Problem 34 - Calculus for Biology and Medicine

Question

Find the equation of the normal line to the curve \(y=\sqrt{x}-1\) at the point \((4,1)\).

Step-by-Step Solution

Answer

The equation of the normal line is \(y = -4x + 17\).

Step 1: Differentiate the Curve

The equation of the normal line involves the derivative of the curve. The original curve is given by \( y = \sqrt{x} - 1 \). First, find the derivative (or slope) of the curve. The derivative of \( y \) with respect to \( x \) is \( y' = \frac{d}{dx}(x^{1/2}) = \frac{1}{2\sqrt{x}} \).

Step 2: Evaluate the Derivative at the Given Point

Substitute the x-coordinate of the given point \((4,1)\) into the derivative to find the slope of the curve at that point. Using \( y' = \frac{1}{2\sqrt{x}} \), evaluate at \( x = 4 \):\[ y' = \frac{1}{2\sqrt{4}} = \frac{1}{4} \]

Step 3: Determine the Slope of the Normal Line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line at \((4,1)\) is \(\frac{1}{4}\), the slope (m) of the normal line is:\[ m = -\frac{1}{\frac{1}{4}} = -4 \]

Step 4: Use the Point-Slope Form to Find the Equation

Use the point-slope form of a line, \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the given point \((4,1)\) and \(m\) is the slope of the normal line. Plug in the values:\[ y - 1 = -4(x - 4) \]

Step 5: Simplify the Equation of the Normal Line

Distribute the slope on the right side:\[ y - 1 = -4x + 16 \]Add 1 to both sides to obtain the equation in slope-intercept form:\[ y = -4x + 17 \]

Key Concepts

Derivative Calculation

To understand the equation of the normal line, we must first grasp the concept of calculating derivatives. The derivative of a function at a point provides the slope of the tangent line to the curve at that point. In this exercise, the function given is \( y = \sqrt{x} - 1 \).

Here’s a quick walkthrough of derivative calculation:

• The type of function \( \sqrt{x} \) translates to \( x^{1/2} \).
  
• By applying the power rule, which states that the derivative of \( x^n \) is \( nx^{n-1} \), we find the derivative of \( x^{1/2} \) to be \( \frac{1}{2}x^{-1/2} \).
  
• This simplifies to \( \frac{1}{2\sqrt{x}} \).

Therefore, the expression \( y' = \frac{1}{2\sqrt{x}} \) becomes essential. It represents how steep the curve is at any point \( x \) along its path.

Point-Slope Form

The point-slope form is a vital equation when you want to write the equation of a line. This formula is particularly useful when you know a specific point on the line and the slope. The equation is expressed as:

• \( y - y_1 = m(x - x_1) \)

This form allows us to easily substitute known values to find the complete equation of a line.

In this exercise, the point-slope form helps us relate the point \((4, 1)\) and the slope of the normal line \(-4\). By substituting \( x_1 = 4 \), \( y_1 = 1 \), and \( m = -4 \), the equation translates into:

• \( y - 1 = -4(x - 4) \)

From here, simple algebraic manipulation will give us the equation in a more familiar slope-intercept form.

Slope of a Tangent Line

The slope of a tangent line is crucial in understanding the contrast between tangent and normal lines. Each line serves different purposes: tangent lines show the curve’s immediate direction, while normal lines are perpendicular.

In this context, using the derivative calculated as \( y' = \frac{1}{4} \) at the point \( (4, 1) \) gives us the slope of the tangent line. Since this slope represents the incline of the curve at that instant:

• The tangent slope \( \frac{1}{4} \) means the curve is gradually increasing at that point.
  
• Finding its perpendicular counterpart (the normal slope) involves taking the negative reciprocal, resulting in a slope of \(-4\).

The negative reciprocal makes sure the normal line and tangent line are at 90 degrees to each other, maintaining their perpendicular nature.