Problem 315 - Calculus Volume 1

Question

For the following exercises, 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 tangent line at \(\left(\frac{\pi}{2}, 0\right)\) is \(y = 0\).

Step 1: Differentiate Implicitly

To find the equation of the tangent line, we first need the derivative of the given function implicitly with respect to \(x\). The given equation is: \(xy + \sin(x) = 1\). Differentiating both sides with respect to \(x\), we have: \[ \frac{d}{dx}(xy) + \frac{d}{dx}(\sin(x)) = \frac{d}{dx}(1) \] Applying the product rule to \(xy\), we get: \[ y + x \frac{dy}{dx} + \cos(x) = 0 \] Rearranging gives the derivative \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = -\frac{y + \cos(x)}{x} \]

Step 2: Evaluate the Derivative at the Given Point

Substitute the point \((\frac{\pi}{2}, 0)\) into the derivative. Using the expression \(\frac{dy}{dx} = -\frac{y + \cos(x)}{x}\), plug in \(x = \frac{\pi}{2}\) and \(y = 0\): \[ \frac{dy}{dx} = -\frac{0 + \cos(\frac{\pi}{2})}{\frac{\pi}{2}} = -\frac{0 + 0}{\frac{\pi}{2}} = 0 \] Thus, the slope of the tangent line at the point \((\frac{\pi}{2}, 0)\) is 0.

Step 3: Write the Equation of the Tangent Line

The equation of a line with slope \(m\) through a point \((x_1, y_1)\) is \(y - y_1 = m(x - x_1)\). We have \(m = 0\), \(x_1 = \frac{\pi}{2}\), and \(y_1 = 0\). Substituting these into the line equation, we get: \[ y - 0 = 0 \times (x - \frac{\pi}{2}) \] Simplifying gives: \[ y = 0 \] This is the equation of the tangent line.

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used to find derivatives of equations not explicitly solved for one variable in terms of another. In the exercise, the function is given as \(xy + \sin(x) = 1\). This equation is not solved for \(y\) in terms of \(x\), making implicit differentiation the right approach.
To perform implicit differentiation, differentiate both sides of the equation concerning \(x\). Remember that when differentiating terms involving \(y\), treat \(y\) as a function of \(x\) and apply the chain rule. For \(xy\), we apply both the product and the chain rules to find its derivative.

• Differentiating \(xy\) gives us \(y + x \frac{dy}{dx}\)
• The derivative of \(\sin(x)\) is \(\cos(x)\)
• Equating to the derivative of 1, which is 0

This results in the equation \(y + x \frac{dy}{dx} + \cos(x) = 0\), showing how implicit differentiation can efficiently handle complex expressions.

Product Rule

The product rule is an essential tool in calculus for finding the derivative of products of two functions. The rule states that the derivative of two functions \(u\) and \(v\) multiplied together is given by: \((uv)' = u'v + uv'\).
In the original exercise, we see the product rule in action with the term \(xy\). Here:

• \(u = x\) and \(v = y\)
• The derivative of \(u\) is \(1\) and of \(v\) is \(\frac{dy}{dx}\)

Applying the product rule gives us \(y + x \frac{dy}{dx}\).
This is combined with the rest of the differentiated terms to set up the implicit differentiation process properly. The product rule is instrumental in accounting for how both variables \(x\) and \(y\) influence the other in the context of the equation.

Derivative Evaluation

Once we have the implicit derivative, the next step is to evaluate it at the specific point given in the problem. This is known as derivative evaluation. In this particular exercise, we substitute the point \((\frac{\pi}{2}, 0)\) into the derived expression \(\frac{dy}{dx} = -\frac{y + \cos(x)}{x}\).
Perform this substitution step-by-step:

• First, replace \(x\) with \(\frac{\pi}{2}\)
• Next, replace \(y\) with \(0\)
• Calculate \(\cos(\frac{\pi}{2})\), which is 0

This simplifies our derivative to \(0\). The result \(0\) means the slope of the tangent line at the point \((\frac{\pi}{2}, 0)\) is zero, indicating a horizontal tangent line.

Equation of a Line

After finding the slope of the tangent line, the next step is writing the line's equation through the given point with the determined slope. This is where the equation of a line formula comes into play. The formula for a line with slope \(m\) passing through a point \((x_1, y_1)\) is:
\[y - y_1 = m(x - x_1)\].
Replacing the slope \(m = 0\), \(x_1 = \frac{\pi}{2}\), and \(y_1 = 0\), we substitute these values into the line equation:

• \(y - 0 = 0 \times (x - \frac{\pi}{2})\)
• This simplifies to \(y = 0\)

Therefore, the equation of the tangent line is \(y = 0\), a horizontal line. Understanding how to correctly use this formula allows you to find tangent lines for various points and slopes efficiently.