Problem 48 - Calculus Early Transcendentals

Question

If \(g(x, y)=x^{2}+y^{2}-4 x,\) find the gradient vector \(\nabla g(1,2)\) and use it to find the tangent line to the level curve \(g(x, y)=1\) at the point \((1,2) .\) Sketch the level curve, the tangent line, and the gradient vector.

Step-by-Step Solution

Answer

The gradient vector is \((-2, 4)\) and the equation of the tangent line is \(y = \frac{1}{2}(x - 1) + 2\).

Step 1: Gradient Calculation

To find the gradient vector \( abla g(x, y) \), we need to calculate the partial derivatives of \( g(x, y) = x^2 + y^2 - 4x \) with respect to \( x \) and \( y \). This yields:- For \( x \): \( \frac{\partial g}{\partial x} = 2x - 4 \)- For \( y \): \( \frac{\partial g}{\partial y} = 2y \)Therefore, the gradient vector at any point \( (x, y) \) is given by \( abla g(x, y) = (2x - 4, 2y) \).

Step 2: Evaluate Gradient at the Point `(1,2)`

Substitute \( x = 1 \) and \( y = 2 \) into the gradient vector equation \( abla g(x, y) = (2x - 4, 2y) \):- \( 2(1) - 4 = -2 \)- \( 2(2) = 4 \)Thus, \( abla g(1,2) = (-2, 4) \).

Step 3: Equation of the Tangent Line

The tangent line at a point is perpendicular to the gradient vector at that point. The point-slope form of a line is given by \( y - y_1 = m(x - x_1) \). Here, the slope \( m \) is the negative reciprocal of the gradient's \( x \)-component divided by its \( y \)-component.Using the gradient \( (-2, 4) \), the slope \( m \) can be calculated as \( -\frac{-2}{4} = \frac{1}{2} \).Substituting the point \( (1, 2) \), the line equation becomes:\[y - 2 = \frac{1}{2}(x - 1)\]

Step 4: Sketch the Diagram

1. **Level Curve:** The equation \( x^2 + y^2 - 4x = 1 \) can be rewritten in standard circle form by completing the square: \((x - 2)^2 + y^2 = 5\), which is a circle centered at \((2, 0)\) with radius \(\sqrt{5}\).2. **Tangent Line:** The tangent line is \( y = \frac{1}{2} (x - 1) + 2 \), which you plot passing through the point \((1, 2)\) and appears tangent to the circle.3. **Gradient Vector:** Draw the vector \((-2, 4)\) starting from the point \((1, 2)\), pointing in the direction of increase, perpendicular to the tangent line.

Key Concepts

Gradient Vector

In multivariable calculus, the gradient vector is a fundamental concept that helps us determine the direction and rate of fastest increase of a function. For a function like \(g(x, y) = x^2 + y^2 - 4x\), the gradient \( abla g(x, y) \) is a vector consisting of partial derivatives of \(g\) with respect to \(x\) and \(y\).
\[abla g(x, y) = \left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}\right) = (2x - 4, 2y)\]
This vector points in the direction of greatest increase of the function, and its length indicates the rate of increase.

To evaluate it specifically at the point \((1, 2)\), substitute \(x = 1\) and \(y = 2\) into the gradient function:

• The partial derivative with respect to \(x\): \(2(1) - 4 = -2\)
• The partial derivative with respect to \(y\): \(2(2) = 4\)

Therefore, \(abla g(1, 2) = (-2, 4)\). This vector is important as it not only shows the direction of steepest ascent but also is perpendicular to the tangent of the level curve at this point.

Tangent Line

The tangent line at a particular point on a curve provides a linear approximation of the curve at that point. It is an essential tool for understanding the local behavior of the curve.
For any curve defined by \(g(x, y)\), the tangent line at a given point is perpendicular to the gradient vector at that point.

In our example, to find the equation of the tangent line at the point \((1, 2)\), we use the gradient vector \((-2, 4)\). The slope of the line is the negative reciprocal of the gradient's \(x\)-component divided by its \(y\)-component:
\[m = -\frac{-2}{4} = \frac{1}{2}\]
This slope shows how the line rises relative to how much it runs. Using the point-slope form of a line:
\[y - y_1 = m(x - x_1)\]
With \(x_1 = 1\), \(y_1 = 2\), and \(m = \frac{1}{2}\), the equation becomes:
\[y - 2 = \frac{1}{2}(x - 1)\]
This line is tangent to the level curve at \((1, 2)\), illustrating how straight-line approximations can be drawn for more complex curves.

Level Curves

Level curves, or contour lines, are a way of visualizing functions of two variables. They represent curves along which the function has a constant value. For example, for a function \(g(x, y) = c\), the level curve would be all points \((x, y)\) satisfying this equation.

In our exercise, the level curve is defined by \(g(x, y) = 1\), or \(x^2 + y^2 - 4x = 1\). By rewriting this equation:

• Complete the square: \((x - 2)^2 + y^2 = 5\)
• This reveals a circle centered at \((2, 0)\) with a radius of \(\sqrt{5}\)

This circular level curve can be plotted easily, providing a visual understanding of the function's behavior.

The tangent line and the gradient vector interact with the level curve at specific points:

• The gradient vector is always perpendicular to the level curve at any point.
• The tangent line, which touches the curve at one point without intersecting, reflects the gradient's perpendicular nature.

Understanding how these elements connect improves our comprehension of multivariable systems and how they can be visually represented.