Problem 62 - Calculus for Biology and Medicine

Question

Find the normal line to $$ f(x)=\frac{x^{3}}{a+1} $$ at \(x=2 a\). Assume that \(a\) is a positive constant.

Step-by-Step Solution

Answer

The normal line at \(x = 2a\) is \( y - \frac{8a^3}{a+1} = -\frac{a+1}{12a^2}(x - 2a) \).

Step 1: Find the derivative of the function

To start, find the derivative of the function, which gives the slope of the tangent line. The function is \( f(x) = \frac{x^3}{a+1} \). To differentiate this, use the power rule. The derivative \( f'(x) \) is given by \( f'(x) = \frac{3x^2}{a+1} \).

Step 2: Evaluate the derivative at x = 2a

Next, substitute \( x = 2a \) into the derivative \( f'(x) = \frac{3x^2}{a+1} \) to find the slope of the tangent line at this point. This gives \( f'(2a) = \frac{3(2a)^2}{a+1} = \frac{12a^2}{a+1} \).

Step 3: Find the equation of the tangent line

Use the point-slope form \(y - y_1 = m(x - x_1)\), where \(m\) is the slope of the tangent line and \((x_1, y_1)\) is the point where the tangent meets the curve. Here, \(x_1 = 2a\) and \(y_1 = f(2a) = \frac{(2a)^3}{a+1} = \frac{8a^3}{a+1}\). The tangent line is \( y - \frac{8a^3}{a+1} = \frac{12a^2}{a+1}(x - 2a) \).

Step 4: Determine the slope of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope \(m_n\) of the normal line is \( m_n = -\frac{a+1}{12a^2} \).

Step 5: Find the equation of the normal line

Using the point-slope form again, the equation of the normal line at \((2a, \frac{8a^3}{a+1})\) with slope \(-\frac{a+1}{12a^2}\) is \( y - \frac{8a^3}{a+1} = -\frac{a+1}{12a^2}(x - 2a) \). Simplifying this equation can provide a more readable format.

Key Concepts

Derivative Calculation

The first step in finding the normal line is calculating the derivative of the given function. Derivatives are key in calculus as they provide the slope of the tangent line at any given point on a curve. For the function

• \( f(x) = \frac{x^3}{a+1} \)

we differentiate it to find \( f'(x) \), which expresses the slope at any \( x \).
Applying the differentiation rules, specifically the power rule, the derivative is:

• \( f'(x) = \frac{3x^2}{a+1} \)

This expression provides the slope of the tangent line at any given point \( x \), setting the stage for finding specific tangent and normal lines.

Tangent Line Equation

The tangent line touches the curve at exactly one point, sharing the slope and direction of the curve at that location. Once the derivative is calculated, you can find the specific slope of the tangent line at a given point by substituting the \( x \)-value into the derivative.
In this case, substituting \( x = 2a \) into our derived \( f'(x) \) gives us:

• \( f'(2a) = \frac{12a^2}{a+1} \)

This represents the slope \( m \) of the tangent line.
To write the tangent line equation, we use the point-slope form:

• \( y - y_1 = m(x - x_1) \)

where \( x_1 = 2a \) and \( y_1 = f(2a) \), which is \( \frac{8a^3}{a+1} \).Putting these together, the equation is:

• \( y - \frac{8a^3}{a+1} = \frac{12a^2}{a+1}(x - 2a) \)

Power Rule Differentiation

The power rule in differentiation is a quick and efficient method to find the derivative of functions raised to a power. If you have a function in the form \( x^n \), the derivative is calculated by multiplying the power \( n \) by the coefficient of \( x \) and then reducing the exponent by one.
For example, for \( x^3 \), the power is 3.

• So the derivative is \( 3x^{2} \).

However, in

• \( f(x) = \frac{x^3}{a+1} \)

we treat \( \frac{1}{a+1} \) as a constant.So applying the power rule, we derive:

• \( f'(x) = \frac{3x^2}{a+1} \).

Understanding the power rule helps simplify the derivative process for polynomial functions, which frequently appear in calculus problems.

Slope of Tangent Line

Once the derivative is known, it gives the slope of the tangent line at any point \( x \) along the curve. The slope indicates how steep the tangent line is and whether it rises or falls.
In our function:

• \( f'(x) = \frac{3x^2}{a+1} \)

Substituting \( x = 2a \) lets us find the slope of the tangent specifically at this point:

• \( f'(2a) = \frac{12a^2}{a+1} \).

Understanding this slope helps us in forming both the tangent line and the normal line.
The steeper the slope, the more dramatically it angles away from the horizontal. Knowing the exact slope is crucial because it directly affects how we determine the normal line.