Problem 30 - Calculus Early Transcendentals

Question

Near a buoy, the depth of a lake at the point with coordinates \((x, y)\) is \(z=200+0.02 x^{2}-0.001 y^{3},\) where \(x, y,\) and \(z\) are measured in meters. A fisherman in a small boat starts at the point \((80,60)\) and moves toward the buoy, which is located at \((0,0) .\) Is the water under the boat getting deeper or shallower when he departs? Explain.

Step-by-Step Solution

Answer

The water under the boat is getting deeper.

Step 1: Define the Gradient of the Depth Function

To determine if the water is getting deeper or shallower, we need to compute the gradient of the depth function at the starting point. The depth function is given by \( z(x, y) = 200 + 0.02x^2 - 0.001y^3 \). We need to find the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).

Step 2: Calculate the Partial Derivatives

Compute the partial derivative with respect to \( x \):\( \frac{\partial z}{\partial x} = 0.04x \).Compute the partial derivative with respect to \( y \):\( \frac{\partial z}{\partial y} = -0.003y^2 \).

Step 3: Evaluate the Gradient at the Starting Point

Evaluate the gradient at the point \((80, 60)\):\[ \left. \frac{\partial z}{\partial x} \right|_{(80, 60)} = 0.04 \times 80 = 3.2 \]\[ \left. \frac{\partial z}{\partial y} \right|_{(80, 60)} = -0.003 \times 60^2 = -10.8 \].

Step 4: Determine the Direction of Motion

The fisherman is moving from \((80, 60)\) to \((0, 0)\). The direction vector for this movement is \( \langle -80, -60 \rangle \). This will help in determining if the gradient is aligned with the direction of motion.

Step 5: Compute the Dot Product with the Gradient

The dot product of the gradient and the direction vector \( \langle -80, -60 \rangle \) is given by:\( 3.2 \times (-80) + (-10.8) \times (-60) = -256 + 648 = 392 \).

Step 6: Interpret the Result

The positive dot product (392) indicates that the depth is increasing as the fisherman moves toward the buoy, meaning the water is getting deeper.

Key Concepts

Partial Derivatives

Partial derivatives are a vital concept when dealing with functions of multiple variables. They essentially help us understand how a function changes as we vary one of its variables while keeping the others constant.

In the exercise, we have a depth function given by:

• \( z(x, y) = 200 + 0.02x^2 - 0.001y^3 \)

To determine how the depth changes with each coordinate, we find the partial derivatives:

• For the variable \(x\), the partial derivative is \( \frac{\partial z}{\partial x} = 0.04x \).
• For the variable \(y\), the partial derivative is \( \frac{\partial z}{\partial y} = -0.003y^2 \).

These derivatives tell us how much \(z\) changes with small changes in \(x\) or \(y\). At the point \((80, 60)\), we get specific values by substituting \(x\) and \(y\) into these derivatives, which forms part of understanding the gradient vector.

Gradient Vector

The gradient vector of a function represents the direction and rate of the steepest ascent of that function. In mathematical terms, for a function \( z = f(x, y) \), the gradient \( abla z \) is a vector composed of the partial derivatives:

• \( abla z = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \right) \)

For our depth function, the gradient at any point \((x, y)\) is:

• \( abla z = (0.04x, -0.003y^2) \)

At the point \((80, 60)\), we substitute these values to find the specific gradient vector:

• \( abla z = (3.2, -10.8) \)

This vector \((3.2, -10.8)\) tells us that the slope of the lake surface changes at specific rates in the \(x\) and \(y\) directions. Knowing the gradient is key to determining how the water depth changes as the fisherman moves through different areas.

Dot Product

The dot product, also known as the scalar product, is a key analytical tool that helps us understand the relationship between vectors. In our exercise, it's used to determine whether the depth is increasing or decreasing.

For two vectors, \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is computed as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)

In the exercise, the gradient vector at the starting point \((80, 60)\) is \((3.2, -10.8)\), and the direction vector the fisherman travels is \((-80, -60)\). Calculating the dot product of these vectors:

• \( 3.2 \times (-80) + (-10.8) \times (-60) = -256 + 648 = 392 \)

A positive result implies that the fisherman moves into an area where the water is deeper, confirming that the depth increases as he approaches the buoy.