Problem 21 - 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. $$x^{2}=2 y,(4,8)$$

Step-by-Step Solution

Answer

Tangent: \( y = 4x - 8 \); Normal: \( y = -\frac{1}{4}x + 9 \). Sketch includes the parabola, tangent, and normal lines.

Step 1: Differentiate the Parabola Equation

Given the parabola equation is \( x^2 = 2y \). We first need to determine the derivative to find the slope of the tangent line. Rewrite the equation as \( y = \frac{x^2}{2} \).Differentiate with respect to \( x \):\[\frac{dy}{dx} = \frac{d}{dx}\left(\frac{x^2}{2}\right) = x\]So, the slope of the tangent line at any point \( (x, y) \) is \( x \).

Step 2: Calculate the Slope of the Tangent Line

We need to find the slope of the tangent at the given point \((4, 8)\). Using the derivative \( x \), substitute \( x = 4 \) into the slope formula:\[ m_{tangent} = 4 \]

Step 3: Find the Tangent Line Equation

Now, use the point-slope form of a line, \( y - y_1 = m(x - x_1) \), with \( m = 4 \) and the point \( (4, 8) \):\[y - 8 = 4(x - 4)\]This simplifies to:\[y = 4x - 8\]This is the equation of the tangent line.

Step 4: Calculate the Slope of the Normal Line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. For the tangent slope \( 4 \), the normal slope \( m_{normal} \) is:\[ m_{normal} = -\frac{1}{4} \]

Step 5: Find the Normal Line Equation

Again using the point-slope form of a line with \( m = -\frac{1}{4} \) and the point \( (4, 8) \):\[y - 8 = -\frac{1}{4}(x - 4)\]Simplify to find:\[y = -\frac{1}{4}x + 9\]This is the equation of the normal line.

Key Concepts

Understanding Parabolas

A parabola is a U-shaped curve that is very important in mathematics, especially when learning about quadratic equations and geometry. Every parabola is symmetrical, which means it's a mirror image on both sides of its vertex. The typical equation for a vertically oriented parabola is in the form of \( y = ax^2 + bx + c \), but sometimes it can be written as \( x^2 = 2y \) to describe another orientation, like in the given problem. Parabolas have a point called the vertex, which is the highest or lowest point of the curve.

• This point can be found by using different mathematical techniques like completing the square or using the formula \( y = \frac{-b}{2a} \) for vertically oriented parabolas.
• Parabolas also open either upwards or downwards, depending on the sign of \( a \) in the equation. Here, the equation \( x^2 = 2y \) describes a parabola that opens upwards.

Understanding these basic properties will help you feel more comfortable working with parabolas.

The Role of Derivatives

A derivative is a fundamental tool in calculus that measures how a function changes as its inputs change. It tells us the slope of a curve at any given point, which means it gives us the rate of change at that point. When working with parabolas or any other curve, knowing the derivative allows us to understand how steep the curve is at any location.

• For the parabola \( y = \frac{x^2}{2} \), the derivative \( \frac{dy}{dx} = x \) tells us that the slope of the tangent line at any point \( (x, y) \) on the parabola is simply \( x \).
• This derivative was found by differentiating \( y = \frac{x^2}{2} \) with respect to \( x \), a key skill in developing an understanding of calculus.

Once the derivative is known, it becomes straightforward to calculate the slope of the tangent line at any specific point by substituting the \( x \)-coordinate into the derivative.

Understanding Slope

The slope of a line is a measure of its steepness, usually represented by the letter \( m \). Slope is calculated as the "rise" over the "run," which is the change in \( y \) divided by change in \( x \) between two points on a line. In the context of the parabola, the slope of the tangent at a specific point can be determined using the derivative.

• At the point \( (4, 8) \) on the parabola, the slope of the tangent line \( m_{tangent} \) is 4, as determined by substituting \( x = 4 \) into the derivative \( \frac{dy}{dx} = x \).
• The slope of the normal line, which is perpendicular to the tangent line at the same point, is the negative reciprocal of the tangent slope. Hence, \( m_{normal} = -\frac{1}{4} \).

Understanding slope is crucial when finding equations of lines because it forms part of the equation through point-slope and slope-intercept forms.

Equation of the Line

To find the equation of any line, understanding its slope and a point it passes through is essential. The point-slope form is handy: \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is the point and \( m \) is the slope. From this, you can derive the equation of both the tangent and normal lines to our parabola at point \( (4, 8) \).

• Using the tangent slope \( 4 \), the equation becomes \( y = 4x - 8 \) after plugging into the point-slope form and simplifying.
• For the normal line, using the slope \(-\frac{1}{4}\), the equation simplifies to \( y = -\frac{1}{4}x + 9 \).

Having these lines helps further in sketching them along with the parabola, emphasizing their geometric relationship: the tangent line just touches the curve while the normal line perpendicularly intersects the point of tangency.