Problem 34 - University Calculus: Early Transcendentals

Question

Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point. $$y^{2}-2 x-4 y-1=0, \quad(-2,1)$$

Step-by-Step Solution

Answer

The point (-2,1) is on the curve. Tangent line: \(y = -x - 1\). Normal line: \(y = x + 3\).

Step 1: Verify the Point on the Curve

Substitute the given point \((-2, 1)\) into the curve's equation: \(y^2 - 2x - 4y - 1 = 0\). Replacing \(x\) with \(-2\) and \(y\) with \(1\), we get: \(1^2 - 2(-2) - 4(1) - 1 = 1 + 4 - 4 - 1 = 0\). Since the equation holds true, the point \((-2, 1)\) is on the curve.

Step 2: Find the Derivative

To find the tangent, we need the derivative of the curve equation with respect to \(x\). Differentiate implicitly: \(\frac{d}{dx}(y^2 - 2x - 4y - 1) = 2y\frac{dy}{dx} - 2 - 4\frac{dy}{dx} = 0\). Simplify the equation: \((2y - 4)\frac{dy}{dx} = 2\).

Step 3: Solve for the Derivative

Further simplify to find \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{2}{2y - 4}\). Substitute \(y = 1\) (from the point (-2,1)) into this equation: \(\frac{dy}{dx} = \frac{2}{2(1) - 4} = \frac{2}{2 - 4} = \frac{2}{-2} = -1\).

Step 4: Find the Tangent Line

The slope \(m\) of the tangent line at \((-2,1)\) is \(-1\). Use the point-slope form \(y - y_1 = m(x - x_1)\). Substitute \(m\), \(x_1 = -2\), \(y_1 = 1\): \(y - 1 = -1(x + 2)\). Simplify to get the equation of the tangent line: \(y = -x - 1\).

Step 5: Find the Normal Line

The normal line is perpendicular to the tangent, so its slope is the negative reciprocal: \(m = 1\). Use the point-slope form for the normal line: \(y - 1 = 1(x + 2)\). Simplify to find the equation of the normal line: \(y = x + 3\).

Key Concepts

Curve Equation

In mathematics, a curve equation is a vital tool that defines a shape in a two-dimensional plane. It forms the basis for plotting and understanding complex shapes. A curve equation can take various forms, like a polynomial, parametric, or implicit equation. In this context, we have an implicit equation:

• \( y^2 - 2x - 4y - 1 = 0 \)

Implicit equations express relationships between \( x \) and \( y \) without having one variable explicitly in terms of the other. To check if a specific point lies on the curve, you substitute the point's \( x \) and \( y \) values into the equation. If the equation holds true (equals zero, in this case), then the point is indeed on the curve.
Verifying points helps in understanding the location of the curve relative to given points, a crucial step before finding the tangent and normal lines.

Tangent Line

The tangent line is a straight line that touches a curve at a single point without crossing it. It has the same slope as the curve at that specific point.
To find the slope of the tangent line, you need to calculate the derivative of the curve equation. This process requires implicit differentiation if the curve is defined implicitly, as in this exercise.
Here, the slope at the point \((-2,1)\) was found using the derivative formula:

• \( \frac{dy}{dx} = \frac{2}{2y - 4} \)

Substituting \( y = 1 \) gives the slope \(-1\).
Next, use the point-slope form of a line equation, \( y - y_1 = m(x - x_1) \), where

• \( m \) is the slope,
• and \((x_1, y_1)\) is the point of tangency.

This yields the tangent line equation:

• \( y = -x - 1 \)

The tangent line's significance lies in its ability to approximate the curve near the point of tangency, providing insights into the curve's local behavior.

Normal Line

A normal line is perpendicular to the tangent line at a point on the curve. The slope of the normal line is the negative reciprocal of the slope of the tangent line.
In this exercise, the slope of the tangent line at \((-2,1)\) is \(-1\). Therefore, the slope of the normal line is

• \( 1 \)

Using the point-slope form, the equation of the normal line is determined as:

• \( y - 1 = 1(x + 2) \)

This equation simplifies to:

• \( y = x + 3 \)

Normal lines are useful in applications involving orthogonal trajectories, reflection angles, and understanding curve geometry. They provide insightful geometric properties about a curve at a given point, enhancing the understanding of spatial relationships in calculus.