Problem 23 - 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. $$ f(x)=5(1-2 x)(x+1)-3, \text { at } x=0 $$

Step-by-Step Solution

Answer

The equation of the normal line is \(y = \frac{1}{5}x + 2\).

Step 1: Differentiate the function using the Product Rule

Given the function,\[ f(x) = 5(1-2x)(x+1) - 3, \]we apply the product rule to \((1-2x)(x+1)\), treating \(u = 1-2x\) and \(v = x+1\). The product rule states \((uv)' = u'v + uv'\), so we first find the derivatives:\[ u' = -2, \quad v' = 1. \]Thus, the derivative \(f'(x)\) is:\[ f'(x) = 5((-2)(x+1) + (1-2x)(1)). \]

Step 2: Simplify the derivative

Now we simplify the expression from Step 1:\[ f'(x) = 5[-2(x+1) + (1-2x)]. \]Expanding the terms gives:\[ = 5[-2x - 2 + 1 - 2x] = 5[-4x - 1]. \]Therefore, the derivative simplifies to:\[ f'(x) = -20x - 5. \]

Step 3: Evaluate the derivative at the given point

We need the slope of the tangent to find the normal. Evaluate the derivative at \(x = 0\):\[ f'(0) = -20(0) - 5 = -5. \]

Step 4: Find the slope of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent:\[ m_{\text{normal}} = -\frac{1}{-5} = \frac{1}{5}. \]

Step 5: Evaluate the function at the given point

Find \(f(0)\) to get the y-coordinate at \(x=0\):\[ f(0) = 5(1-2\cdot0)(0+1) - 3 = 5(1)(1) - 3 = 2. \]So the point is \((0, 2)\).

Step 6: Write the equation of the normal line in slope-intercept form

Using the point-slope form \(y - y_1 = m(x - x_1)\), where \(m = \frac{1}{5}\) and point \((0, 2)\):\[ y - 2 = \frac{1}{5}(x - 0). \]Simplifying to slope-intercept form gives:\[ y = \frac{1}{5}x + 2. \]

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus used to find the derivative of the product of two functions. In our problem, we have the function components as \(u = 1-2x\) and \(v = x+1\). The product rule states that the derivative of a product \(uv\) is given by:

• \((uv)' = u'v + uv'\)

We first need to compute the derivatives of \(u\) and \(v\):

• \(u' = -2\)
• \(v' = 1\)

Applying the product rule gives us the derivative of our expression:

• \(f'(x) = 5((-2)(x+1) + (1-2x)(1))\)

Simplifying these terms helps in finding the desired critical points, making it easier to work with the function.

Derivative

In calculus, the derivative represents the rate of change of a function with respect to a variable. A derivative provides the slope of the tangent line to the curve of a function at any given point. In our exercise, differentiating the function

• \(f(x) = 5(1-2x)(x+1) - 3\)

requires using the product rule, which was done in the previous step.
After applying the product rule and simplifying, we obtained the derivative:

• \(f'(x) = -20x - 5\)

This function represents the slope of the tangent at any point \(x\). Evaluating \(f'(0)\) gives us the slope at \(x=0\), which determines how steep or flat the tangent at that specific point is.

Tangent Line

The tangent line is a straight line that touches a curve at a single point without crossing it, providing an instantaneous rate of change at that point. In practical terms, it approximates the curve near the point of tangency.
To find the equation of the tangent line, we need its slope and a point it passes through. From our derivative:

• \(f'(0) = -5\)

This \(-5\) tells us the slope of the tangent line at the point where \(x=0\).
Next, we find the specific point by evaluating the function at \(x=0\), \(f(0) = 2\), giving the point \((0,2)\).

• The tangent line equation in the slope-intercept form is: \(y = -5x + 2\), using point-slope form for calculation.

Normal Line

The normal line is perpendicular to the tangent line at the point of tangency on a curve. The slope of the normal line is the negative reciprocal of the tangent line's slope.
Since we found the slope of the tangent line to be \(-5\), the slope \(m_{\text{normal}}\) of the normal line is:

• \(m_{\text{normal}} = \frac{1}{5}\)

The normal line also passes through the same point as the tangent line, which we found to be \((0, 2)\).
Using the point-slope formula, we derive the normal line equation:

• \(y - 2 = \frac{1}{5}(x - 0)\)
• Thus, in slope-intercept form, it becomes: \(y = \frac{1}{5}x + 2\)

With these calculations, we've successfully identified the normal line that maintains a 90-degree angle to the tangent at the point of interest.