Problem 40 - 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. $$x^{2} \cos ^{2} y-\sin y=0, \quad(0, \pi)$$

Step-by-Step Solution

Answer

Tangent line: \(y = \pi\), Normal line: \(x = 0\).

Step 1: Verify the Point is on the Curve

First, substitute the given point \((0, \pi)\) into the equation \(x^{2} \cos^{2} y - \sin y = 0\). Since \(x = 0\) and \(y = \pi\), we calculate:\[0^{2} \cos^{2}(\pi) - \sin(\pi) = 0 \times 1 - 0 = 0.\]The left-hand side equals the right-hand side, so the point \((0, \pi)\) is indeed on the curve.

Step 2: Differentiate the Curve Implicitly

Differentiate the curve equation with respect to \(x\). The differentiated form of \(x^{2} \cos^{2} y - \sin y = 0\) is:\[2x \cos^{2} y - x^{2} \cdot 2 \cos y \cdot (-\sin y) \frac{dy}{dx} - \cos y \frac{dy}{dx} = 0.\]Simplify this to find \(\frac{dy}{dx}\) by rearranging the equation in terms of \(\frac{dy}{dx}\).

Step 3: Solve for dy/dx at the Given Point

At the point \((0, \pi)\), we substitute into the simplified derivative:\[0 - 0 \cdot 2 \cdot (-1) \cdot \frac{dy}{dx} - (-1) \cdot \frac{dy}{dx} = 0.\]This simplifies to \(\frac{dy}{dx} = 0\) since both terms involving \(x\) vanish.

Step 4: Find the Equation of the Tangent Line

The slope of the tangent line is given by \(\frac{dy}{dx}\), which is \(0\) at the point \((0, \pi)\). Using the point-slope form of a line equation \(y - y_1 = m(x - x_1)\), and substituting \(m = 0\) and the point \((0, \pi)\), we get the tangent line equation:\[y = \pi.\]

Step 5: Find the Equation of the Normal Line

The normal line is perpendicular to the tangent. Its slope is the negative reciprocal of the tangent slope, so it is undefined (vertical line) since the tangent slope is \(0\). Thus, the normal line is:\[x = 0.\]

Key Concepts

Tangent Line

In calculus, the tangent line plays a crucial role as it represents the linear approximation of a curve at a particular point. It is essentially the straight line that "just touches" the curve at that particular spot without crossing it. To find the tangent line to a curve at a certain point, one typically uses derivative mathematics. The derivative at that point gives us the slope of the tangent line.

In the given problem, the derivative of the curve at point \( (0, \pi) \) was calculated using implicit differentiation. The result, \ \frac{dy}{dx} = 0 \ at this point, indicates that the tangent line is horizontal. Applying the point-slope form of a line, \( y - y_1 = m(x - x_1) \), with \( m = 0 \) (zero-slope) and point \( (0, \pi) \), the tangent line becomes \( y = \pi \).

This result highlights the connection between derivatives and tangent lines in calculus problem-solving. The zero slope denotes that the tangent line is flat and does not tilt upward or downward, aligning perfectly parallel to the x-axis.

Normal Line

The normal line to a curve at a given point is the line perpendicular to the tangent line at that same point. It represents the direction in which the curve shifts fastest away from the tangent line. To find the normal line, one generally uses the concept of perpendicular slopes.

The key characteristic of perpendicular lines is that the product of their slopes is \ -1 \. When the slope of the tangent line is \ 0 \ (as in the exercise), a normal line theoretically has an undefined slope. This is because the slope of the normal line is \ \frac{1}{0} \ which is an undefined or vertical slope.

In the provided calculus problem, since the tangent line's slope was determined to be \ 0 \ at point \( (0, \pi) \), the normal line at this point is a vertical line. Its equation is \( x = 0 \). This vertical line crosses the x-axis at \ x = 0 \ and maintains the perpendicular directionality expected from a normal line. This concept helps students comprehend how tangent and normal lines interact with curves visually and mathematically.

Calculus Problem Solving

Calculus problem-solving requires a clear understanding of several core concepts, including derivatives, slopes, and algebraic manipulation. When tackling problems like the one given, using implicit differentiation is key. This technique is employed when dealing with equations where \ y \ is not isolated or directly expressed as a function of \ x \.

In the exercise, the curve equation was implicitly differentiated to determine the slope at the point \( (0, \pi) \). Successful calculus problem-solving involves:

• Verifying that a given point lies on the curve by substituting the values back into the original equation and ensuring both sides equalize.
• Using derivatives to find tangent slopes at specific points.
• Translating mathematical results into the geometry of tangent and normal lines.

Understanding these steps enables students to visualize the curve's behavior at specific points and develop a deeper grasp of how calculus can describe real-world relationships. Moreover, practicing these techniques enhances problem-solving skills and confidence in handling both common and complex calculus challenges.