Problem 32 - Calculus for Biology and Medicine

Question

Find the equation of the normal line to the curve \(y=4 / x\) at the point \((-1,-4)\).

Step-by-Step Solution

Answer

The equation of the normal line is \(y = \frac{1}{4}x - \frac{15}{4}\).

Step 1: Find the Derivative

Start by finding the derivative of the function \(y = \frac{4}{x}\) to determine the slope of the tangent line at the point \((-1, -4)\). Using the power rule, \(\frac{d}{dx}\left(\frac{4}{x}\right) = \frac{d}{dx}(4x^{-1}) = -4x^{-2}\). So the derivative is \(y' = -\frac{4}{x^2}\).

Step 2: Calculate Slope at Given Point

Substitute \(x = -1\) into the derivative to find the slope of the tangent line at \((-1, -4)\). This gives \(y'(-1) = -\frac{4}{(-1)^2} = -4\). So the slope of the tangent line is \(-4\).

Step 3: Find Slope of the Normal Line

The normal line is perpendicular to the tangent line. The slope of the normal line is the negative reciprocal of the tangent slope. If the tangent slope is \(-4\), the normal slope is \(\frac{1}{4}\).

Step 4: Use Point-Slope Form to Write Equation

Use the point-slope form of a line, \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is \((-1, -4)\) and the slope \(m\) is \(\frac{1}{4}\). The equation becomes \(y - (-4) = \frac{1}{4}(x - (-1))\).

Step 5: Simplify the Equation

Simplify the equation from Step 4: \(y + 4 = \frac{1}{4}(x + 1)\). Distribute the \(\frac{1}{4}\) on the right side to get \(y + 4 = \frac{1}{4}x + \frac{1}{4}\). Solve for \(y\) to get the final equation, \(y = \frac{1}{4}x - \frac{15}{4}\).

Key Concepts

Derivative Calculation

The concept of a derivative lies at the heart of calculus as it provides the rate at which a function is changing at any given point. For the function given by the equation \(y = \frac{4}{x}\), we want to determine the slope of the tangent line at a particular point by calculating its derivative.
The function can be expressed in power form as \(4x^{-1}\). To find the derivative, we use the power rule of differentiation, which states that \(\frac{d}{dx}(x^n) = nx^{n-1}\). Applying this rule, we find:

• \(\frac{d}{dx}(4x^{-1}) = -4x^{-2}\)

This simplifies to \(y' = -\frac{4}{x^2}\). The derivative, \(y'\), represents the slope of the tangent line at any point \(x\) on the curve.
This is the preliminary step in finding the equation of the normal line, as the slope of this tangent line gives us crucial information for further calculations.

Tangent Line

Once we have computed the derivative, the next step is to calculate the slope of the tangent line at the specific point of interest, here being \((-1, -4)\). By substituting \(x = -1\) into the derivative function \(y' = -\frac{4}{x^2}\), we can find the exact slope of the tangent at that point.

• Substitute \(x = -1\) into \(y'\): \(y'(-1) = -\frac{4}{(-1)^2} = -4\)

Therefore, the slope of the tangent line at the point \((-1, -4)\) is \(-4\). This tangent line is essentially the best straight-line approximation to the curve at \((-1, -4)\), directly representing how steep the curve is at this point.

Slope of a Line

Understanding the slope is critical in determining the direction and steepness of a line. The slope of a line is generally expressed as \(m = \frac{\Delta y}{\Delta x}\), representing the change in \(y\) for a given change in \(x\). In this exercise, the slope of the tangent line has already been calculated as \(-4\).
However, since we are dealing with the normal line, which is perpendicular to the tangent line, we need to find the slope of the normal line through a simple transformation:

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

For a tangent slope of \(-4\), the slope for the normal line turns out to be \(\frac{1}{4}\). This value tells us that the normal line rises 1 unit for every 4 units it moves horizontally to the right.

Point-Slope Form

The point-slope form is a convenient way to express the equation of a line when you know a point on the line and its slope. It is structured as \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope of the line.
For our problem, the normal line passes through the point \((-1, -4)\) with a slope of \(\frac{1}{4}\). Inserting these values into the point-slope form gives us:

• \(y + 4 = \frac{1}{4}(x + 1)\)

To express this in a more traditional \(y = mx + b\) format, distribute and simplify:
After simplification, we arrive at the equation \(y = \frac{1}{4}x - \frac{15}{4}\), which clearly represents the equation of the normal line to the curve at the given point.