Question

In Exercises 17-26, find the lines that are (a) tangent and (b) normal to the curve at the given point. $$2 x y+\pi \sin y=2 \pi, \quad(1, \pi / 2)$$

Short answer

The line tangent to the curve at the point (1, \( \pi /2 \)) is \( y = -x + \pi - 1/2 \), and the line normal to the curve at the same point is \( y = x + \pi - 3/2 \)

Step-by-step solution

Calculation of derivative using implicit differentiation

First, take the derivative of both sides of the curve equation. This gives us \(2x dy/dx + 2y + \pi y' \cos y = 0\). To solve for \(dy/dx\), isolate it on one side of the equation: \(dy/dx = -\frac{(2y + \pi \cos y)}{(2x + \pi \sin y)}\).

Apply the given point

Next, substitute the coordinates of the given point (1, \(\pi/2\)) into the formula for the derivative. Simplify to find the slope of the tangent: \(dy/dx = -\frac{(2(\pi/2) + \pi \cos(\pi/2))}{(2*1 + \pi \sin(\pi/2))} = -1\).

Equations of tangent and normal lines

Finally, use the slope-point form of the equation of a line to write the equations of the tangent and normal lines. The slope of the normal line is the negative reciprocal of the slope of the tangent line, so it is 1. The tangent line is \(y - \pi/2 = -1(x - 1)\), which simplifies to \(y = -x + \pi - 1/2\). The normal line is \(y - \pi/2 = 1(x - 1)\), which simplifies to \(y = x + \pi - 3/2\).

Key concepts

Tangent and Normal Lines

Understanding tangent and normal lines to a curve at a certain point is crucial for many applications in mathematics and physics. A tangent line is a straight line that touches a curve at only one point and has the same slope as the curve at that point. It effectively mirrors the direction in which the curve is headed at that specific location. Conceptualizing a tangent line is useful for visualizing instantaneous rates of change and is a cornerstone of differential calculus.On the flip side, a normal line to a curve at a given point is perpendicular to the tangent line at that point. It represents a direction of zero change in the value of the function, pointing straight outwards from the curved path. In geometric terms, normal lines are crucial when constructing orthogonal trajectories and in various engineering applications involving force vectors.When given an implicit equation like the exercise's curve, we use implicit differentiation to find the slopes of these lines at a point. Once the slope is determined, writing out the equations for both the tangent and normal lines requires us to utilize the slope-point form of a line, ensuring our lines pass through the specific point on the curve. The convenience of this method is that it is broadly applicable to curves described by various types of equations, not just functions in the form of y=f(x).

Derivative Calculation

The process of derivative calculation, particularly in the context of implicit differentiation, is essential for situations where it is difficult or impossible to solve for y explicitly as a function of x. Implicit differentiation involves taking the derivative of both sides of an equation with respect to x, which cleverly allows us to find the derivative of y with respect to x (denoted as dy/dx) even when y is mingled with x in the equation.The Steps:

• Differentiate each term with respect to x, treating y as an implicitly defined function of x.
• Whenever you differentiate a term involving y, tack on a dy/dx to account for the derivative of y.
• Rearrange the differentiated equation to solve for dy/dx.
• Substitute the coordinates of the specific point you are interested in to find the numerical slope value at that point.

The result is the slope of the tangent line to the curve at the point in question. This calculated derivative acts like a snapshot, capturing the instantaneous rate of change of y with respect to x at that point, which is directly translated into the gradient of the tangent. By understanding this method, students can tackle a wide variety of problems involving rates of change and the behavior of curves.

In the exercise, this process yielded a slope of -1 for the tangent line at the point (1, \(\frac{\pi}{2}\)).

Slope-Point Form of a Line

The last piece in the puzzle of finding tangent and normal lines is writing their equations correctly. For this, we use the slope-point form of the equation of a line. This form is incredibly useful as it gives us an intuitive way of crafting the equation of a line when we have a point which it passes through and its slope.General form:\[ y - y_1 = m(x - x_1) \]where \(m\) is the slope of the line, and \((x_1, y_1)\) is the point through which the line passes. In constructing a solution for the exercise, the point \((1, \frac{\pi}{2})\) and the slopes -1 (tangent) and 1 (normal) are substituted into the formula.For the tangent line, with slope -1, we have:\[ y - \frac{\pi}{2} = -1(x - 1) \]Simplifying, this becomes the equation \( y = -x + \frac{3\pi}{2} - 1 \).Similarly, for the normal line with slope 1, we use:\[ y - \frac{\pi}{2} = 1(x - 1) \]Which simplifies to \( y = x + \frac{\pi}{2} - \frac{3}{2} \).Using the slope-point form makes it straightforward to switch between the concept of slope and the actual equation of a line on a graph. It's a fundamental tool for students studying algebra and calculus. By mastering this, students can graph lines, understand their direction and steepness, and solve various geometric problems involving lines and curves.