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

Sketch the parabola \( y = 2x^2 \) and draw \( \nabla f(1,2) = (-4, 1) \) at (1,2), pointing perpendicular to the curve.

Step 1: Identify the Level Curve

The function is given by \( f(x, y) = \frac{y}{x^2} \). To find the equation of the level curve that passes through \( \mathbf{p} = (1,2) \), calculate \( f(1,2) \), which equals \( \frac{2}{1^2} = 2 \). The level curve equation is thus \( \frac{y}{x^2} = 2 \), or equivalently, \( y = 2x^2 \).

Step 2: Calculate the Gradient Vector

The gradient \( abla f(x, y) \) of \( f(x, y) = \frac{y}{x^2} \) is obtained by partial differentiation.ewline\( \frac{\partial f}{\partial x} = -\frac{2y}{x^3} \) and \( \frac{\partial f}{\partial y} = \frac{1}{x^2} \).ewlineTherefore, \( abla f(x, y) = \left(-\frac{2y}{x^3}, \frac{1}{x^2}\right) \).ewlineSubstitute \( \mathbf{p} = (1,2) \) to find \( abla f(1,2) = \left(-\frac{2 \times 2}{1^3}, \frac{1}{1^2}\right) = (-4, 1) \).

Step 3: Draw the Level Curve and Gradient Vector

The level curve \( y = 2x^2 \) is a parabola opening upwards. On the Cartesian plane, plot this parabola and the point \( \mathbf{p} = (1,2) \).ewlineTo draw the gradient vector \( abla f(\mathbf{p}) = (-4, 1) \), place its initial point at \( \mathbf{p} \). The vector points in the direction of \((-4, 1)\).

Step 4: Understand the Significance of the Gradient

The gradient vector \( abla f(\mathbf{p}) = (-4, 1) \) is perpendicular to the level curve \( y = 2x^2 \) at the point \( \mathbf{p} \). This indicates that the direction of the steepest ascent of the function at a point is normal to the level curve at that point.

Key Concepts

Level Curves

Imagine level curves as the footprints of a hill on a map. They represent points where the function maintains a constant value. In this problem, the function is given by \( f(x, y) = \frac{y}{x^2} \). To find the specific level curve that goes through the point \( \mathbf{p} = (1, 2) \), we first compute \( f(1,2) \), giving us \( 2 \). Therefore, the equation of the level curve is \( y = 2x^2 \).
This results in a parabola that depicts all (x, y) pairs where \( f(x, y) \) equals 2. Such curves help us view how the function behaves across different regions of the plane.
Level curves are a fundamental concept in multivariable calculus, allowing visualization of how a function changes and holds specific values across its domain.

Gradient Vector

The gradient vector \( abla f(x, y) \) of a function like \( f(x, y) = \frac{y}{x^2} \) provides a powerful tool for understanding directional change. By computing its partial derivatives, we find \( \frac{\partial f}{\partial x} = -\frac{2y}{x^3} \) and \( \frac{\partial f}{\partial y} = \frac{1}{x^2} \).
Thus, the gradient vector becomes \( abla f(x, y) = \left(-\frac{2y}{x^3}, \frac{1}{x^2}\right) \). This gradient tells us how much the function changes as we move away from a point in x and y directions.

• The gradient points in the direction of the greatest rate of increase of the function.
• Its length indicates the rate of this increase.

By specifying \( \mathbf{p} = (1, 2) \), we arrive at \( abla f(1,2) = (-4, 1) \), providing crucial insights into how the function behaves near this point.

Partial Differentiation

Partial differentiation is like focusing on one variable at a time while keeping others constant. For the function \( f(x, y) = \frac{y}{x^2} \), we explored this by deriving \( \frac{\partial f}{\partial x} = -\frac{2y}{x^3} \) and \( \frac{\partial f}{\partial y} = \frac{1}{x^2} \).
These derivatives tell us:

• How much \( f(x, y) \) changes with respect to x while keeping y fixed.
• How \( f(x, y) \) alters with respect to y, holding x constant.

This view is valuable in multivariable calculus as it simplifies analysis by handling one dimension at a time. By understanding each partial derivative, the overall change in the function with respect to all its variables can be predicted.

Steepest Ascent

The steepest ascent is all about finding the quickest path upwards on a "mountain" represented by a function. It is indicated by the direction of the gradient vector. At any point, \( abla f(x, y) \) not only points in the direction of this steepest climb but also lies perpendicular to the level curve at that point.
In our problem, at the point \( \mathbf{p} = (1, 2) \), the gradient vector \( abla f(1,2) = (-4, 1) \) shows where the function is increasing the fastest. The steepest ascent means moving along the direction given by our gradient vector.

• It helps predict how a small step in that direction will affect the function's value.
• Understanding this direction is critical for navigating through changes in multivariable functions.

This connection between the gradient and level curves paints a clear picture of the terrain formed by the function, helping predict how a function will behave if you move from one point to another.