Problem 39 - Single Variable Calculus

Question

(a) find an equation of the tangent line to the graph of the function at the indicated point, and (b) use a graphing utility to plot the graph of the function and the tangent line on the same screen. \(y=\frac{e^{x}}{1+x} ; \quad\left(1, \frac{1}{2} e\right)\)

Step-by-Step Solution

Answer

The equation of the tangent line to the function \(y = \frac{e^x}{1 + x}\) at the point \((1, \frac{1}{2} e)\) is \(y = \frac{e}{4}(x - 1) + \frac{1}{2} e\). Use a graphing utility to plot the function and the tangent line on the same screen to verify the result.

Step 1: Find the derivative of the function

To find the slope of the tangent line at the given point, we need to calculate the derivative of the function: $$y = \frac{e^x}{1 + x}$$ Using the quotient rule, we can find the derivative:\( y'=\frac{v \cdot u'-u \cdot v'}{v^{2}} \) where \(u=e^{x}\) and \(v=1+x\). First, find the derivative of \(u\) and \(v\): \(u' = \frac{d(e^x)}{dx} = e^x\) \(v' = \frac{d(1+x)}{dx} = 1\) Now apply the quotient rule: \(y' = \frac{(1+x)(e^x)-(e^x)}{(1+x)^2}\)

Step 2: Evaluate the derivative at the given point

Now we need to evaluate the derivative at the given point to find the slope of the tangent line: \((1, \frac{1}{2} e)\Rightarrow x = 1\) \(y'(1) = \frac{(1+1)(e^1)-(e^1)}{(1+1)^2} = \frac{2e-e}{4} = \frac{e}{4}\) So the slope of the tangent line at \((1, \frac{1}{2} e)\) is \(\frac{e}{4}\).

Step 3: Use point-slope form to find the equation of the tangent line

Now we can use the point-slope form to find the equation of the tangent line: $$y-y_{1} = m(x-x_{1})$$ where \((x_{1}, y_{1})=(1, \frac{1}{2} e)\) and \(m=\frac{e}{4}\). \begin{align*} y - \frac{1}{2} e &= \frac{e}{4}(x - 1) \\ y &= \frac{e}{4}(x - 1) + \frac{1}{2} e \end{align*} So the equation of the tangent line is: $$y = \frac{e}{4}(x - 1) + \frac{1}{2} e$$

Step 4: Use a graphing utility to plot the function and the tangent line

Finally, use a graphing utility to plot the function: $$y = \frac{e^x}{1 + x}$$ and the tangent line: $$y = \frac{e}{4}(x - 1) + \frac{1}{2} e$$ on the same screen. Check that the tangent line touches the curve at the point \((1, \frac{1}{2} e)\).

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing it. This means it only brushes against the curve at that point, matching its slope exactly. If you imagine holding a ruler on that curve, it would lie flat against it just at that point, perfectly aligned.

To find the equation of a tangent line, you need:

• The point where the line touches the curve.
• The slope of the curve at that point, which is given by the derivative of the function.

In our exercise, the tangent line was computed using the point-slope form and the slope found from the derivative. Understanding how to compute the derivative and use it to find the tangent line is key in this concept.

Derivative

The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. It provides the slope of the tangent line to the function's graph at any given point. This is crucial for understanding rates of change and patterns in various functions.

In the exercise, the function we had was a fraction, which made it ideal for applying the quotient rule—a method used to find the derivative when dealing with quotients of two functions. The derivative lets us find the exact slope of the tangent line at the specified point. This value was vital in determining the tangent line equation in the problem.

Quotient Rule

The quotient rule is a guideline in calculus for finding the derivative of a function that is the ratio of two other functions. The rule is particularly useful because many real-world problems involve ratios, and understanding how these behave when their individual components change is essential.

The formula for applying the quotient rule is \( y' = \frac{v \cdot u' - u \cdot v'}{v^2} \), where \( u \) is the numerator function and \( v \) is the denominator function.
For our exercise, the equation \( y = \frac{e^x}{1+x} \) involved taking derivatives of both the numerator \( e^x \) and the denominator \( 1+x \) and then applying the quotient rule to get the derivative. This process allowed us to find the correct slope for the tangent line.

Graphing Utility

A graphing utility is a powerful tool used in mathematics to visualize functions and their behaviors. It helps confirm analytical solutions by showing how equations look as graphs. This visual representation allows students to better understand the relationship between functions and their derivatives.

Using a graphing utility involves several steps:

• Enter the equation of the function.
• Enter the equation of the tangent line.
• Adjust the viewing window if necessary to see both clearly.

By graphing the function and its tangent line, students can verify that our calculated tangent line closely matches the expected behavior of the curve at the given point \((1, \frac{1}{2} e)\). This hands-on approach makes abstract concepts more concrete and easier to understand.