Problem 24 - Calculus for Biology and Medicine

Question

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

The derivative at \(x=2\) is \(-\frac{1}{4}\), and the normal line is \(y=4x-7.5\).

Step 1: Define the Derivative Formally

The formal definition of the derivative of a function \(y = f(x)\) is given by \[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]In our case, \(f(x) = \frac{1}{x}\). Therefore, the derivative is \[f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}\]

Step 2: Simplify the Difference Quotient

To simplify the expression \(\frac{\frac{1}{x+h} - \frac{1}{x}}{h}\), find a common denominator:\[\frac{\frac{1}{x+h} - \frac{1}{x}}{h} = \frac{x - (x+h)}{h(x+h)x} = \frac{-h}{h(x+h)x}\]Cancel \(h\) in the numerator and the denominator:\[= \frac{-1}{(x+h)x}\]

Step 3: Evaluate the Limit

Evaluate the limit as \(h\) approaches 0:\[f'(x) = \lim_{h \to 0} \frac{-1}{(x+h)x} = \frac{-1}{x^2}\]Therefore, the derivative of \(y = \frac{1}{x}\) is\[f'(x) = -\frac{1}{x^2}\]

Step 4: Calculate the Derivative at \(x=2\)

Substitute \(x = 2\) into the derivative:\[f'(2) = -\frac{1}{2^2} = -\frac{1}{4}\]Therefore, the derivative at \(x=2\) is \(-\frac{1}{4}\).

Step 5: Confirm the Point is on the Graph

Verify that the point \(\left(2, \frac{1}{2}\right)\) is on the graph of \(y = \frac{1}{x}\) by substituting \(x=2\):\[y = \frac{1}{2} \quad \text{(which matches with the y-coordinate of the point)}\]Therefore, the point \(\left(2, \frac{1}{2}\right)\) is on the graph.

Step 6: Find the Normal Line Equation

The slope of the normal line is the negative reciprocal of the slope of the tangent line, which is \(-\frac{1}{4}\). Therefore, the slope of the normal line is \(4\).Using point-slope form for the point \(\left(2, \frac{1}{2}\right)\):\[y - \frac{1}{2} = 4(x - 2)\]Simplify to get the normal line equation:\[y = 4x - 7.5\]

Step 7: Graph the Function and Lines

Graph \(y = \frac{1}{x}\), marking the point \(\left(2, \frac{1}{2}\right)\). Draw the tangent line using the slope \(-\frac{1}{4}\) through the same point. Also, plot the normal line \(y = 4x - 7.5\) on the same coordinate system to visualize their relationships.

Key Concepts

Formal Definition of Derivative

In mathematics, the concept of a derivative is fundamental for understanding rates of change. The derivative can be seen as a tool that describes how a function behaves as its input values change. This is often visualized as the slope of the tangent line to the function at a given point.

The formal definition of a derivative is given by:

• \[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]

Here are the steps to find a derivative using this definition:

• Find \(f(x + h)\) by substituting \(x + h\) into the function.
• Calculate the difference quotient \(\frac{f(x+h) - f(x)}{h}\).
• Simplify the expression to make taking the limit as \(h\) approaches 0 easier.
• Evaluate the limit to find the derivative \(f'(x)\).

Using the formal definition of the derivative helps us understand exactly how we mathematically compute the slope of the tangent line to a function, which is key for deeper insights into its behavior.

Tangent Line

The tangent line is a straight line that touches a curve at exactly one point, representing the best linear approximation to the curve at that point. This line gives us the slope of the function at that specific location. For the function \(y = \frac{1}{x}\) at \(x = 2\), it played a crucial role in understanding how the function changes around that point.

To find the equation of the tangent line at a point:

• First, compute the derivative of the function at that point. For instance, if we have the derivative \(f'(2) = -\frac{1}{4}\), this value is the slope of the tangent line.
• Use the point-slope form: \[y - y_1 = m(x - x_1)\] where \(m\) is the slope (\(-\frac{1}{4}\)) and \((x_1, y_1)\) is the point \((2, \frac{1}{2})\).

Plugging the values, we get the equation of the tangent line as:

• \[y - \frac{1}{2} = -\frac{1}{4}(x - 2)\]

Simplifying further yields the exact linear equation representing the tangent line at that specific point.

Normal Line

The normal line to a curve at a given point is perpendicular to the tangent line at that same point. Understanding the normal line is important because it helps in analyzing how the function's graph is bending at a certain location.

If the slope of the tangent line is known, say \(-\frac{1}{4}\), the normal line's slope is the negative reciprocal of this slope. This is calculated as follows:

• Slope of normal line = \(-\frac{1}{\left(-\frac{1}{4}\right)} = 4\)

Using the point-slope form once again, the equation for the normal line at point \((2, \frac{1}{2})\) becomes:

• \[y - \frac{1}{2} = 4(x - 2)\]

Simplifying, we get:

• \[y = 4x - 7.5\]

This gives an equation for the normal line, showcasing its perpendicular relationship with the tangent line and highlighting a key feature of the curve at the point \((2, \frac{1}{2})\).