Chapter 3: Problem 81
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
Expert verified
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
01
1. Find the Derivative
Find the derivative of \(y=e^{2x}\). Using the chain rule, we have:
\[
y'(x) = \dfrac{dy}{dx} = \dfrac{d}{dx} \left(e^{2x}\right) = e^{2x} \cdot 2 = 2e^{2x}
\]
02
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.
03
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 \(\left(\frac{1}{2}\ln 3, 3\right)\).
04
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.
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.