Problem 24 - Calculus for Biology and Medicine

Question

(a) Use the formal definition to find the derivative of \(y=\frac{1}{x}\) at \(x=2\). (b) Show that the point \(\left(2, \frac{1}{2}\right)\) is on the graph of \(y=\frac{1}{x}\), and find the equation of the normal line at the point \(\left(2, \frac{1}{2}\right)\). (c) Graph \(y=\frac{1}{x}\) and the tangent line at the point \(\left(2, \frac{1}{2}\right)\) in the same coordinate system.

Step-by-Step Solution

Answer

Derivative at \(x=2\) is \(-\frac{1}{4}\), normal line: \(y=4x-7.5\). Point \((2, \frac{1}{2})\) is on curve.

Step 1: Apply the definition of derivative

To find the derivative of the function \( y = \frac{1}{x} \) at \( x = 2 \), we use the limit definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]First, compute \( f(2+h) \) and \( f(2) \):For \( y = \frac{1}{x} \), \[ f(2+h) = \frac{1}{2+h} \quad \text{and} \quad f(2) = \frac{1}{2} \]Substitute these values into the limit definition:\[ f'(2) = \lim_{h \to 0} \frac{\frac{1}{2+h} - \frac{1}{2}}{h} \]

Step 2: Simplify the limit expression

Simplify the expression by finding a common denominator:\[ \frac{1}{2+h} - \frac{1}{2} = \frac{2 - (2 + h)}{(2+h)(2)} = \frac{-h}{2(2+h)} \]This makes the overall expression:\[ f'(2) = \lim_{h \to 0} \frac{-h}{h(2)(2+h)} = \lim_{h \to 0} \frac{-1}{2(2+h)} \]Now take the limit as \( h \to 0 \):\[ f'(2) = \frac{-1}{4} \]

Step 3: Verify the point on the curve

Substitute \( x = 2 \) into the function \( y = \frac{1}{x} \) to verify the point:\[ y = \frac{1}{2} \]Thus, the point \((2, \frac{1}{2})\) is on the graph of \( y = \frac{1}{x} \).

Step 4: Find the equation of the normal line

The derivative at \( x = 2 \) gives the slope of the tangent line, \( m_t = \frac{-1}{4} \). The slope of the normal line is the negative reciprocal of the tangent's slope:\[ m_n = -\frac{1}{m_t} = 4 \]Using the point-slope form of a line, \( y - y_1 = m_n(x - x_1) \), and the point \((2, \frac{1}{2})\), the equation of the normal line is:\[ y - \frac{1}{2} = 4(x - 2) \]Simplify to find:\[ y = 4x - 7.5 \]

Step 5: Prepare to graph the function and the tangent line

Plot the function \( y = \frac{1}{x} \) and calculate the equation of the tangent line at \( x = 2 \): The tangent line's equation at the point \( (2, \frac{1}{2}) \) using the slope \( \frac{-1}{4} \) derived in Step 2 is:\[ y - \frac{1}{2} = -\frac{1}{4}(x - 2) \]Simplify:\[ y = -\frac{1}{4}x + 1 \]This line and the function should be graphed in the same coordinate system along with the normal line found in Step 4.

Key Concepts

Derivative

A derivative in calculus represents the rate at which a function is changing at any given point. It can be thought of as the slope of the tangent line to the graph of the function at a specific point. To find this slope for a function like \( y = \frac{1}{x} \), we use the limit definition of the derivative. This involves calculating \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \). By computing this limit as \( h \) approaches zero, we can determine the instantaneous rate of change of the function at \( x = 2 \). In this exercise, the derivative \( f'(2) \) was found to be \(-\frac{1}{4}\), indicating the slope of the tangent line at \( x = 2 \) on the graph of \( y = \frac{1}{x} \). Calculating derivatives this way is crucial for understanding how functions behave and change.

Tangent Line

The tangent line to a curve at a given point is a straight line that just "touches" the curve at that point without crossing it. The slope of this line is given by the derivative of the function at that specific point. For the function \( y = \frac{1}{x} \) at \( x = 2 \), the slope of the tangent line is \(-\frac{1}{4}\). This slope tells us how steep the line is and in which direction it's going.

Simple steps help us write the equation of the tangent line:

• Use the point \((2, \frac{1}{2})\), which lies on the curve.
• Apply the point-slope form: \( y - y_1 = m(x - x_1) \), where \( m \) is the slope.
• The equation becomes \( y - \frac{1}{2} = -\frac{1}{4}(x - 2) \), simplifying to \( y = -\frac{1}{4}x + 1 \).

Understanding tangent lines helps us approximate functions around particular points and is a foundation for more advanced calculus concepts.

Normal Line

A normal line to a curve at a given point is a line perpendicular to the tangent line at that point. Finding the equation for the normal line requires knowing the slope of the tangent line because the normal line's slope is the negative reciprocal of that tangent slope. For instance, the tangent slope determined was \(-\frac{1}{4}\), making the slope of the normal line \(4\).

Here's how you form the equation:

• Use the point \((2, \frac{1}{2})\) on the curve.
• Recall the negative reciprocal slope: \( m_n = 4 \).
• Incorporate this into the point-slope form \( y - y_1 = m_n(x - x_1) \).
• Derive the equation: \( y - \frac{1}{2} = 4(x - 2) \).
• After simplification, it becomes \( y = 4x - 7.5 \).

Understanding normal lines is particularly important in physics and engineering, where perpendicular directions often represent important physical concepts like forces.

Graphing Functions

Graphing functions is an essential skill for visualizing mathematical concepts. For the function \( y = \frac{1}{x} \), with peculiar behavior due to its asymptotes, graphing shows its general properties across different regions of the coordinate plane.

When graphing:

• Note that \( y = \frac{1}{x} \) has vertical asymptote at \( x = 0 \) and horizontal asymptote at \( y = 0 \).
• Plot important points such as \( (2, \frac{1}{2}) \) to use as guides.

Adding the tangent line from the calculated slope \(-\frac{1}{4}\) generates

• Equation \( y = -\frac{1}{4}x + 1 \).
• Shows the tangent only "touches" the curve at \( (2, \frac{1}{2}) \), confirming the derivative work.

Graphing these elements together helps see how functions behave locally (at specific points) and globally (throughout their domains), thus solidifying one's understanding of functional behavior.