Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equations of the tangent and the normal lines to the given parabola at the given point. Sketch the parabola, the tangent line, and the normal line. $$x^{2}=-10 y,(2 \sqrt{5},-2)$$

Step-by-Step Solution

Answer

Tangent line: \(y = -\frac{2\sqrt{5}}{5}x + 2\); Normal line: \(y = \frac{\sqrt{5}}{2}x - 5\).

Step 1: Find the derivative of the parabola

The equation of the parabola is given by \(x^2 = -10y\). To find the slope of the tangent line, we need its derivative with respect to \(x\). Rewriting the equation in terms of \(y\), we get \(y = -\frac{x^2}{10}\). Thus, the derivative of \(y\) with respect to \(x\) is \(\frac{dy}{dx} = -\frac{2x}{10} = -\frac{x}{5}\).

Step 2: Evaluate the derivative at the given point

The given point is \((2\sqrt{5}, -2)\). Plug the \(x\)-value into the derivative \(\frac{dy}{dx} = -\frac{x}{5}\) to find the slope of the tangent line at this point: \(m_{tangent} = -\frac{2\sqrt{5}}{5}\).

Step 3: Find the equation of the tangent line

Use the point-slope form of the line equation, \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = (2\sqrt{5}, -2)\) and \(m = -\frac{2\sqrt{5}}{5}\). Substitute to get \(y + 2 = -\frac{2\sqrt{5}}{5}(x - 2\sqrt{5})\). Simplify to obtain the equation: \(y = -\frac{2\sqrt{5}}{5}x + 2\).

Step 4: Find the slope of the normal line

The normal line is perpendicular to the tangent line. The slope of the normal line \(m_{normal}\) is the negative reciprocal of the tangent slope \(-\frac{2\sqrt{5}}{5}\), which gives \(m_{normal} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2} \) after rationalization.

Step 5: Find the equation of the normal line

Using the point-slope form again, with \(m = \frac{\sqrt{5}}{2}\), and the point \((2\sqrt{5}, -2)\), we have \(y + 2 = \frac{\sqrt{5}}{2}(x - 2\sqrt{5})\). Simplify this equation to obtain \(y = \frac{\sqrt{5}}{2}x - 5\).

Key Concepts

Derivatives

Derivatives are a core concept in calculus. They describe the rate at which a function changes. When we talk about the derivative of a function, we often refer to the slope of the tangent line at any given point on the function. This slope tells us how steeply the line is rising or falling at that particular point.

To find the derivative of a parabola, like the one given here \( x^2 = -10y \), we first express \( y \) in terms of \( x \). For our exercise, the equation becomes \( y = -\frac{x^2}{10} \).

• Differentiate with respect to \( x \): \( \frac{dy}{dx} = -\frac{2x}{10} \).
• Simplify: \( \frac{dy}{dx} = -\frac{x}{5} \).

This derivative \( -\frac{x}{5} \) represents the slope of the tangent line for any \( x \) value on the curve of the parabola.

Equation of a Line

An equation of a line typically is written in slope-intercept form as \( y = mx + b \), where \( m \) is the slope, and \( b \) is the y-intercept. Another popular form is the point-slope form: \( y - y_1 = m(x - x_1) \). This form is helpful when you have a point on the line and know the slope.

For our problem, the equation of the tangent line is found by using the point \( (2\sqrt{5}, -2) \) and the slope \( m = -\frac{2\sqrt{5}}{5} \). Plugging these into the equation gives:

• Substitute into the point-slope formula: \( y + 2 = -\frac{2\sqrt{5}}{5}(x - 2\sqrt{5}) \).
• Simplify to achieve the final form: \( y = -\frac{2\sqrt{5}}{5}x + 2 \).

This is the equation of the tangent line at the given point on the parabola.

Slope of Tangent

The slope of a tangent line to a curve at any point is the derivative evaluated at that point. It gives us a precise idea of the line's steepness at that exact location. It's like taking a snapshot of the curve's tilt at one singular point.

For the parabola \( x^2 = -10y \), the derivative \( \frac{dy}{dx} = -\frac{x}{5} \) gives the slope. At the point \( (2\sqrt{5}, -2) \), substituting \( x = 2\sqrt{5} \):

• Calculate: \( m_{tangent} = -\frac{2\sqrt{5}}{5} \).

This value indicates the incline of the tangent line at this specific location of the parabola.

Parabola

A parabola is a symmetrical curve that is a result of a quadratic function. Parabolas can open up, down, left, or right. The given parabola in our problem opens to the left, as shown by its equation \( x^2 = -10y \).

Here are some key features:

• The axis of symmetry: Vertical in this case.
• Vertex: This is where the curve changes direction.
• Direction based on coefficient: \( -10y \) suggests a horizontal opening to the left.

Understanding these characteristics helps identify the shape and direction of the parabola quickly, allowing us to work out related tasks, like finding tangent and normal lines at specific points.