Problem 48 - Brief Applied Calculus

Question

Find the equation for the tangent line to the curve \(y=f(x)\) at the given \(x\) -value. $$ f(x)=\sqrt{x^{2}+3} \text { at } x=1 $$

Step-by-Step Solution

Answer

The equation for the tangent line is \( y = \frac{1}{2}x + \frac{3}{2} \).

Step 1: Find the derivative of the function

To find the equation of the tangent line, we need the derivative of the function. The function is given as \( f(x) = \sqrt{x^2 + 3} \). Using the chain rule: \( f'(x) = \frac{1}{2}\sqrt{x^2 + 3}(2x) = \frac{x}{\sqrt{x^2 + 3}} \).

Step 2: Evaluate the derivative at the given point

Substitute \( x = 1 \) into the derivative to find the slope of the tangent line at this point. \( f'(1) = \frac{1}{\sqrt{1^2 + 3}} = \frac{1}{2} \). Thus, the slope of the tangent line is \( \frac{1}{2} \).

Step 3: Calculate the value of the function at the given x-value

The value of the function at \( x = 1 \) is \( f(1) = \sqrt{1^2 + 3} = \sqrt{4} = 2 \). Hence, the point on the curve is \((1, 2)\).

Step 4: Use the point-slope formula to find the tangent line

The equation of a line using the point-slope formula is \( y - y_1 = m(x - x_1) \). With the slope \( m = \frac{1}{2} \) and point \( (1, 2) \), the equation becomes: \( y - 2 = \frac{1}{2}(x - 1) \).

Step 5: Simplify the equation

Simplifying \( y - 2 = \frac{1}{2}(x - 1) \), we get \( y - 2 = \frac{1}{2}x - \frac{1}{2} \). Adding 2 to both sides, \( y = \frac{1}{2}x + \frac{3}{2} \).

Key Concepts

Tangent Line

A tangent line to a curve at a particular point is a straight line that touches the curve precisely at that point. Importantly, the tangent line has the same slope as the curve at the point of contact. This means that it closely approximates the curve near this point. If you zoom in infinitely, the curve will appear to be a straight line, which is why the tangent line gives the best linear approximation to the function at that point.
For instance, with the function given as \( f(x) = \sqrt{x^2 + 3} \) at \( x = 1 \), the task is to find this tangent line. Calculating the slope of the tangent line involves finding the derivative, indicating how the curve changes at that point. The tangent line then becomes a tool for simplifying complex curves by allowing us to work with simple straight lines, especially useful in calculus.

Derivative

The derivative of a function measures how the function's output changes as the input changes. It is a fundamental concept in calculus and helps us determine the slope of a tangent line at any given point on a curve.
Consider the function \( f(x) = \sqrt{x^2 + 3} \). To find the tangent line, we first need its derivative. Using the chain rule, the derivative \( f'(x) \) is calculated as \( \frac{x}{\sqrt{x^2 + 3}} \). This expression gives us a function describing the slope of the curve at any point. By evaluating the derivative at specific \( x \)-values, you can find how steep the tangent line is at those points. In this case, the derivative at \( x = 1 \) determines the slope of the tangent line that meets the curve \( y = f(x) \) at precisely \( x = 1 \).

Point-Slope Formula

The point-slope formula is a reliable method for finding the equation of a line when you know a point on the line and its slope. The formula is expressed as \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \( (x_1, y_1) \) is the known point on the line.
To apply the point-slope formula for the function \( f(x) = \sqrt{x^2 + 3} \) at the point \( x = 1 \):

• We previously determined the slope \( m \) as \( \frac{1}{2} \) at \( x = 1 \) from the derivative.
• The point on the curve is \((1, 2)\) since \( f(1) = 2 \).

Substituting these into the point-slope formula, we obtain \( y - 2 = \frac{1}{2}(x - 1) \). This equation represents the line tangent to the curve at the point \( x = 1 \).

Chain Rule

The chain rule is a fundamental technique in calculus for differentiating composite functions. When encountering a function made of multiple nested functions, the chain rule simplifies the process of finding the derivative.
In our example, the function \( f(x) = \sqrt{x^2 + 3} \) is composite, as it involves the square root function of \( x^2 + 3 \).

• The outer function is the square root, \( \sqrt{u} \), where \( u = x^2 + 3 \).
• The inner function is \( u = x^2 + 3 \).

The chain rule states that the derivative of a composite function \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \). So, applying this to our function:

• The derivative of the outer function, \( \sqrt{u} \), is \( \frac{1}{2\sqrt{u}} \).
• The derivative of the inner function, \( x^2 + 3 \), is \( 2x \).

Thus, the chain rule gives us the derivative \( f'(x) = \frac{1}{2\sqrt{x^2 + 3}} \cdot 2x = \frac{x}{\sqrt{x^2 + 3}} \), crucial for finding the tangent line's slope.