Problem 288 - Calculus 3

Question

[T] Use technology to sketch the level curve of \(f(x, y)=4 x-2 y+3\) that passes through \(P(1,2)\) and draw the gradient vector at \(P .\)

Step-by-Step Solution

Answer

The level curve is \(y = 2x\), and the gradient vector at \(P(1, 2)\) is \((4, -2)\).

Step 1: Determine the Level Curve Equation

Start by substituting the point \(P(1,2)\) into the equation for \(f(x, y) = 4x - 2y + 3\). Calculate \(f(1, 2)\) to determine the level curve. \[ f(1, 2) = 4(1) - 2(2) + 3 = 4 - 4 + 3 = 3. \] This indicates that the level curve is given by the equation \(4x - 2y + 3 = 3\). Simplify this equation. \[ 4x - 2y = 0 \] \[ 2x = y. \]

Step 2: Sketch the Level Curve

Sketch the line \(y = 2x\) on a coordinate plane. This is a straight line that passes through the origin (0,0) and is the level curve of the function \(f(x, y) = 4x - 2y + 3\) at \(f(x, y)=3\). It also passes through the point \(P(1, 2)\).

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

The gradient \(abla f(x, y)\) is a vector of partial derivatives: \(abla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\). Calculate each derivative: \[ \frac{\partial f}{\partial x} = 4, \] \[ \frac{\partial f}{\partial y} = -2. \] Therefore, \(abla f(x, y) = (4, -2)\).

Step 4: Draw the Gradient Vector at Point P

Draw the gradient vector \((4, -2)\) starting from point \(P(1, 2)\). This vector points in the direction of the greatest rate of increase of the function \(f(x, y)\) and is perpendicular to the level curve \(y = 2x\) at \(P(1, 2)\).

Key Concepts

Gradient Vector

The gradient vector, often denoted as \( abla f(x, y) \), is a fundamental concept in multivariable calculus. It is a vector that contains partial derivatives of a function. This vector points in the direction where the function increases most steeply.
One key property of the gradient vector is its perpendicularity to level curves. If you imagine a hill, the gradient vector shows the steepest path upward. In our exercise, the function is given as \( f(x, y) = 4x - 2y + 3 \).
To find the gradient vector, calculate the partial derivatives:

• Partial derivative with respect to \( x \): \( \frac{\partial f}{\partial x} = 4 \)
• Partial derivative with respect to \( y \): \( \frac{\partial f}{\partial y} = -2 \)

Thus, the gradient vector is \( (4, -2) \). At any point, such as \( P(1, 2) \), this vector points in the direction where \( f \) increases the fastest, providing valuable insight into the behavior of the function.

Partial Derivatives

Partial derivatives are used to understand how a function changes as each variable varies while others are kept constant. They are the building blocks of the gradient vector and provide a deeper insight into the behavior of multivariable functions.
For a function of two variables, like \( f(x, y) \), the partial derivative with respect to \( x \), \( \frac{\partial f}{\partial x} \), measures how \( f \) changes as \( x \) changes, holding \( y \) constant. Similarly, \( \frac{\partial f}{\partial y} \) assesses the change as \( y \) changes.
In the function \( f(x, y) = 4x - 2y + 3 \):

• To find \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant, resulting in: \( 4 \).
• For \( \frac{\partial f}{\partial y} \), treat \( x \) as constant, resulting in: \( -2 \).

Partial derivatives are essential in forming the gradient vector and understanding the directional derivatives, providing clues about the function's behavior in different directions.

Coordinate Plane

The coordinate plane is a two-dimensional space defined by the \( x \) and \( y \) axes. This plane allows us to visualize functions, like the function \( f(x, y) = 4x - 2y + 3 \) from our exercise, and plot important features such as level curves and vectors.
Level curves represent points where the function has the same value. In our exercise, the level curve \( y = 2x \) represents all points where the function equals 3. Sketching level curves helps visualize the terrain of a function across the plane, similar to contour lines on a topographic map.
The coordinate plane also accommodates vectors, such as the gradient vector. In this context, the gradient vector \( (4, -2) \) drawn from point \( P(1, 2) \) illustrates the steepest increase direction of the function on the plane. The interplay of these elements on the coordinate plane is crucial to understanding calculus concepts, providing a complete picture of a function’s behavior across a two-dimensional space.