Problem 19 - 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. $$y^{2}=16 x,(1,-4)$$

Step-by-Step Solution

Answer

Tangent: \(y = -2x - 2\); Normal: \(y = \frac{1}{2}x - \frac{9}{2}\).

Step 1: Understand the Parabola Equation

The given equation of the parabola is \(y^2 = 16x\). This is a sideways opening parabola with its vertex at the origin \((0, 0)\) and it opens to the right because the \(x\) variable is positive.

Step 2: Determine the Derivative

To find the slope of the tangent at the point \((1,-4)\), we need the derivative. Differentiate both sides of \(y^2 = 16x\) with respect to \(x\), using implicit differentiation:\[\frac{d}{dx} (y^2) = \frac{d}{dx} (16x)\]This gives us:\[2y \frac{dy}{dx} = 16\]Solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{16}{2y} = \frac{8}{y}\].

Step 3: Calculate Slope of Tangent Line

Substitute the ordinate \(y = -4\) of the given point into the expression for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{8}{-4} = -2\].The slope of the tangent line at \((1,-4)\) is \(-2\).

Step 4: Write the Equation of Tangent Line

Using the point-slope form \(y - y_1 = m(x - x_1)\), where \(m\) is the slope:\[y - (-4) = -2(x - 1)\]Simplify:\[y + 4 = -2x + 2\]\[y = -2x - 2\].The equation of the tangent line is \(y = -2x - 2\).

Step 5: Calculate Slope of Normal Line

The slope of the normal line is the negative reciprocal of the slope of the tangent. So the slope of the normal is:\[m_{normal} = \frac{1}{-(-2)} = \frac{1}{2}\].

Step 6: Write the Equation of Normal Line

Using the point-slope form for the normal line:\[y - (-4) = \frac{1}{2}(x - 1)\]Simplify:\[y + 4 = \frac{1}{2}x - \frac{1}{2}\]\[y = \frac{1}{2}x - \frac{9}{2}\].The equation of the normal line is \(y = \frac{1}{2}x - \frac{9}{2}\).

Step 7: Sketch the Parabola and Lines

Sketch the parabola \(y^2 = 16x\), which opens to the right. Plot the point \((1, -4)\) where the tangent and normal lines intersect the parabola. Draw the tangent line \(y = -2x - 2\) and the normal line \(y = \frac{1}{2}x - \frac{9}{2}\) through this point.

Key Concepts

Tangent Line

When analyzing curves in calculus, it's often useful to explore the concept of a tangent line. A tangent line is a straight line that touches a curve at just one point. At this point, the tangent line has the same direction (i.e., the same slope) as the curve itself. This means that the tangent line provides a linear approximation of the curve at that specific location.

Tangent lines have key characteristics:

• They touch the curve at only one point, although they may intersect the curve elsewhere if extended.
• They have the same slope as the curve at the point of tangency.

To find the equation of a tangent line, we use the derivative, which gives us the slope of the curve at a particular point. In our example, we used implicit differentiation to determine the derivative of the given parabola equation. After finding the slope, we used the point-slope form to construct the equation of the tangent line.

Normal Line

A normal line is another important concept related to curves. While the tangent line gives us a sense of how the curve behaves at a point, the normal line gives a perpendicular insight. A normal line to a curve at a specific point is the line perpendicular to the tangent line at that point. Therefore, if you know the slope of the tangent line, you can easily find the slope of the normal line.

Here are some important features of the normal line:

• It is perpendicular to the tangent line at the point of contact on the curve.
• The slope of the normal line is the negative reciprocal of the slope of the tangent line.

In our problem, the tangent line had a slope of -2. Therefore, the normal line, being perpendicular, had a slope of 1/2, as negative reciprocal of -2 is 1/2. This allowed us to find the equation of the normal line using the point-slope form just as with the tangent line.

Implicit Differentiation

Implicit differentiation is a powerful technique for finding derivatives of functions that are not given explicitly. When an equation involves both dependent and independent variables in a tangled relation, explicit functions can't easily be applied. This is often the case with equations like those of parabolas, circles, and ellipses, where one variable is squared or otherwise mixed with the other.

Implicit differentiation lets us find the derivative of these equations by differentiating both sides with respect to one variable, while treating all other variables implicitly as functions of that variable:

• This technique requires us to apply the chain rule, often multiple times and solve for the derivative.
• It allows us to find the rate at which one variable changes with respect to another when variables are intertwined.

In the provided problem, we used implicit differentiation to differentiate the equation of the parabola. This helped us obtain the slope needed for the tangent line, demonstrating the utility of this method in calculus.