Problem 24 - Calculus for Biology and Medicine

Question

Apply the product rule to find the normal line, in slope-intercept form, of \(y=f(x)\) at the specified point. \(f(x)=\frac{(2-x)(3-x)}{4}\), at \(x=-1\)

Step-by-Step Solution

Answer

The normal line is \( y = \frac{4}{7}x + \frac{25}{7} \).

Step 1: Find the Expression for the Derivative Using the Product Rule

Since the function is in the form of a product, let \( u(x) = 2-x \) and \( v(x) = 3-x \). Hence, \( f(x) = \frac{u(x) \cdot v(x)}{4} \). The product rule states that \( (uv)' = u'v + uv' \). First, compute the derivatives: \( u'(x) = -1 \) and \( v'(x) = -1 \). Therefore, \( (uv)' = (-1)(3-x) + (2-x)(-1) = -3+x -2 +x = 2x - 5 \). Hence, \( f'(x) = \frac{1}{4}(2x - 5) \).

Step 2: Compute the Slope of the Tangent Line at the Specified Point

Substitute \( x = -1 \) into \( f'(x) \): \( f'(-1) = \frac{1}{4}(2(-1) - 5) = \frac{1}{4}(-2 - 5) = -\frac{7}{4} \). This gives the slope of the tangent line at \( x = -1 \).

Step 3: Find the Y-Coordinate of the Function at the Specified Point

Substitute \( x = -1 \) into \( f(x) = \frac{(2-x)(3-x)}{4} \): \[ f(-1) = \frac{(2-(-1))(3-(-1))}{4} = \frac{(2+1)(3+1)}{4} = \frac{3 \times 4}{4} = 3 \]. Thus, the point on the function is \((-1, 3)\).

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. The slope of the tangent line is \(-\frac{7}{4}\), so the slope of the normal line is \( \frac{4}{7} \).

Step 5: Use Point-Slope Form to Find the Equation of the Normal Line

The point-slope form equation is \( y - y_1 = m(x - x_1) \). Using the point \((-1, 3)\) and the slope \( \frac{4}{7} \), substitute to find: \( y - 3 = \frac{4}{7}(x + 1) \).

Step 6: Convert to Slope-Intercept Form

Solve \( y - 3 = \frac{4}{7}(x + 1) \) for \( y \): Distribute \( \frac{4}{7} \): \( y - 3 = \frac{4}{7}x + \frac{4}{7} \). Add 3 to both sides to isolate \( y \):\( y = \frac{4}{7}x + \frac{4}{7} + 3 \). Convert 3 to a fraction with a denominator of 7: \( y = \frac{4}{7}x + \frac{4}{7} + \frac{21}{7} \). Combine terms: \( y = \frac{4}{7}x + \frac{25}{7} \).

Key Concepts

Product Rule

When dealing with functions that are presented as products, the product rule is an essential tool in calculus. Its purpose is to help find the derivative of a product of two functions. If you have a function in the form of \(u(x) \cdot v(x)\), the product rule is applied to calculate the derivative, denoted as \((uv)'\).

• The product rule formula is: \((uv)' = u'v + uv'\).
• "\(u'\)" and "\(v'\)" are the derivatives of \(u(x)\) and \(v(x)\) respectively.

For example, in this exercise, we identified \(u(x) = 2-x\) and \(v(x) = 3-x\). By computing the derivatives \(u'(x) = -1\) and \(v'(x) = -1\), and then substituting them into the product rule formula, we derived \((uv)' = 2x - 5\). This streamlining showcases why the product rule is indispensable when performing derivative calculations involving products of functions.
Understanding the product rule allows you to confidently address more complex calculus problems.

Normal Line

In calculus, once you have a function's derivative, you can determine the slopes of tangent and normal lines at particular points. A normal line to a curve at a given point is perpendicular to the tangent line at the same point. The significance of the normal line lies in its orthogonal relationship to the tangent.

• The slope of the normal line is the negative reciprocal of the slope of the tangent line.

Consider the given exercise. After computing the tangent line's slope at \(x = -1\), found to be \(-\frac{7}{4}\), we find the normal line's slope by taking the negative reciprocal: \(\frac{4}{7}\).
This conversion from tangent slope to normal slope simplifies finding equations for lines orthogonal to tangents on a curve, deepening understanding of the geometric relationships in calculus.

Derivative Calculation

Derivative calculation is fundamental in understanding the rate of change of a function. Through derivatives, you can predict how a small change in input alters the output of a function.

• In this context, derivatives help determine slopes of tangent lines, crucial for understanding a function's behavior at specific points.

For our problem, the function \(f(x)\) was a quotient, simplified into a product form for easier derivative calculation. By applying the product rule, we found \(f'(x) = \frac{1}{4}(2x - 5)\).
Calculating this derivative and evaluating it at \(x = -1\) provided the slope \(-\frac{7}{4}\) of the tangent line at this point, allowing further steps to derive the normal line equation.
Understanding how to manipulate derivatives empowers students to dissect functions and predict their behavior at any given point.