Problem 40 - 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 \(f\). $$f(x, y)=-4+6 x^{2}+3 y^{2} ; P(-1,-2)$$

Step-by-Step Solution

Answer

Answer: The level curve passing through point P(-1, -2) is \(-4 + 6x^2 + 3y^2 = 14\). At point P, the direction of maximum increase is along the gradient vector (-12, -12), the direction of maximum decrease is opposite to the-gradient vector (12, 12), and directions of no change are orthogonal to the gradient vector, such as (12, -12) and (-12, 12).

Step 1: 1. Find the level curve through point P for function f

To find the level curve through point P, we need to substitute the given point into the function and solve for the resulting constant value. The level curve is defined by the equation: $$f(-1, -2) = C$$ where C is a constant. Substituting the point P(-1, -2), we get: $$C = -4 + 6(-1)^2 + 3(-2)^2 = -4 + 6 + 12 = 14$$ So, the level curve going through point P is: $$f(x, y) = -4 + 6x^2 + 3y^2 = 14$$

Step 2: 2. Find the gradient of f at point P

To find the gradient of f, we first compute the partial derivatives of f with respect to x and y: $$\frac{\partial f}{\partial x} = 12x$$ $$\frac{\partial f}{\partial y} = 6y$$ Now we evaluate these at point P(-1, -2): $$\nabla f(-1, -2) = (12(-1), 6(-2)) = (-12, -12)$$

Step 3: 3. Sketch the xy-plane, including point P, level curve, and directions

To complete the sketch, follow these steps: a. Plot point P(-1, -2) on the xy-plane. b. Sketch the level curve: \(-4 + 6x^2 + 3y^2 = 14\) (which is an ellipse). c. Add the gradient vector \(\nabla f(-1, -2) = (-12, -12)\) at point P. d. Direction of maximum increase: along the gradient vector (-12, -12). e. Direction of maximum decrease: opposite to the gradient vector (12, 12). f. Direction of no change: orthogonal to the gradient vector, e.g., (12, -12) and (-12, 12). In the end, you should have a graph showing the elliptical level curve through point P, with arrows indicating the directions of maximum increase, maximum decrease, and no change for the function f.

Key Concepts

Level Curves

In calculus, level curves are an essential way to visually represent functions of two variables, like our function \(f(x, y) = -4 + 6x^2 + 3y^2\). A level curve is essentially a cross-section of a 3D surface at a particular height. To find the level curve through a given point, you substitute that point into the function and solve for a constant, which we call \(C\).

Think of level curves as contour lines on a map that represent places with the same height. In this exercise, substituting the point \((-1, -2)\) into the function gives a level curve corresponding to the equation \(-4 + 6x^2 + 3y^2 = 14\). This equation reveals an ellipse positioned in the \(xy\)-plane.

The importance of level curves?

• They help visualize how a function behaves in two dimensions.
• They can indicate areas of steep or gentle incline on the surface.

Gradient Vectors

The gradient vector of a function gives us a critical insight into how the function increases or decreases. It’s like a compass pointing in the direction of the steepest ascent on a map. For the function \(f(x, y)\), the gradient, denoted \(abla f\), is a vector composed of partial derivatives with respect to each variable.

For the function \(-4 + 6x^2 + 3y^2\), calculating the gradient involves taking the derivative of \(f\) with respect to \(x\) and \(y\):

• \(\frac{\partial f}{\partial x} = 12x\)
• \(\frac{\partial f}{\partial y} = 6y\)

Substitute the point \((-1, -2)\) to find the gradient vector at that point: \((-12, -12)\). This vector shows us:

• The direction of greatest increase, which is along the gradient \((-12, -12)\).

Directional Derivatives

Directional derivatives help us understand how a function changes in a specific direction. Unlike the gradient, which tells us where the function increases most, a directional derivative can tell us how the function increases or decreases in any chosen direction.

Mathematically, the directional derivative of a function \(f\) at a point \(P\) in the direction of a vector \(\vec{v}\) is denoted as \(D_{\vec{v}}f(P)\) and can be calculated using the gradient vector:

• \(D_{\vec{v}}f(P) = abla f(P) \cdot \vec{v}\)

Here, "\(\cdot\)" represents the dot product. This measure shows how much \(f\) changes as you move slightly in the direction of \(\vec{v}\). In the directions of the gradient vector \((-12, -12)\), the directional derivative is maximized, indicating the steepest ascent. Conversely, moving opposite the gradient yields the steepest descent.

Partial Derivatives

Partial derivatives extend the concept of differentiation to multivariable functions, showing how a function changes with respect to one variable while keeping others constant.

For the function \(-4 + 6x^2 + 3y^2\), the partial derivatives are the building blocks for obtaining the gradient vector:

• \(\frac{\partial f}{\partial x} = 12x\): This derivative indicates how \(f\) changes as \(x\) changes, while \(y\) stays constant.
• \(\frac{\partial f}{\partial y} = 6y\): Indicates how \(f\) changes as \(y\) changes, with \(x\) held constant.

At point \((-1, -2)\), these derivatives are \(-12\) for \(x\) and \(-12\) for \(y\).

Partial derivatives provide insight into the "slope" of a function along just one axis and are crucial for understanding the overall behavior of multivariable functions.