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 \((5,2)\).

Step-by-Step Solution

Answer

The equation of the normal line is \( y = -4x + 22 \).

Step 1: Understand the Problem

We need to find the equation of the normal line to the curve given by \( y = \sqrt{x-1} \) at the specific point \( (5, 2) \). The normal line is perpendicular to the tangent line at that point.

Step 2: Find the Derivative

The first step in finding the equation of the tangent line is to compute the derivative \( y' \) of the curve. If \( y = \sqrt{x-1} \), then using the chain rule, the derivative is: \[ y' = \frac{d}{dx}(x-1)^{1/2} = \frac{1}{2}(x-1)^{-1/2}(1) = \frac{1}{2\sqrt{x-1}}. \]

Step 3: Evaluate the Derivative at the Given Point

Next, evaluate the derivative at \( x = 5 \). This will give us the slope of the tangent line at the point \( (5, 2) \). \[ y'(5) = \frac{1}{2\sqrt{5-1}} = \frac{1}{2\cdot2} = \frac{1}{4}. \]

Step 4: Determine the Slope of the Normal Line

Since the normal line is perpendicular to the tangent line, its slope is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is: \[ m_{\text{normal}} = -4. \]

Step 5: Use the Point-Slope Form

The equation of a line with slope \( m \) passing through a point \((x_1, y_1)\) is given by the point-slope form \( y - y_1 = m(x - x_1) \). Here, \( m = -4 \), \( x_1 = 5 \), and \( y_1 = 2 \). Use this to form the equation of the normal line: \[ y - 2 = -4(x - 5). \]

Step 6: Simplify the Equation of the Normal Line

Distribute and simplify the equation from the previous step to express it in the slope-intercept form \( y = mx + b \):\[ \begin{align*}y - 2 & = -4x + 20, \y & = -4x + 22.\end{align*} \]

Key Concepts

Tangent Line

A tangent line to a curve at a given point is a straight line that just 'touches' the curve at that point without crossing it. It represents the instantaneous direction of the curve at that specific point, akin to how a cyclist's direction is tangent to the path they're currently riding on. In mathematical terms, the tangent line can be used to understand the rate at which the curve is changing at that exact spot.

Knowing how to find a tangent line is essential. To solve it, we need a point on the curve and, importantly, the slope of the tangent line at that point.

• The slope of the tangent line can be directly obtained from the derivative of the curve's equation evaluated at the given point.
• It helps us predict the behavior of the curve around that point, similar to predicting the trajectory of a bicycle without making massive shifts in its path.

Derivative

In calculus, the derivative tells us how a function changes at any given point, basically giving us the 'instantaneous rate of change'. It's like checking the speedometer in a car to know its exact speed at a specific moment. For the exercise, the derivative of the curve's function, namely\( y = \sqrt{x-1} \), is calculated to \( y' = \frac{1}{2\sqrt{x-1}} \), using chain rule.

When you compute the derivative at a particular point (like \((5,2)\) in the exercise), it provides the slope of the tangent line at that point. This slope value is crucial because:

• It helps to determine the linear approximation of the curve around that point.
• It provides foundational information essential for finding both tangent and normal lines.

Perpendicular Lines

Perpendicular lines in geometry are two lines that intersect each other at a 90-degree angle. In the context of curves and calculus problems, this concept is vital because any normal line to a curve at a given point is perpendicular to the tangent line at that same point.

What does this mean practically?

• To find the slope of the perpendicular line, which in this case is the normal line, we use the negative reciprocal of the slope of the tangent line.
• This property ensures that the tangent and normal lines meet forming perfect right angles, mirroring how axes in a coordinate plane are connected to each other.

Point-Slope Form

The point-slope form is a way to write the equation of a straight line when you know both a specific point on the line and the line's slope. It's represented as \( y - y_1 = m(x - x_1) \), where:

• \( m \) is the slope.
• \( (x_1, y_1) \) are the coordinates of the given point.

In the exercise, after finding the slope of the normal line to be \( -4 \), the point-slope form helps us easily draft the equation since our line passes through \((5,2)\). Using this form simplifies the math and allows easy conversion to other forms, such as the slope-intercept form. It acts as a stepping stone to understanding and solving more complex geometric problems.