Question

Find the equation of the line tangent to \(y=e^{2 x}\) at \(x=\frac{1}{2} \ln 3 .\) Graph the function and the tangent line.

Short answer

Based on the given step-by-step solution, determine the equation of the tangent line to the curve \(y=e^{2x}\) at \(x=\frac{1}{2}\ln 3\). The equation of the tangent line is \(y = 6x - 3\ln 3 + 3\).

Step-by-step solution

1. Find the Derivative

Find the derivative of \(y=e^{2x}\). Using the chain rule, we have: \[ y'(x) = \dfrac{dy}{dx} = \dfrac{d}{dx} \left(e^{2x}\right) = e^{2x} \cdot 2 = 2e^{2x} \]

2. Evaluate the Derivative at Given Point

Evaluate the derivative of the function at \(x=\frac{1}{2}\ln 3\) to find the slope of the tangent line: \[ y'\left(\frac{1}{2}\ln 3\right) = 2e^{2\cdot\left(\frac{1}{2}\ln 3\right)} = 2e^{\ln 3} = 2\cdot3 = 6 \] So, the slope of the tangent line is 6.

3. Find the Corresponding Point on the Curve

Now, we need to find the point on the curve where \(x = \frac{1}{2}\ln 3\): \[ y\left(\frac{1}{2}\ln 3\right) = e^{2\cdot\left(\frac{1}{2}\ln 3\right)} = e^{\ln 3} = 3 \] So, the point at which the tangent line intersects the curve is \(\left(\frac{1}{2}\ln 3, 3\right)\).

4. Find the Equation of the Tangent Line

We found that the tangent line has a slope of 6 and passes through the point \(\left(\frac{1}{2}\ln 3, 3\right)\). We can use the point-slope form to find the equation of the tangent line: \[ y - y_1 = m (x - x_1) \] Where \((x_1, y_1) = \left(\frac{1}{2}\ln 3, 3\right)\) and \(m = 6\): \[ y - 3= 6 \left(x - \frac{1}{2}\ln 3\right) \quad\Rightarrow\quad y = 6x - 3\ln 3 + 3 \] This is the equation of the tangent line.

5. Graph the Function and the Tangent Line

Graph \(y=e^{2x}\) (in blue) and its tangent line \(y=6x - 3\ln 3 + 3\) (in red) at \(x=\frac{1}{2}\ln 3\). We can also plot the point \(\left(\frac{1}{2}\ln 3, 3\right)\) (in green) where the tangent line intersects the curve. Please refer to a graphing tool or software to plot these functions and visualize the result.

Key concepts

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. It's especially valuable when dealing with functions that are nested or appear in a sequence, like exponential functions in this exercise. To put it simply, if you have a function inside another function, the chain rule helps you find the derivative of the whole expression. The formula for the chain rule is:\[(f(g(x)))' = f'(g(x)) \cdot g'(x)\]Here's how it works:

• You first differentiate the outer function and leave the inner function intact.
• Then, multiply this result by the derivative of the inner function.

In our example, we differentiated the exponential function \(e^{2x}\) by first finding the derivative of the outer exponential function \(e^u\) concerning \(u\), which remains \(e^u\). The inner function \(u = 2x\) is then derived to get \(2\). So, by applying the chain rule, we could express the derivative as \(2e^{2x}\). This technique is crucial for tackling a wide array of differentiation problems, not just in this scenario but any time you're dealing with composite functions.

Tangent Line

A tangent line to a curve at a given point is a straight line that "just touches" the curve at that point. It has the same slope as the curve at the point of tangency, making it a powerful tool for understanding the behavior of functions at specific instances.To find a tangent line, you need two main ingredients:

• The slope of the tangent line at the point, which is the derivative's value at that point.
• The exact point where the line touches the curve.

In this problem with the function \(y = e^{2x}\), we located the point where the tangent line makes contact with the curve by inserting \(x = \frac{1}{2} \ln 3\) into the function, revealing the point \((\frac{1}{2} \ln 3, 3)\).Using the derivative calculated as \(6\) at this point, we applied the point-slope form to find our tangent line's equation: \(y = 6x - 3\ln 3 + 3\). This equation tells us how the line behaves in relation to \(x\) and matches the slope and position of the curve exactly where they meet.Understanding tangent lines helps you graphically represent how fast and steeply functions change, and they provide linear approximations for more complex curves.

Exponential Functions

Exponential functions are a key type of mathematical function where a constant base is raised to a variable exponent. They appear frequently across different fields, including sciences and finance, due to their unique growth properties.In this exercise, our focus is on the function \(y = e^{2x}\), which is an exponential function with base \(e\), where \(e\) is Euler's number (approximately 2.71828). Exponential functions have some distinguishing features:

• They grow rapidly, increasing not arithmetically (like a sum) but multiplicatively.
• They have a consistent rate of growth. In the case of base \(e\), this growth rate is the derivative of \(e^x\), a unique property of exponential functions.

When differentiating exponential functions, particularly those with \(e\) as the base, they retain their form. For instance, differentiating \(e^{2x}\) gives \(2e^{2x}\), a reflection of their consistent growth factor.Exponential functions form the foundation of our understanding of phenomena involving rapid and continuous change, making them vital to explore in learning calculus and analyzing real-world problems.