Problem 58 - Calculus and its applications

Question

Find an equation for the tangent line to the graph of \(y=\sqrt{x^{2}+3 x}\) at the point (1,2)

Step-by-Step Solution

Answer

The tangent line equation is \( y = \frac{5}{4}x + \frac{3}{4} \).

Step 1: Find the derivative

First, determine the derivative of the function to find the slope of the tangent line at the point (1,2). The function is given by \( y = \sqrt{x^2 + 3x} \). Let \( u = x^2 + 3x \), then \( y = \sqrt{u} \). By chain rule, the derivative \( \frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} \). Find \( \frac{du}{dx} = 2x + 3 \). Thus, substitute back to get \( \frac{dy}{dx} = \frac{1}{2\sqrt{x^2 + 3x}} \cdot (2x + 3) \).

Step 2: Evaluate the derivative at the given point

To find the slope of the tangent line at \( x = 1 \), substitute \( x = 1 \) into the derivative \( \frac{dy}{dx} = \frac{(2 \cdot 1 + 3)}{2\sqrt{1^2 + 3 \cdot 1}} \). Simplify \( \sqrt{1^2 + 3 \cdot 1} = \sqrt{4} = 2 \). Therefore, \( \frac{dy}{dx} = \frac{5}{4} \). This slope is \( \frac{5}{4} \).

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

Now that we have the slope \( m = \frac{5}{4} \) and the point (1,2), we can use the point-slope form of a line equation: \( y - y_1 = m(x - x_1) \). Substitute \( (x_1, y_1) = (1, 2) \) and \( m = \frac{5}{4} \): \( y - 2 = \frac{5}{4}(x - 1) \).

Step 4: Simplify the equation

Distribute the slope in the equation: \( y - 2 = \frac{5}{4}x - \frac{5}{4} \). Add 2 to both sides to solve for \( y \): \( y = \frac{5}{4}x - \frac{5}{4} + \frac{8}{4} \). Simplify to get \( y = \frac{5}{4}x + \frac{3}{4} \).

Key Concepts

Derivative Calculation

To find the equation of the tangent line, we first need to calculate the derivative of the given function. The function is defined as \( y = \sqrt{x^2 + 3x} \). This requires derivative calculation to determine the slope of the tangent line at a specific point, which in this case is \((1,2)\).

Calculating the derivative gives the slope of the function at any given point. For \( y = \sqrt{x^2 + 3x} \), we let \( u = x^2 + 3x \) so that \( y = \sqrt{u} \).
Using the chain rule, the derivative \( \frac{dy}{dx} \) is calculated as:

\[ \frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} \]

Here, \( \frac{du}{dx} = 2x + 3 \). This gives us:

\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x^2 + 3x}} \cdot (2x + 3) \]

With this formula, we can now find the slope of the tangent line, specifically at the point where \( x = 1 \).

Substituting \( x = 1 \) results in:

• Calculate \( \sqrt{x^2 + 3x} \) at \( x = 1 \), so \( \sqrt{1^2 + 3 \times 1} = \sqrt{4} = 2 \).
• Substitute into the formula: \( \frac{dy}{dx} = \frac{(2 \times 1 + 3)}{2 \times 2} = \frac{5}{4} \).

This results in the slope \( \frac{5}{4} \) at the point \( (1, 2) \).

Point-Slope Form

Once we found the slope of the tangent line, we use the point-slope form to derive the equation of the tangent line itself. The point-slope form is a standard way of expressing the equation of a straight line given a point on the line and its slope.

The general formula for a point-slope equation is given by:

\[ y - y_1 = m(x - x_1) \]

In this exercise, the slope \( m \) is \( \frac{5}{4} \), and the point \( (x_1, y_1) \) is \( (1, 2) \). We substitute these values into the point-slope form equation:

• \( y - 2 = \frac{5}{4}(x - 1) \)

To simplify further and express it in the slope-intercept form \( y = mx + b \), the equation can be expanded and rearranged.

• Distribute \( \frac{5}{4} \) within the parentheses to get \( \frac{5}{4}x - \frac{5}{4} \)
• Add 2 to both sides to solve for \( y \):

\[ y = \frac{5}{4}x - \frac{5}{4} + \frac{8}{4} \]

Simplifying gives \( y = \frac{5}{4}x + \frac{3}{4} \), making it clear and concise.

Chain Rule

The chain rule is a fundamental technique in calculus for differentiating composite functions. In cases where a function is nested inside another, such as \( y = \sqrt{x^2 + 3x} \), the chain rule is invaluable. It allows us to handle the complexity by breaking down the derivative into manageable parts.

Here's how we used the chain rule in this problem:

• First, redefine the complex part of the function into a simple variable: set \( u = x^2 + 3x \).
• This gives us the function \( y = \sqrt{u} \).
• According to the chain rule, the derivative \( \frac{dy}{dx} \) is the product of \( \frac{dy}{du} \) and \( \frac{du}{dx} \).
• Calculate \( \frac{dy}{du} = \frac{1}{2\sqrt{u}} \) because of the square root of \( u \).

Next, find \( \frac{du}{dx} \):

• By differentiating \( u = x^2 + 3x \), \( \frac{du}{dx} = 2x + 3 \).

The chain rule provides the part-by-part differentiation and multiplication:

\[ \frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot (2x + 3) \]

Bringing it all together, the chain rule simplifies working with composite functions.