Problem 61 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The normal line is \(y - \frac{4a}{a+1} = -\frac{a+1}{4a}(x - 2)\).

Step 1: Find the Derivative of the Function

First, find the derivative of the function to get the slope of the tangent line. The function is given as \[f(x) = \frac{a x^2}{a+1}.\]\[\text{Differentiating using the rule } \frac{d}{dx}(x^n) = nx^{n-1}, \text{ we get:}\]\[f'(x) = \frac{d}{dx}\left(\frac{a}{a+1} x^2\right) = \frac{2ax}{a+1}.\]

Step 2: Evaluate the Derivative at the Given Point

Next, substitute \(x = 2\) into the derivative to find the slope of the tangent line at this point.\[f'(2) = \frac{2a(2)}{a+1} = \frac{4a}{a+1}.\]

Step 3: Calculate the Perpendicular Slope

The slope of the normal line is the negative reciprocal of the tangent slope.\[m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{\frac{4a}{a+1}} = -\frac{a+1}{4a}.\]

Step 4: Find the Point on the Curve

Calculate the y-coordinate of the point where the tangent contacts the curve, i.e., at \(x = 2\).\[f(2) = \frac{a(2)^2}{a+1} = \frac{4a}{a+1}.\]The point is \((2, \frac{4a}{a+1})\).

Step 5: Write the Equation of the Normal Line

Now use the point-slope form of a line's equation, \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = (2, \frac{4a}{a+1})\) and \(m = -\frac{a+1}{4a}\).Substitute in:\[y - \frac{4a}{a+1} = -\frac{a+1}{4a}(x - 2).\]

Key Concepts

Derivative of a Function

In calculus, the derivative of a function measures how the function's value changes as its input changes. It's essentially the function's rate of change or slope. For the function given, \( f(x) = \frac{a x^2}{a+1} \), finding the derivative involves applying the power rule. The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \). Here, the function is a quadratic equation where \( a \) is a constant multiplier. To differentiate, we treat \( \frac{a}{a+1} \) as a constant coefficient, giving us the derivative:

• \( f'(x) = \frac{2ax}{a+1} \)

This resulting derivative tells us the slope of the tangent line at any point \( x \). Evaluating this at a specific \( x \) value, such as \( x = 2 \), provides the slope at this particular point on the curve.

Tangent and Normal Lines

Tangent and normal lines are key concepts in understanding the geometry of curves. The tangent line touches the curve at exactly one point and has the same slope as the curve at that point. For example, the tangent line to our function \( f(x) \) at \( x = 2 \) has a slope obtained from the derivative:

• The slope of the tangent line, \( m_{\text{tangent}} = \frac{4a}{a+1} \).

Normal lines, on the other hand, are perpendicular to the tangent lines at the point of contact. This means the slope of the normal line is the negative reciprocal of the tangent line's slope. Thus, for our example, the normal line slope is:

• \( m_{\text{normal}} = -\frac{a+1}{4a} \)

Recognizing these relationships helps in visualizing how curves behave locally at different points.

Point-Slope Form of a Line

The point-slope form is a common way to express the equation of a line using a known point and the line's slope. The formula is:

• \( y - y_1 = m(x - x_1) \)

Where \((x_1, y_1)\) is a point on the line, and \(m\) is the slope of the line. This form is particularly useful in calculus when working with tangent and normal lines. For the provided function, after finding the point of contact on the curve, which is \((2, \frac{4a}{a+1})\), and knowing the slope of the normal line as \(-\frac{a+1}{4a} \), the point-slope form allows us to express the equation of the normal line as follows:

• \( y - \frac{4a}{a+1} = -\frac{a+1}{4a}(x - 2) \)

This representation simplifies to a line equation passing through the calculated point with the specified normal slope, making it straightforward to derive the line's specific properties.