Problem 315 - Calculus Volume 1

Question

Find the equation of the tangent line to the graph of the given equation at the indicated point. Use a calculator or computer software to graph the function and the tangent line. $$ x y+\sin (x)=1,\left(\frac{\pi}{2}, 0\right) $$

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = 0 \).

Step 1: Differentiate the equation implicitly

Given the equation \( xy + \sin(x) = 1 \), we need to find its derivative, which represents the slope of the tangent line. To do so, differentiate both sides with respect to \( x \), taking into account that \( y \) is a function of \( x \). This gives \( x \frac{dy}{dx} + y + \cos(x) = 0 \). Solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{-y - \cos(x)}{x} \).

Step 2: Evaluate the derivative at the given point

Substitute \( x = \frac{\pi}{2} \) and \( y = 0 \) into the derivative \( \frac{dy}{dx} = \frac{-y - \cos(x)}{x} \). This yields \( \frac{dy}{dx} = \frac{0 - 0}{\frac{\pi}{2}} = \frac{-\cos\left(\frac{\pi}{2}\right)}{\frac{\pi}{2}} = \frac{0}{\frac{\pi}{2}} = 0 \). The slope at this point is 0.

Step 3: Write the equation of the tangent line

The equation of a line is given by \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. We have \( m = 0 \) from Step 2 and the point \( \left(\frac{\pi}{2}, 0\right) \) lies on the tangent line. Substitute these values into the line equation: \( y = 0 \cdot x + b \). Since \( y = 0 \) when \( x = \frac{\pi}{2} \), it follows that \( b = 0 \). Thus, the equation of the tangent line is \( y = 0 \).

Step 4: Graph the function and tangent line

Using a graphing calculator or software, plot the function \( xy + \sin(x) = 1 \) and the horizontal line \( y = 0 \). The tangent at the point \( \left(\frac{\pi}{2}, 0\right) \) appears as a horizontal line on the graph, touching the curve at this point.

Key Concepts

Tangent Line

The concept of a tangent line is fundamental in calculus, particularly when working with curves in the plane. A tangent to a curve at a given point is a straight line that touches the curve at that point but does not cross it at that local vicinity. The tangent line represents the instantaneous rate of change of the curve at that point, akin to the concept of a slope in linear equations.

In the context of the exercise, the tangent line to the curve defined by the equation \(xy + \sin(x) = 1\) at the point \(\left(\frac{\pi}{2}, 0\right)\) is what we seek to find. Knowing that a tangent line reflects the behavior of the curve precisely at that point leads us to use derivatives, which capture the instantaneous rate of change, enabling us to determine the slope of the tangent line.

In simpler terms, if you imagine a tightrope walker crossing a bridge, the tangent line at any point of the bridge tells us exactly how steeply the bridge is inclined there, allowing the tightrope walker to maintain balance and direction.

Derivative Evaluation

Evaluating a derivative is a crucial step in finding a tangent line. It involves calculating the slope of a function at a specific point. This slope is represented by the derivative of a function, often denoted as \(\frac{dy}{dx}\) when dealing with single-variable calculus.

The exercise employs implicit differentiation because the function is not explicitly solved for \(y\). Implicit differentiation is a powerful tool for finding derivatives of more complex equations like \(xy + \sin(x) = 1\).

• Differentiate each term with respect to \(x\), remembering to treat \(y\) as a function of \(x\).
• Apply the chain rule where necessary.
• Rearrange the result to solve for \(\frac{dy}{dx}\).

In this case, after differentiating and substituting the given point \(\left(\frac{\pi}{2}, 0\right)\), we determine that the slope is \(0\). This indicates that at this particular point, the curve is neither inclining nor declining; it is perfectly flat.

Equations of Lines

Constructing the equation of a line is an essential skill in algebra and calculus. The general form of a line's equation is given by \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.

In this exercise, after finding that the derivative (or slope at the point \(\left(\frac{\pi}{2}, 0\right)\)) is \(0\), the equation of the tangent line simplifies significantly. The slope \(m\) of \(0\) tells us that the tangent line is horizontal.

• Given the point where the tangent line touches the curve, \(x = \frac{\pi}{2}\) and \(y = 0\), it directly follows that \(b\) will also be \(0\).
• This results in a very simple equation: \(y = 0\).

The tangent line's equation is a horizontal line running through the x-axis at \(y = 0\). This showcases one of the most straightforward instances of finding a line equation in calculus.