Problem 18 - Precalculus: Mathematics for Calculus

Question

Find an equation of the tangent line to the curve at the given point. Graph the curve and the tangent line. $$y=\sqrt{1+2 x}, \quad \text { at }(4,3)$$

Step-by-Step Solution

Answer

Tangent line: \( y = \frac{1}{3}x + \frac{5}{3} \).

Step 1: Determine the derivative of the function

To find the equation of the tangent line, we first need the derivative of the given function, which gives us the slope. The function is given as \( y = \sqrt{1+2x} \). We differentiate this with respect to \( x \) using the chain rule. Let \( u = 1 + 2x \), then \( y = u^{1/2} \).Thus, \( \frac{dy}{dx} = \frac{1}{2} u^{-1/2} \cdot \frac{du}{dx} \). We know \( \frac{du}{dx} = 2 \), thus the derivative \( \frac{dy}{dx} = \frac{1}{2} \cdot (1+2x)^{-1/2} \cdot 2 = (1+2x)^{-1/2} \).

Step 2: Evaluate the derivative at the point

Now we evaluate the derivative at the given point \((4,3)\). Substitute \( x = 4 \) into the derivative we found: \[ \frac{dy}{dx} \bigg|_{x=4} = (1 + 2 \times 4)^{-1/2} = (1 + 8)^{-1/2} = 9^{-1/2} = \frac{1}{3} \]. So, the slope of the tangent line at \((4,3)\) is \( \frac{1}{3} \).

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

With the slope \( m = \frac{1}{3} \) and the point \((4, 3)\), we use the point-slope form of the line equation: \( y - y_1 = m(x - x_1) \). Substituting the values, we get: \[ y - 3 = \frac{1}{3}(x - 4) \].Simplify the equation: \[ y - 3 = \frac{1}{3}x - \frac{4}{3} \]. Add 3 to both sides to solve for \( y \), resulting in: \[ y = \frac{1}{3}x + \frac{5}{3} \].

Step 4: Graph the function and the tangent line

Plot the original curve \( y = \sqrt{1+2x} \) and plot the tangent line \( y = \frac{1}{3}x + \frac{5}{3} \) on the same graph. At \( x = 4 \), check that both the curve and the tangent line intersect the point \( (4, 3) \). This visual confirms both that the tangent line correctly touches the curve at the specified point and aligns with the calculated slope.

Key Concepts

Derivative Calculation

To find the equation of a tangent line, we must start by calculating the derivative of the given function. Derivatives represent the slope of the tangent line to a curve at any point. In this exercise, the function is given as \( y = \sqrt{1+2x} \).
To find the derivative, we apply the Chain Rule, which is a method used to differentiate composite functions. Let's break it down:

• Identify the inner function, which is \( u = 1 + 2x \).
• The outer function becomes \( y = u^{1/2} \).

Taking the derivative of the outer function with respect to \( u \), we get \( \frac{dy}{du} = \frac{1}{2} u^{-1/2} \).
Next, differentiate the inner function \( u = 1 + 2x \) with respect to \( x \) to get \( \frac{du}{dx} = 2 \).

• Finally, apply the chain rule: \( \frac{dy}{dx} = \frac{1}{2} \cdot u^{-1/2} \cdot \frac{du}{dx} = (1+2x)^{-1/2} \).

This derivative tells us the slope of the tangent line at any point \( x \) on the curve.

Point-Slope Form

The point-slope form is a valuable tool for discovering the equation of a tangent line when you know a point on the line and the slope. This format is simple and easy to use with the structure:

• \( y - y_1 = m(x - x_1) \)

Here, \( m \) represents the slope, and \((x_1, y_1)\) is the specific point the line passes through.
In our exercise, we've computed the slope from the derivative, which is \( \frac{1}{3} \) at the point \((4, 3)\).
Substitute these values into the point-slope equation: \( y - 3 = \frac{1}{3}(x - 4) \).
Next, solve for \( y \) to get the equation in the form \( y = mx + b \), resulting in: \( y = \frac{1}{3}x + \frac{5}{3} \).
Point-slope form is flexible: it easily adapts to incorporate specific data points for accurate line modeling.

Chain Rule

The Chain Rule is an essential tool in calculus for differentiating composite functions, which are functions made up of two or more other functions. This technique simplifies finding the derivative without delving into overly complex calculations.
Consider, for example, our function \( y = \sqrt{1+2x} \). This is a composite function where:

• The outside function is the square root \( y = u^{1/2} \)
• The inside function is \( u = 1 + 2x \)

To apply the Chain Rule:1. Differentiate the outer function: \( \frac{dy}{du} = \frac{1}{2} u^{-1/2} \).2. Differentiate the inner function: \( \frac{du}{dx} = 2 \).3. Multiply these derivatives to find \( \frac{dy}{dx} \):

• \( \frac{dy}{dx} = \frac{1}{2} u^{-1/2} \cdot 2 = (1 + 2x)^{-1/2} \)

This provides the compound derivative, giving us the slope of the curve at any point \( x \). The Chain Rule efficiently calculates these slopes, especially in situations involving nested functions.

Graphing Functions

Graphing functions is a core skill necessary for visualizing mathematical relationships and understanding how various equations behave. In this scenario, we will graph both the original function and its tangent line.
Start with the function \( y = \sqrt{1+2x} \):

• Plot various points to understand its curve—this function will have a gentle arching shape since it involves a square root.

Next, add the tangent line equation we found, \( y = \frac{1}{3}x + \frac{5}{3} \).
This line should just touch the curve at point \((4, 3)\), confirming our calculations.

• Graphing these together shows not just their intersection but reveals visually how the tangent line is the instantaneous direction the curve is heading at \( x=4 \).

Seeing both the curve and the tangent line on a graph supports the conceptual understanding of these components and their connections.