Problem 77 - Calculus and its applications

Question

Find an equation of the line tangent to the graph of \(f(x)=e^{2 x}\) at the point (0,1)

Step-by-Step Solution

Answer

The equation of the tangent line is \(y = 2x + 1\).

Step 1: Understand Tangent Line Formula

The equation of a line tangent to a function at a given point is given by the point-slope form: \[ y - y_1 = m(x - x_1) \]where \((x_1, y_1)\) is the point at which the tangent touches the graph, and \(m\) is the slope of the tangent at that point.

Step 2: Determining the Point of Tangency

The problem states that the tangent line is at the point \((0,1)\). This means we have \((x_1, y_1) = (0,1)\), which will be used in the tangent line formula.

Step 3: Calculating the Slope of the Tangent

The slope of the tangent line is the value of the derivative of the function \(f(x)\) at the given point. First, find the derivative of \(f(x) = e^{2x}\), which is \[ f'(x) = \frac{d}{dx}[e^{2x}] = 2e^{2x}. \]Evaluate it at \(x = 0\):\[ f'(0) = 2e^{2 imes 0} = 2 imes 1 = 2. \]Thus, the slope \(m = 2\).

Step 4: Substitute Values into Tangent Line Equation

Using the point-slope form from Step 1 and the values from Steps 2 and 3, substitute \(x_1 = 0\), \(y_1 = 1\), and \(m = 2\):\[ y - 1 = 2(x - 0) \].

Step 5: Simplify the Equation

Simplify the equation derived in Step 4 to obtain the tangent line equation:\[ y - 1 = 2x \]\[ y = 2x + 1. \]This is the equation for the tangent line at the given point.

Key Concepts

Point-Slope Form

The point-slope form is a fundamental concept in understanding how to find the equation of a line, including tangent lines. It is represented as \( y - y_1 = m(x - x_1) \) where:

• \((x_1, y_1)\) are the coordinates of a specific point through which the line passes.
• \(m\) is the slope of the line.

This form is especially useful in calculus, particularly for tangents, because it allows us to quickly plug in the necessary values - the point and the slope - to find an equation of the line.
For instance, when we have a tangent line touching the curve, the coordinates \((x_1, y_1)\) are the point of tangency, and the slope \(m\) is the derivative of the function at that point. After substituting these values into the point-slope equation, we can easily convert to the slope-intercept form if necessary, providing a comprehensive view of the line's behavior.

Derivative

The concept of a derivative is pivotal in calculus as it describes how a function changes at any given point, known as the rate of change. For the problem at hand, calculating the derivative of the function \( f(x) = e^{2x} \) helps us determine the slope of the tangent line.
The derivative, expressed as \( f'(x) = \frac{d}{dx}[e^{2x}] = 2e^{2x} \), is derived using the chain rule, a method necessary for differentiating composite functions such as exponentials. This particular step computes how tipping the input of the function slightly affects its output.

• At \(x = 0\), the slope, or \( f'(0) \), is obtained by substituting \(0\) into the derivative, resulting in \(2\).

The derivative thus guides us to the necessary slope for the tangent line, which in turn becomes a crucial component in the point-slope form.

Exponential Function

Exponential functions, represented by expressions like \( e^{2x} \), exhibit growth or decay where the rate is proportional to its current value. Such functions are pivotal in modeling real-world phenomena like population growth or radioactive decay.
In the given function, \( e^{2x} \), the base \(e\) is a constant approximately equal to 2.71828, recognized for its appearance in natural logarithms and continuous growth processes.

• The exponent \(2x\) indicates a faster rate of growth compared to simpler exponents; each unit increase in \(x\) drastically sharpens the curve's rise.
• In calculus, exponential functions are known for the feature that their rates of change (derivatives) are proportional to their current value, highlighted by the derivative \( f'(x) = 2e^{2x} \).

Understanding exponential functions, alongside point-slope form and derivatives, equips students with the necessary tools to analyze and draw tangents to curves effectively.