Problem 34 - Calculus Early Transcendentals

Question

Suppose you are climbing a hill whose shape is given by the equation \(z=1000-0.005 x^{2}-0.01 y^{2},\) where \(x, y,\) and \(z\) are measured in meters, and you are standing at a point with coordinates \((60,40,966) .\) The positive \(x\) -axis points east and the positive \(y\) -axis points north. (a) If you walk due south, will you start to ascend or descend? At what rate? (b) If you walk northwest, will you start to ascend or descend? At what rate? (c) In which direction is the slope largest? What is the rate of ascent in that direction? At what angle above the horizontal does the path in that direction begin?

Step-by-Step Solution

Answer

(a) Ascend at 0.8 m/m. (b) Descend at ~0.141 m/m. (c) Steepest ascent rate is 1 m/m, angle 45°.

Step 1: Find the Gradient Vector

The gradient \( abla z \) of the function \( z = 1000 - 0.005x^2 - 0.01y^2 \) gives us the direction of the steepest ascent. Calculate the partial derivatives with respect to \( x \) and \( y \):\[ \frac{\partial z}{\partial x} = -0.01x, \quad \frac{\partial z}{\partial y} = -0.02y. \]This gives us the gradient vector \( abla z = (-0.01x, -0.02y) \).

Step 2: Evaluate Gradient at Given Point

Evaluate the gradient vector at the point \( (60, 40) \):\[ abla z (60, 40) = (-0.01 \times 60, -0.02 \times 40) = (-0.6, -0.8). \]This gradient indicates the direction of steep ascent from \( (60, 40) \).

Step 3: Ascend or Descend due to Southward Movement

Walking due south implies moving in the negative \( y \)-direction, i.e., \( (0, -1) \). The rate of change of \( z \) is the dot product of the gradient and the direction vector:\[ abla z \cdot (0, -1) = -0.8 \times (-1) = 0.8. \]Since the rate is positive, you will ascend by 0.8 meters per meter.

Step 4: Ascend or Descend when Walking Northwest

Walking northwest implies moving in the direction \( (-1, 1) \). Normalizing this vector, we get \( \frac{1}{\sqrt{2}}(-1, 1) \). The rate of change is:\[ abla z \cdot \left( \frac{1}{\sqrt{2}}(-1, 1) \right) = (-0.6, -0.8) \cdot \left( \frac{1}{\sqrt{2}}(-1, 1) \right) = \frac{1}{\sqrt{2}}(0.2). \]This results in a descent rate of approximately 0.141 meters per meter.

Step 5: Direction of Steepest Ascent and its Rate

The direction of the steepest ascent is along the gradient vector \((-0.6, -0.8)\). The magnitude or rate in this direction is:\[ \|abla z\| = \sqrt{(-0.6)^2 + (-0.8)^2} = 1. \]Hence, the rate of ascent in this direction is 1 meter per meter.

Step 6: Calculate Angle with Horizontal Plane

To find the angle \( \theta \) with the horizontal, we use:\[ \tan \theta = \frac{1}{1} = 1 \Rightarrow \theta = \tan^{-1}(1) = 45^\circ. \]Thus, the slope rises at an angle of 45 degrees above the horizontal.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They represent the rate of change of a function with respect to one variable while keeping the others fixed. In the context of our hill-shaped function, partial derivatives help determine how steeply the hill rises or falls in specific directions. By calculating these for each coordinate - in our exercise, with respect to \(x\) and \(y\) - we understand how the hill's shape changes:

• For \( \frac{\partial z}{\partial x} = -0.01x \), it shows how steep the incline or decline is when moving along the \(x\)-axis (east-west direction).
• For \( \frac{\partial z}{\partial y} = -0.02y \), it reflects the slope change when moving along the \(y\)-axis (north-south direction).

Together, these partial derivatives form the gradient vector, which plays a crucial role in analyzing the terrain's slope.

Directional Derivatives

Directional derivatives extend the concept of partial derivatives to any direction, not just the ones aligned with the axes. They measure the rate of change of a function in a specified direction. In this exercise, it helps us understand what happens to the altitude when walking in other directions on the hill.

• For example, to determine whether you will ascend or descend when walking southwest, you compute the directional derivative using the direction vector of movement.
• This rate of change is given by the dot product between the gradient vector and the direction's unit vector.
• The direction of walking due south corresponds to the vector \((0, -1)\), leading to a positive rate of change, meaning an ascent.

This versatile derivative provides a powerful tool for navigating multitudes of pathways on a surface.

Steepest Ascent

The steepest ascent refers to the direction in which a function's value increases most rapidly. This direction aligns directly with the gradient vector you calculated. For the hill function, this provides a clear path to climb upwards most effectively:

• The gradient vector \((-0.01x, -0.02y)\), when evaluated at the given point \((60, 40)\), results in \((-0.6, -0.8)\).
• This vector offers the most rapid increase in height per meter traveled, effectively guiding the steepest climb possible.
• Calculating the vector's magnitude, we find a maximum steepest ascent rate of 1 meter per meter, giving a straightforward assessment of incline intensity.

Thus, knowledge of the gradient assures one finds the most efficient path uphill.

Rate of Change

The rate of change concept is central to understanding how the value of a function varies with movement or influence over time or space. This metric here specifically refers to how the hill's height changes as one moves in different directions:

• The positive rate of change when walking south signifies ascending, whereas a negative rate signifies descending.
• The specific magnitude of this rate is determined by the directional derivatives calculated earlier.
• In the case of northwest movement, this direction results in a descent rate of about 0.141 meters per meter.

By determining these rates from partial and directional derivatives, you can make predictions about travel over any terrain or surface.