2 + xy + y^2 + 7 = 16 $$ After sketching the plane with the contour curve, place point P(-3, 3) on the curve. Then draw the arrows representing the directions of maximum increase, maximum decrease, and no change based on the information in Step 3.

Key Concepts

Gradient

The gradient is a vital concept when working with level curves in multivariable calculus. It is a vector that contains all the partial derivatives of a function. For a function of two variables, such as \( f(x, y) = x^2 + xy + y^2 + 7 \), the gradient at any point \( (x, y) \) is calculated using:

• \( \frac{\partial f}{\partial x} = 2x + y \)
• \( \frac{\partial f}{\partial y} = x + 2y \)

Together, these make the gradient vector \( abla f(x, y) = (2x + y, x + 2y) \). This vector points in the direction of the steepest ascent. At the specific point \( P(-3, 3) \), the gradient \( abla f(-3, 3) = (-3, 3) \) provides us with detailed information on how the function behaves:

• Maximum increase is in the direction of \( (-3, 3) \).
• Maximum decrease, being the opposite direction, is \( (3, -3) \).
• No change is perpendicular to the gradient.

The gradient not only helps identify these directions but also underlines how rapidly the function is changing at a point.

Directional Derivatives

Directional derivatives allow us to explore how quickly a function changes in any given direction from a particular point. If we have a function \( f(x, y) \) and a direction vector \( \mathbf{v} = (a, b) \), the directional derivative of \( f \) in the direction of \( \mathbf{v} \) is computed as:\[ D_{\mathbf{v}} f(x, y) = abla f(x, y) \cdot \mathbf{u} \]where \( \mathbf{u} \) is the unit vector in the same direction as \( \mathbf{v} \). The unit vector is obtained by normalizing \( \mathbf{v} \) as \( \mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} \).For example, at the point \( P(-3,3) \) with a gradient \( abla f(-3, 3) = (-3, 3) \), we can assess the function's change in any direction by computing the dot product between the gradient and the desired direction vector. This shows which directions yield maximum change and which yield no change. The directional derivative provides insight into how the function rises or falls as we move in various paths.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions that have more than one variable, such as \( f(x, y) \), which depends on both \( x \) and \( y \). This section of calculus is crucial for analyzing surfaces and curves in three-dimensional space. The essential tools include:

• Partial Derivatives: These are the derivatives of a multivariable function with respect to each variable, treating others as constants.
• Gradient: A vector of all partial derivatives, indicating the direction of steepest ascent for the function.
• Level Curves: These are curves along which the function has a constant value, helping to visualize the "shape" of the function.
• Directional Derivatives: Indicate how the function changes as we move in a specific direction from a point.

Using these tools, we can effectively study and visualize multivariable functions, determining behavior over surfaces and volumes rather than just linear paths. Understanding these concepts is key for complex problem-solving in physics, engineering, and beyond.