Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the level curve of \(f(x, y)=y / x^{2}\) that goes through \(\mathbf{p}=(1,2)\). Calculate the gradient vector \(\nabla f(\mathbf{p})\) and draw this vector, placing its initial point at \(\mathbf{p}\). What should be true about \(\nabla f(\mathbf{p})\) ?

Step-by-Step Solution

Answer

The level curve is \(y = 2x^2\), and the gradient \(\nabla f(1,2) = (-4, 1)\) is perpendicular to this curve at \(\mathbf{p} = (1, 2)\).

Step 1: Determine the Level Curve Equation

The given function is \(f(x, y) = \frac{y}{x^2}\). To find the level curve that passes through the point \(\mathbf{p} = (1, 2)\), we substitute these coordinates into the function to find the level value \(c\).\[ f(1, 2) = \frac{2}{1^2} = 2 \].So the level curve is determined by fixing \(f(x, y) = 2\). This gives us the equation \(\frac{y}{x^2} = 2\), or \(y = 2x^2\).

Step 2: Sketch the Level Curve

To sketch the level curve \(y = 2x^2\), recognize that it describes a parabola opening upwards with vertex at the origin. Plot a few points, such as \((0, 0)\), \((1, 2)\), and \((-1, 2)\), and draw the symmetric parabola through these points.

Step 3: Calculate the Gradient \(\nabla f(x, y)\)

The gradient vector \(abla f(x, y)\) is given by the partial derivatives with respect to \(x\) and \(y\). So calculate:\[ f_x(x, y) = \frac{\partial}{\partial x}\left(\frac{y}{x^2}\right) = -\frac{2y}{x^3} \] \[ f_y(x, y) = \frac{\partial}{\partial y}\left(\frac{y}{x^2}\right) = \frac{1}{x^2} \].Thus, the gradient is \(abla f(x, y) = \left(-\frac{2y}{x^3}, \frac{1}{x^2}\right)\).

Step 4: Evaluate \(\nabla f(\mathbf{p})\)

Evaluate the gradient vector at the point \(\mathbf{p} = (1, 2)\):\(abla f(1, 2) = \left(-\frac{4}{1^3}, \frac{1}{1^2}\right) = (-4, 1)\).

Step 5: Draw the Gradient Vector

Place the initial point of the gradient vector at \(\mathbf{p} = (1, 2)\). From this point, draw the vector \((-4, 1)\). This means starting at \(\mathbf{p}\) and moving 4 units to the left and 1 unit up.

Step 6: Analyze the Gradient's Direction

The gradient vector \(abla f(\mathbf{p}) = (-4, 1)\) is perpendicular to the level curve at the point \(\mathbf{p} = (1, 2)\). This is a general property of gradient vectors, indicating the direction of steepest ascent.

Key Concepts

Gradient Vector

In multivariable calculus, the gradient vector is an essential concept. It gives us a way to understand the direction and rate of change of a multivariable function. Specifically, for a function like \( f(x, y) \), the gradient vector \( abla f(x, y) \) consists of partial derivatives with respect to each variable.

• The gradient vector \( abla f(x, y) \) is defined as \( \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
• It always points in the direction of the greatest rate of increase of the function.
• On a level curve, the gradient is perpendicular to the curve itself.

Understanding the gradient of \( f(x, y) = \frac{y}{x^2} \), we calculate the gradient using the partial derivatives. At the point \( \mathbf{p} = (1, 2) \), this gradient vector is \((-4, 1)\). This indicates the direction in which \( f(x, y) \) increases most rapidly at \( \mathbf{p} \). The negative -4 suggests a strong movement in the negative x-direction, while 1 indicates a slight increase in the y-direction.

Partial Derivatives

Partial derivatives are the building blocks of the gradient vector in multivariable calculus. They measure how a function changes as each input variable is varied while keeping the others constant.

• For a function \( f(x, y) = \frac{y}{x^2} \), the partial derivative with respect to x is found by differentiating \( f \) treating \( y \) as a constant. It results in \( f_x(x, y) = -\frac{2y}{x^3} \).
• The partial derivative with respect to \( y \) considers \( x \) as constant, yielding \( f_y(x, y) = \frac{1}{x^2} \).

By substituting the point \( \mathbf{p} = (1, 2) \) into these derivatives, we get the components of the gradient vector: \( -4 \) from \( f_x \) and \( 1 \) from \( f_y \). Partial derivatives thus give detailed information about the local behavior of the function along each axis.

Multivariable Calculus

Multivariable calculus extends the basic principles of differentiation and integration to functions of two or more variables. It opens up a world of possibilities beyond single-variable calculus, accommodating real-world phenomena that depend on multiple factors.

• Functions of multiple variables, such as \( f(x, y) = \frac{y}{x^2} \), depend on more than one input.
• Instead of a single derivative, we explore partial derivatives in each direction.
• We use the concept of gradients to find the steepest ascent or descent of a function.

Level curves like \( y = 2x^2 \), derived from these multivariable functions, offer a visual representation of constant values and provide an insightful tool for analyzing the behavior of the function in a plane. This branch of calculus is essential for fields such as physics, engineering, and economics, where systems are inherently multidimensional.

Parabola

When we talk about a parabola in mathematics, we are usually referring to a curve that represents a quadratic function, and it is characterized by its U-shape. In this exercise, the level curve \( y = 2x^2 \) forms a parabola, a very familiar shape in algebra.

• The equation \( y = 2x^2 \) implies a parabola that opens upwards, with its vertex at the origin \( (0, 0) \).
• The coefficient 2 in front of \( x^2 \) affects how "narrow" or "wide" the parabola is.
• By plotting points like \((0, 0)\), \((1, 2)\), and \((-1, 2)\), we can visualize this parabola on a coordinate plane.

The parabola represents a particular level curve of the function \( f(x, y) = \frac{y}{x^2} \), where \( f(x, y) \) has a constant value, in this case, 2. It's a powerful example of how multidimensional calculus can translate into simple 2D graphics that are easy to interpret.