Problem 289 - Calculus Volume 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

The level curve is an ellipse defined by \( x^2 + 4y^2 = 4 \), and the gradient vector at \( P(-2, 0) \) is \((-4, 0)\).

Step 1: Understanding the Level Curve

A level curve of a function is the set of all points where the function has a constant value. Here, we need the level curve of the function \( f(x, y) = x^2 + 4y^2 \) that passes through \( P(-2,0) \). To find the constant value, plug point \( P(-2,0) \) into the function.

Step 2: Calculate the Constant Value

Substitute the coordinates of point \( P(-2,0) \) into the function: \[ f(-2, 0) = (-2)^2 + 4(0)^2 = 4 + 0 = 4. \] This means the level curve that passes through \( P(-2, 0) \) is the curve along which \( x^2 + 4y^2 = 4. \)

Step 3: Sketch the Level Curve

The equation \( x^2 + 4y^2 = 4 \) represents an ellipse centered at the origin. Using technology (like graphing software), you can sketch this curve to visualize it. The major axis is along the y-direction (vertical), and the semi-axes lengths can be determined as 2 and 1.

Step 4: Calculate the Gradient Vector

The gradient of a function \( f(x, y) \) is \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Calculate the partial derivatives: \( \frac{\partial f}{\partial x} = 2x \) and \( \frac{\partial f}{\partial y} = 8y \). Therefore, \( abla f = (2x, 8y) \).

Step 5: Evaluate the Gradient Vector at Point P

Now substitute point \( P(-2,0) \) into the gradient: \[ abla f(-2,0) = (2(-2), 8(0)) = (-4, 0). \] The gradient vector at point \( P(-2,0) \) is \((-4, 0)\).

Key Concepts

Level Curve

A level curve is a fascinating concept in calculus. It represents an imaginary contour line on a surface, much like a topographic line on a map. This line indicates where the function maintains a constant value. In the context of the function\[f(x, y) = x^2 + 4y^2,\]we find the level curve by setting the function to a constant value that passes through a specific point, such as point \(P(-2,0)\). To find this constant, insert the coordinates of point \(P\) into the function:\[f(-2, 0) = (-2)^2 + 4(0)^2 = 4.\]Thus, the level curve equation is \(x^2 + 4y^2 = 4.\)
The result is an ellipse, meaning every point on this curve will have the same output value of 4 from the function. Using graphing technology can help sketch this ellipse, illuminating the level curve visually.

Gradient Vector

The gradient vector is an essential concept in multivariable calculus. It pushes us to contemplate how changes in \(x\) and \(y\) affect the function. For the function \(f(x, y) = x^2 + 4y^2\), the gradient is a vector with components derived from the partial derivatives with respect to each variable.
- The partial derivative with respect to \(x\) is \( \frac{\partial f}{\partial x} = 2x \). - The partial derivative with respect to \(y\) is \( \frac{\partial f}{\partial y} = 8y \).
The gradient vector, \(abla f\), can be written as:\[abla f = (2x, 8y). \]
It essentially shows the direction in which the function increases most rapidly. Calculate this vector at our point of interest, \(P(-2,0)\), to get \((-4, 0)\). This gradient vector illustrates that at \(P\), any movement in the negative \(x\)-direction increases the value of \(f(x, y)\) the fastest.

Ellipse

Ellipses are curves in which the sum of distances from any point on the curve to two fixed points (foci) is constant. Here, the level curve equation \(x^2 + 4y^2 = 4\) resembles this classical form of an ellipse.The equation indicates the ellipse is centered at the origin, \((0,0)\). Since it can be rearranged to the form:\[\frac{x^2}{2^2} + \frac{y^2}{1^2} = 1,\]it gets easier to understand the geometrical dimensions.

• The semi-major axis (longest) lies along the y-axis with a length of 2.
• The semi-minor axis (shortest) lies along the x-axis with a length of 1.

Ellipses can offer tremendous insights in calculus as they represent varying rates of change along different directions. Such an understanding aids in visualizing constraints and functional consistency within specific bounds.

Partial Derivatives

Partial derivatives give us the rate of change of a multivariable function with respect to one variable at a time, holding others constant. They're the foundation of many calculus concepts, including the gradient vector.
In our function \(f(x, y) = x^2 + 4y^2\), you need to differentiate with respect to both variables:

• With respect to \(x\), treating \(y\) as constant:\[\frac{\partial f}{\partial x} = 2x.\]
• With respect to \(y\), treating \(x\) as constant:\[\frac{\partial f}{\partial y} = 8y.\]

These derivatives measure how sensitive the function \(f\) is to changes in \(x\) or \(y\). Understanding partial derivatives is essential for comprehending how functions behave locally around a point, becoming fundamental in various applications ranging from physics to economics.