Problem 289 - Calculus 3

Question

[T] Use technology to sketch the level curve of \(f(x, y)=x^{2}+4 y^{2}\) that passes through \(P(-2,0)\) and draw the gradient vector at \(P .\)

Step-by-Step Solution

Answer

Level curve: \(x^2 + 4y^2 = 4\). Gradient at \((-2, 0)\) is \((-4, 0)\).

Step 1: Understanding the Level Curve

The exercise asks for the level curve of the function \(f(x, y) = x^2 + 4y^2\) that passes through the point \(P(-2, 0)\). A level curve is a set of points \((x, y)\) such that \(f(x, y) = c\) for some constant \(c\). To find \(c\), substitute \((x, y) = (-2, 0)\) into the function: \(f(-2, 0) = (-2)^2 + 4(0)^2 = 4\). Thus, the level curve is given by the equation \(x^2 + 4y^2 = 4\).

Step 2: Sketching the Level Curve

The level curve \(x^2 + 4y^2 = 4\) can be rewritten as \(\frac{x^2}{4} + \frac{y^2}{1} = 1\), which is the equation of an ellipse centered at the origin with semi-major axis along the \(x\)-axis of length 2 and semi-minor axis along the \(y\)-axis of length 1. Use a graphing tool to sketch this ellipse.

Step 3: Finding the Gradient Vector

The gradient vector \(abla f(x, y)\) of a function \(f(x, y)\) is given by \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\). For \(f(x, y) = x^2 + 4y^2\), compute the partial derivatives: \(\frac{\partial f}{\partial x} = 2x\) and \(\frac{\partial f}{\partial y} = 8y\). Therefore, \(abla f(x, y) = (2x, 8y)\).

Step 4: Evaluating the Gradient at Point P

Substitute \((x, y) = (-2, 0)\) into the gradient vector: \(abla f(-2, 0) = (2(-2), 8(0)) = (-4, 0)\). This is the gradient vector at the point \(P(-2, 0)\).

Step 5: Drawing the Gradient Vector

The gradient vector \((-4, 0)\) indicates a direction on the \(x\)-axis from the point \((-2, 0)\). Draw an arrow starting at \(P(-2, 0)\) and extending 4 units to the left along the \(x\)-axis, since the gradient points in the direction of the greatest increase of the function.

Key Concepts

ellipse

An ellipse is a significant shape in geometry and mathematics, often compared to the shape of an elongated circle. It is defined as the set of all points \( (x, y) \), such that the sum of the distances from two distinct points, known as foci, is constant. For the equation \( x^2 + 4y^2 = 4 \), by rewriting it as \( \frac{x^2}{4} + \frac{y^2}{1} = 1 \), we identify it as the equation of an ellipse. In this form, it illustrates that we have an ellipse centered at the origin \( (0, 0) \).

• The semi-major axis lies along the x-axis, with a length of 2.
• The semi-minor axis lies along the y-axis, with a length of 1.

This specific arrangement makes the ellipse look broad horizontally and narrow vertically. Understanding the characteristics of an ellipse helps in recognizing its properties like orientation and extent, pivotal when dealing with level curves in functions.

gradient vector

The gradient vector is a powerful tool in calculus, representing the direction and rate of greatest increase of a function. For a function \( f(x, y) \), the gradient vector \( abla f \) is symbolized as \( (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) \). In this exercise, the function \( f(x,y) = x^2 + 4y^2 \) has been used to find the gradient vector.

• The partial derivative with respect to \( x \) is \( 2x \).
• The partial derivative with respect to \( y \) is \( 8y \).

Therefore, the gradient vector is \( (2x, 8y) \). At any point, substituting the values yields a vector that shows the steepest ascent direction. At point \( P(-2,0) \), the gradient becomes \( (-4,0) \), indicating movement to the left along the x-axis. This insight is crucial in multivariable calculus as it reveals where the function is increasing or decreasing most swiftly.

partial derivatives

In calculus, partial derivatives play an essential role in understanding changes in multivariable functions. A partial derivative of a function with multiple variables measures how the function changes as one variable changes, holding the others constant. For the function \( f(x,y) = x^2 + 4y^2 \),

• The partial derivative with respect to \( x \) is calculated as \( \frac{\partial f}{\partial x} = 2x \).
• The partial derivative with respect to \( y \) is determined as \( \frac{\partial f}{\partial y} = 8y \).

These derivatives tell us how the function \( f \) changes when either \( x \) or \( y \) are tweaked slightly. Such calculations are instrumental in forming the gradient vector and finding level curves. Analyzing partial derivatives helps in understanding the local behavior of a function, a foundational concept across various fields like economics, physics, and engineering.