Problem 21 - Calculus for Biology and Medicine

Question

Use the formal definition of the derivative to find the derivative of \(y=5 x^{2}\) at \(x=-1\). (b) Show that the point \((-1,5)\) is on the graph of \(y=5 x^{2}\), and find the equation of the tangent line at the point \((-1,5)\). (c) Graph \(y=5 x^{2}\) and the tangent line at the point \((-1,5)\) in the same coordinate system.

Step-by-Step Solution

Answer

The derivative at \(x = -1\) is \(-10\). At \((-1,5)\), the tangent line equation is \(y = -10x - 5\).

Step 1: Apply the Formal Definition of the Derivative

The formal definition of the derivative of a function \( f(x) \) at a point \( x = a \) is given by \( f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \). For the function \( y = 5x^2 \), substitute \( f(x) = 5x^2 \) and find \( f'(x) \) at \( x = -1 \).

Step 2: Calculate Using the Limit Definition

First, calculate \( f(-1+h) = 5(-1+h)^2 \). Expand the square to get \( f(-1+h) = 5(1 - 2h + h^2) = 5 - 10h + 5h^2 \). Now, calculate \( f(-1) = 5(-1)^2 = 5 \). Substitute these into the limit: \[ f'(-1) = \lim_{h \to 0} \frac{5 - 10h + 5h^2 - 5}{h} = \lim_{h \to 0} \frac{-10h + 5h^2}{h}. \]

Step 3: Simplify the Expression

Factor \( h \) out of the numerator: \( -10h + 5h^2 = h(-10 + 5h) \). Cancel \( h \) from the numerator and denominator: \[ f'(-1) = \lim_{h \to 0} (-10 + 5h) \]. As \( h \to 0 \), the expression simplifies to \( f'(-1) = -10 \). The derivative of \( y = 5x^2 \) at \( x = -1 \) is \( -10 \).

Step 4: Check Point on the Graph

To show that the point \((-1, 5)\) is on the graph, substitute \( x = -1 \) into the equation \( y = 5x^2 \). Calculate \( y = 5(-1)^2 = 5 \). This confirms that \((-1, 5)\) is on the graph.

Step 5: Find the Equation of the Tangent Line

The equation of the tangent line at a point \((x_0, y_0)\) with slope \( m \) is \( y - y_0 = m(x - x_0) \). With \( (x_0, y_0) = (-1, 5) \) and the slope \( m = -10 \), the equation becomes: \[ y - 5 = -10(x + 1) \]. Simplify this to \( y = -10x - 5 \).

Step 6: Final Step: Graph the Function and Tangent Line

To graph \( y = 5x^2 \), plot the parabola opening upwards with the vertex at the origin. Add the tangent line with the equation \( y = -10x - 5 \), which should touch the graph at the point \((-1,5)\), confirming the slope and point of tangency.

Key Concepts

Formal Definition of Derivative

The formal definition of the derivative is a fundamental concept in calculus that helps us understand how a function changes at any given point. Essentially, the derivative gives us the slope of the tangent line at a particular point on the graph of the function. The mathematical expression for the formal definition of the derivative of a function \( f(x) \) at a point \( x = a \) is given by the formula: \( f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \).

Here's a breakdown of this formula:

• \( h \): Approaches zero, representing an infinitely small change in \( x \).
• \( f(a+h) - f(a) \): Represents the change in the function's value over the interval \( [a, a+h] \).
• \( \frac{f(a+h) - f(a)}{h} \): This is the average rate of change, or the slope, over the small interval.
• \( \lim_{h \to 0} \): The limit process ensures we find the exact slope as the interval becomes infinitesimally small.

By using the above formula, you can determine the instantaneous rate of change of the function at any point.

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing it. This line represents the instantaneous rate of change of the curve at that specific point. In other words, it shows the slope of the curve at that particular location.

For any function \( y = f(x) \), the equation of the tangent line at a point \((x_0, y_0)\) is given by:

• Slope \( m \): The slope of the tangent line, which is also the derivative of the function at the point, expressed as \( f'(x_0) \).
• Point-slope form: The equation is written as \( y - y_0 = m(x - x_0) \).

Using the example from the exercise, for the function \( y = 5x^2 \), the slope of the tangent line at \( x = -1 \) is \( -10 \). This leads to the equation \( y - 5 = -10(x + 1) \), which simplifies to: \( y = -10x - 5 \).

This tangent line provides a linear approximation to the function at the touchpoint and is essential in understanding how the curve behaves near the specified point.

Limit Definition of Derivative

The limit definition of derivative is similar to the formal definition and is crucial when calculating derivatives of functions. It helps illustrate how the rate of change of a function comes about by examining what happens as we get closer and closer to the point of interest.

Consider our problem with the function \( y=5x^2 \). Here's how the limit definition of the derivative is applied:

• Calculate \( f(-1+h) \): Begin by substituting \( x = -1+h \) into the function to find its value near \( -1 \).
• Expand and simplify: For \( f(-1+h) = 5(-1+h)^2 \), this expands to \( 5 - 10h + 5h^2 \).
• Substitute in the limit formula: \( f'(-1) = \lim_{h \to 0} \frac{f(-1+h) - f(-1)}{h} = \lim_{h \to 0} \frac{5 - 10h + 5h^2 - 5}{h} \).
• Simplify the expression: You typically factor out an \( h \) from the numerator and then simplify.

Finally, as \( h \) approaches 0, the expression simplifies to the derivative \( -10 \), indicating the slope of the tangent line at \( x = -1 \). This direct approach demystifies how rates of change are determined using limits.