Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

The elevation of a mountain above sea level at \((x, y)\) is \(3000 e^{-\left(x^{2}+2 y^{2}\right) / 100}\) meters. The positive \(x\) -axis points east and the positive \(y\) -axis points north. A climber is directly above \((10,10)\). If the climber moves northwest, will she ascend or descend and at what slope?

Step-by-Step Solution

Answer

The climber will descend with a slope of \( \frac{-60e^{-3}}{\sqrt{2}} \).

Step 1: Understand the Problem

We need to analyze the elevation function given by \( z(x, y) = 3000e^{-\frac{x^2 + 2y^2}{100}} \). The climber's current position is \((10, 10)\). We want to find out the direction of change in elevation as the climber moves northwest, and calculate the slope of her path.

Step 2: Calculate the Gradient

The gradient \( abla z \) gives the direction of the steepest ascent. We calculate partial derivatives: \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \). \[ abla z = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \right) = \left( -\frac{6000x}{100} e^{-\frac{x^2 + 2y^2}{100}}, -\frac{12000y}{100} e^{-\frac{x^2 + 2y^2}{100}} \right) \].

Step 3: Evaluate Gradient at Current Position

Substitute \( x = 10 \) and \( y = 10 \) into the gradient. \[ abla z(10, 10) = \left( -60 e^{-\frac{300}{100}}, -120 e^{-\frac{300}{100}} \right) = \left( -60 e^{-3}, -120 e^{-3} \right) \].

Step 4: Determine the Direction of Motion

The northwest direction implies moving along the vector \((-1, 1)\) which is normalized to \( \frac{1}{\sqrt{2}}(-1, 1) \).

Step 5: Calculate the Directional Derivative

The directional derivative in the direction of a unit vector \( \mathbf{u} = (-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \) is: \[ D_{\mathbf{u}}z = abla z(10, 10) \cdot \mathbf{u} = \left( -60 e^{-3}, -120 e^{-3} \right) \cdot \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \].

Step 6: Compute the Dot Product

Calculate the dot product: \[ D_{\mathbf{u}}z = \frac{1}{\sqrt{2}}(60e^{-3}) + \frac{1}{\sqrt{2}}(-120e^{-3}) = \frac{60e^{-3} - 120e^{-3}}{\sqrt{2}} = \frac{-60e^{-3}}{\sqrt{2}} \].

Step 7: Interpret the Result

Since \( D_{\mathbf{u}}z = \frac{-60e^{-3}}{\sqrt{2}} \) is negative, the elevation decreases as the climber moves northwest. The path has a slope of \( \frac{-60e^{-3}}{\sqrt{2}} \).

Key Concepts

Gradient

The gradient is a crucial concept when understanding how a function changes in different spatial directions. It is expressed as a vector that points in the direction of the greatest rate of increase of the function. In mathematical terms, for a function \( z(x, y) \), the gradient is denoted by \( abla z \) and is composed of the partial derivatives with respect to its variables: \( \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \right) \).

For our elevation function, \( z(x, y) = 3000e^{-\frac{x^2 + 2y^2}{100}} \), the gradient helps us determine how steep the mountain is at any given point. Simply put, the gradient points us in the direction that climbs the mountain the fastest. Understanding the gradient can significantly aid in finding the best route for ascent or deciding which way to avoid if descending safely is the goal.

In our exercise, we have calculated the gradient at \((10, 10)\), which resulted in \( \left( -60e^{-3}, -120e^{-3} \right) \). This vector gives us an insight into how the mountain slopes at that specific position, indicating the strongest incline towards which the value of the elevation function increases.

Partial Derivatives

Partial derivatives are the building blocks for understanding change in multi-variable functions. They measure how a function varies with respect to one variable while keeping others constant. In the context of a surface represented by an elevation function \( z(x, y) \), the partial derivative \( \frac{\partial z}{\partial x} \) indicates how the elevation changes as you move east or west, while \( \frac{\partial z}{\partial y} \) shows the change as you move north or south.

Calculating these derivatives for our function, \( z(x, y) = 3000e^{-\frac{x^2 + 2y^2}{100}} \), gives us:

• \( \frac{\partial z}{\partial x} = -\frac{6000x}{100} e^{-\frac{x^2 + 2y^2}{100}} \)
• \( \frac{\partial z}{\partial y} = -\frac{12000y}{100} e^{-\frac{x^2 + 2y^2}{100}} \)

Evaluating these at \( x=10 \) and \( y=10 \) allows us to understand how the elevations are changing due to shifts in the east-west and north-south directions in that specific location.

Elevation Function

Elevation functions describe the height or altitude of a surface over a plane, often in geographic contexts. In this exercise, the elevation function \( z(x, y) = 3000e^{-\frac{x^2 + 2y^2}{100}} \) defines how high the mountain stands above sea level over a geographic grid.

The structure of this function includes a constant multiplier, \( 3000 \), representing the maximum elevation. The exponential term \( e^{-\frac{x^2 + 2y^2}{100}} \) reduces this elevation depending on the \( x \) and \( y \) coordinates, creating a shape that is highest at the center, here at the origin \((0, 0)\), and descends as one moves farther away.

This form is typical for elevation functions as terrains often have a central peak with decreasing heights outwards. The negative exponent indicates regions of rapid change in elevation, which are best captured by analyzing the gradient and partial derivatives. Using this function, you can predict the mountain's shape, calculate changes, and anticipate the steepness of slopes encountered at various points.

Slope Calculation

Slope calculation is essential in determining how steep a path is as you move across a landscape. The slope in a specific direction can be derived using the directional derivative, which measures how a function changes as you travel along a given vector. In our case, the climber moves northwest, which is represented by the vector \((-1, 1)\).

To find the slope in this direction, we use a unit vector \( \mathbf{u} = \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \). The directional derivative \( D_{\mathbf{u}}z \) then becomes the dot product between the gradient at \((10, 10)\), \( \left( -60e^{-3}, -120e^{-3} \right) \), and this unit vector. The result \( \frac{-60e^{-3}}{\sqrt{2}} \) tells us the rate at which elevation changes as the climber moves northwest.

Since this value is negative, it confirms that the climber descends as she goes northwest, offering a precise indication of the upward or downward steepness she encounters. Understanding this slope is crucial for climbers to gauge how challenging or easy the ascent may be.