Problem 80 - Calculus Single Variable

Question

Show that there is a value \(c\) such that the tangent lines to the graphs of \(y=e^{x}\) and \(y=\ln (x)\) at \(\left(c, e^{c}\right)\) and \((c, \ln (c))\) are parallel.

Step-by-Step Solution

Answer

The value \(c \approx 0.567143\) makes the tangent lines parallel.

Step 1: Understand Parallel Tangent Lines

Two tangent lines are parallel if their slopes are equal. For this problem, we need to find a common point, \(c\), where the derivative of \(y = e^{x}\) and the derivative of \(y = \ln(x)\) at \(x = c\) are equal.

Step 2: Find the Derivatives

Calculate the derivative of \(y = e^{x}\), which is \(y' = e^{x}\). Now calculate the derivative of \(y = \ln(x)\), which is \(y' = \frac{1}{x}\). We need to find where these derivatives are equal.

Step 3: Set the Slopes Equal

To find the value of \(c\), set the derivatives equal to each other: \(e^{c} = \frac{1}{c}\). This equation needs to be solved for \(c\).

Step 4: Solve for c

Rewrite the equation as \(e^{c}c = 1\). To solve for \(c\), consider the function \(f(c) = e^{c}c - 1\) and find its root. Through numerical methods or estimation, we find that \(c \approx 0.567143\).

Step 5: Verify Tangency

At \(c \approx 0.567143\), the slope of \(y = e^{x}\) is \(e^{0.567143} \approx 1.567143\) and the slope of \(y = \ln(x)\) is \(\frac{1}{0.567143} \approx 1.567143\). Thus, the tangent lines are parallel at this point.

Key Concepts

Derivative

The concept of a derivative is central to understanding tangent lines. To find the slope of a tangent line to a function at a particular point, you calculate the derivative of that function. A derivative essentially tells you how a function changes as its input changes.

• For most functions, the formula for the derivative will depend on the specific form of the function.
• The derivative of a function at a point gives the slope of the tangent line to the graph of that function at that point.
• In practical terms, it's like zooming in really close to the curve to look at just that little piece of the function and approximating it with a line.

In this exercise, calculating the derivative helps us find where the slopes of the tangent lines to two different functions are equal, which means the lines are parallel.

Exponential Function

An exponential function is one where a constant base is raised to a variable exponent. The most famous example is the function given by the formula \(f(x) = e^x\), where \(e\) is Euler's number, approximately 2.71828.

• The exponential function \(e^x\) is unique because it is its own derivative. This means the slope of the function at any point is simply \(e^x\).
• It continuously grows at a rate proportional to its current value, which is a key feature of exponential growth.

In the context of this exercise, the function \(e^x\) is used to find a value \(c\) so that the tangent lines at points \((c, e^c)\) have slopes that match those of another function.

Logarithmic Function

Logarithmic functions are the inverses of exponential functions. The natural logarithm function, \(\ln(x)\), is the inverse of the exponential function \(e^x\). This means if \(y = e^x\), then \(x = \ln(y)\).

• The derivative of \(\ln(x)\) is \(\frac{1}{x}\), which describes the slope of the tangent line to \(y = \ln(x)\) at any point \(x\).
• Logarithmic functions increase, but at a decreasing rate, unlike exponential functions which increase at an increasing rate.

In this task, finding a \(c\) where the derivatives are equal involves equating \(e^c\) and \(\frac{1}{c}\), and solving for \(c\) gives us the point where the tangent lines are parallel.

Parallel Lines

Parallel lines are lines in a plane that do not intersect; they have the same slope. Understanding the slopes of lines is crucial when determining if they are parallel.

• For two tangent lines to be parallel, their slopes must be the same at the points of tangency.
• In our exercise, we set the derivative of the exponential and logarithmic functions equal to each other to find a \(c\) that gives the same slope, hence parallel lines.

The exercise objective was to find such a \(c\) where the tangent lines are parallel, highlighting the geometric implication of functions' derivatives.