Problem 28 - Calculus for Biology and Medicine

Question

Find the equation of the tangent line to the curve \(y=2 / x\) at the point \((2,1)\).

Step-by-Step Solution

Answer

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

Step 1: Determine the Derivative

To find the equation of the tangent line, we first need to determine the derivative of the function, since it gives us the slope of the tangent line at any point. The given function is \( y = \frac{2}{x} \). Let's rewrite it as \( y = 2x^{-1} \). The derivative, using the power rule, is \( y' = -2x^{-2} = -\frac{2}{x^2} \).

Step 2: Find the Slope of the Tangent Line at the Given Point

To find the slope at the point \((2, 1)\), substitute \( x = 2 \) into the derivative \( y' = -\frac{2}{x^2} \). So, \( y'(2) = -\frac{2}{2^2} = -\frac{2}{4} = -\frac{1}{2} \). Thus, the slope of the tangent line at this point is \(-\frac{1}{2}\).

Step 3: Use the Point-Slope Form to Find the Equation

The point-slope form of a line is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the point on the line. We have \( m = -\frac{1}{2} \) and the point \((2, 1)\). Substitute these into the formula: \( y - 1 = -\frac{1}{2}(x - 2) \).

Step 4: Simplify the Equation

Now we'll simplify the equation from Step 3. Distribute the slope on the right-hand side: \( y - 1 = -\frac{1}{2}x + 1 \). Add 1 to both sides to get: \( y = -\frac{1}{2}x + 2 \). This is the equation of the tangent line.

Key Concepts

Derivative

Derivatives play a crucial role in calculus as they measure how a function's value changes with respect to changes in its input variable. This concept is often visualized as the slope of a curve at a particular point. Consider the function given in the exercise: \( y = \frac{2}{x} \).
To find the slope of the tangent line at a specific point, we first need the function's derivative. We rewrite the function as \( y = 2x^{-1} \) to simplify differentiation using the power rule, which states that the derivative of \( x^n \) is \( nx^{n-1} \).

• Applying the power rule, the derivative becomes \( y' = -2x^{-2} \), which simplifies to \( y' = -\frac{2}{x^2} \).

Now, by plugging in the x-value from the point of interest, \(x = 2\), into this derivative, we find the instantaneous rate of change or the slope of the curve at that point. This slope is essential for constructing the tangent line equation.

Tangent Line

The tangent line to a curve at a given point touches the curve at exactly one point and has the same slope as the curve at that point. It is essentially the best linear approximation of the curve near that point. In this exercise, we are tasked with finding the tangent line to the curve represented by \( y = \frac{2}{x} \) at the point \((2, 1)\).
Once we have determined the derivative, we can find the slope of this tangent line. By evaluating the derivative at \( x = 2 \), we calculate:

• \( y'(2) = -\frac{1}{2} \).

This value represents the slope of our tangent line. A tangent line with slope \(-\frac{1}{2}\) means that for every unit increase in \( x \), the value of \( y \) decreases by \( \frac{1}{2} \) units. This understanding helps us form a precise equation of the tangent line.

Point-Slope Form

The point-slope form is an algebraic equation of a line that is particularly useful when you know a point on the line and the slope. The formula is given by \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \( (x_1, y_1) \) is the known point on the line.
In the given exercise, we've determined:

• The slope, \( m = -\frac{1}{2} \).
• The point on the tangent line, \((x_1, y_1) = (2, 1)\).

Substituting these values into the point-slope formula, we have:
\( y - 1 = -\frac{1}{2}(x - 2) \).
After simplifying, by distributing \(-\frac{1}{2}\) and solving for \( y \), we obtain the final equation:

• \( y = -\frac{1}{2}x + 2 \).

This equation represents the precise linear tangent to the original curve at the point \((2, 1)\), confirming our understanding of these foundational calculus principles.