Problem 23 - 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. $$y^{2}=-15 x,(-3,-3 \sqrt{5})$$

Step-by-Step Solution

Answer

Tangent line: \(y + 3\sqrt{5} = \frac{\sqrt{5}}{2}(x + 3)\); Normal line: \(y + 3\sqrt{5} = -\frac{2}{\sqrt{5}}(x + 3)\).

Step 1: Identify the Parabola Equation

The given equation of the parabola is \(y^2 = -15x\). This is a standard form of a parabola that opens to the left.

Step 2: Find the Derivative

To find the slope of the tangent line, we first need the derivative of the parabola equation. The equation is \(y^2 = -15x\), so implicitly differentiating with respect to \(x\) gives:\[2y \frac{dy}{dx} = -15\]. Solving for \(\frac{dy}{dx}\), we get: \[\frac{dy}{dx} = \frac{-15}{2y}\].

Step 3: Calculate the Slope of the Tangent Line

Substitute the point \((-3, -3\sqrt{5})\) into the derivative:\[\frac{dy}{dx} = \frac{-15}{2(-3\sqrt{5})} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2}\]. This is the slope of the tangent line at the given point.

Step 4: Write the Equation of the Tangent Line

Using point-slope form of a line: \(y - y_1 = m(x - x_1)\), where \(m\) is the slope, and \((x_1, y_1)\) is the point \((-3, -3\sqrt{5})\), substitute the values to get the equation of the tangent line: \[y + 3\sqrt{5} = \frac{\sqrt{5}}{2}(x + 3)\].

Step 5: Calculate the Slope of the Normal Line

The normal line is perpendicular to the tangent line, hence its slope is the negative reciprocal of the tangent's slope. Thus, the slope of the normal line is \(-\frac{2}{\sqrt{5}}\).

Step 6: Write the Equation of the Normal Line

Using the point-slope form again for the normal line: \[y + 3\sqrt{5} = -\frac{2}{\sqrt{5}}(x + 3)\].

Step 7: Graph the Parabola, Tangent and Normal Lines

Plot the parabola \(y^2 = -15x\), which opens to the left. Mark the point \((-3, -3\sqrt{5})\) on the curve. Draw the tangent line through this point with slope \(\frac{\sqrt{5}}{2}\), and the normal line with slope \(-\frac{2}{\sqrt{5}}\), ensuring it is perpendicular to the tangent line.

Key Concepts

Parabola Equation

Let's start by understanding the **parabola equation** given in the exercise. The equation we have is \(y^2 = -15x\). This is a standard form of a parabola that opens towards the left. In general, a parabola can open up, down, left, or right, and its direction is determined based on the squared term and its coefficient.
The equation \(y^2 = ax\) indicates a parabola that opens horizontally. Specifically, if \(a\) is negative, as in this case with \(-15\), the parabola opens to the left. Analyzing a parabola from its equation helps to understand its shape and direction, which are crucial for sketching and analyzing its properties, like calculating tangent lines.

Implicit Differentiation

**Implicit differentiation** is a technique used when it is difficult to solve an equation for one variable in terms of another. In this case, the parabola equation \(y^2 = -15x\) doesn't solve neatly for \(y\) as a function of \(x\), necessitating implicit differentiation.
To differentiate implicitly, we treat \(y\) as a function of \(x\) and use the chain rule. Differentiating both sides of \(y^2 = -15x\) with respect to \(x\) yields the equation:

• \(2y \frac{dy}{dx} = -15\)

This gives us the slope of the curve at any point, noted as \(\frac{dy}{dx}\). Solving this equation for \(\frac{dy}{dx}\) helps us find the slope of the tangent line at a given point, which is essential for later calculations.

Slope Calculation

**Slope calculation** is crucial for finding the tangent line. The slope of a line gives its steepness and direction. In calculus, the derivative of a function at a point provides this slope for curves.
Starting with our implicit differentiation result:

• \(\frac{dy}{dx} = \frac{-15}{2y}\)

We substitute the given point \((-3, -3\sqrt{5})\) into the derivative to find the slope at this particular point.

• \(\frac{dy}{dx} = \frac{-15}{2(-3\sqrt{5})} = \frac{\sqrt{5}}{2}\)

This value tells us the slope of the tangent line at the point \((-3, -3\sqrt{5})\), which is critical for writing the equation of both tangent and normal lines.

Point-Slope Form

The **point-slope form** of a line is a straightforward way to write the equation of a line when you have a specific point on the line and the slope. The formula is:

• \(y - y_1 = m(x - x_1)\)

where \((x_1, y_1)\) are the coordinates of the given point and \(m\) is the slope of the line.
For the tangent line through the point \((-3, -3\sqrt{5})\), we use the slope \(\frac{\sqrt{5}}{2}\), leading to:

• \(y + 3\sqrt{5} = \frac{\sqrt{5}}{2}(x + 3)\)

For the normal line, important because it is perpendicular to the tangent, the slope is the negative reciprocal of the tangent slope, \(-\frac{2}{\sqrt{5}}\). Thus, similarly, the normal line's equation is:

• \(y + 3\sqrt{5} = -\frac{2}{\sqrt{5}}(x + 3)\)

These forms allow you to find the precise equations of lines that are essential in calculus and geometry for understanding curve behaviors and properties.