Problem 18 - University Calculus: Early Transcendentals

Question

Differentiate the functions. Then find an equation of the tangent line at the indicated point on the graph of the function. $$w=g(z)=1+\sqrt{4-z}, \quad(z, w)=(3,2)$$

Step-by-Step Solution

Answer

The tangent line equation is \( w = -\frac{1}{2}z + \frac{7}{2} \).

Step 1: Differentiate the function

Given the function \( w = g(z) = 1 + \sqrt{4-z} \). We first find the derivative with respect to \( z \). Use the chain rule for differentiation. Let \( u = 4-z \), then \( \sqrt{u} \) is our function. The derivative of \( \sqrt{u} \) is \( \frac{1}{2\sqrt{u}} \). Since \( u = 4-z \), the derivative of \( u \) is \( -1 \). Apply the chain rule: \[ \frac{dw}{dz} = \frac{d}{dz}(\sqrt{4-z}) = \frac{1}{2\sqrt{4-z}} \times (-1) = -\frac{1}{2\sqrt{4-z}} \]

Step 2: Evaluate the derivative at the point

Evaluate the derivative at the given point \( z = 3 \):\[ \frac{dw}{dz}\bigg|_{z=3} = -\frac{1}{2\sqrt{4-3}} = -\frac{1}{2\sqrt{1}} = -\frac{1}{2} \]

Step 3: Write the equation of the tangent line

The equation of the tangent line is given by the point-slope form \( w - w_1 = m(z - z_1) \), where \( m \) is the slope and \((z_1, w_1)\) is the point \((3, 2)\). Substitute \( m = -\frac{1}{2} \), \( z_1 = 3 \), and \( w_1 = 2 \):\[ w - 2 = -\frac{1}{2}(z - 3) \]Simplify:\[ w - 2 = -\frac{1}{2}z + \frac{3}{2} \]\[ w = -\frac{1}{2}z + \frac{3}{2} + 2 \]\[ w = -\frac{1}{2}z + \frac{7}{2} \]

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at exactly one point, reflecting the curve's slope at that precise position. Imagine drawing a line that skims a curve without cutting through or creating a second point of intersection. The equation of a tangent line is useful for approximating the values of a function near the point of tangency. By knowing the function's behavior is smooth, you create a local linear approximation changing with the curve.In the context of the problem, the tangent line correlates to the point \((z, w) = (3, 2)\). Calculating its slope and writing its equation lays the foundation for deeper analysis. This method essentially combines understanding with calculation to give further insight into the graph's behavior.

Chain Rule

The chain rule is a fundamental tool for differentiating composite functions in calculus. When faced with a function composed of multiple layers, the chain rule helps us unpack one layer at a time to compute the overall derivative.Consider the given function \(w = g(z) = 1 + \sqrt{4-z}\). The inner function is \(u = 4-z\), and the outer function is \(\sqrt{u}\). The derivative of \(\sqrt{u}\) is \(\frac{1}{2\sqrt{u}}\), but we must multiply by the derivative of \(u\), which is \(-1\).The chain rule therefore allows us to take the derivative step-by-step, building from the inside out:

• Derive the inner layer: \(\frac{du}{dz} = -1\).
• Derive the overall function: \(\frac{d}{dz}(\sqrt{4-z}) = \frac{1}{2\sqrt{4-z}} \times (-1)\).

Thus, comprehending the chain rule significantly simplifies handling complex functions.

Point-Slope Form

The point-slope form is a formula specifically designed for writing the equation of a line given a point and a slope. This form is especially useful in calculus when deriving equations of tangent lines after finding the derivative.The formula looks like this:\[ w - w_1 = m(z - z_1) \]Here, \(m\) represents the slope of the line, and \((z_1, w_1)\) represents the given point the line will pass through.In this exercise, once the derivative has been evaluated to find the slope at \(z = 3\), you use the point-slope form to construct the equation of the tangent line:

• Slope (\(m\)): \(-\frac{1}{2}\)
• Point: \((3, 2)\)

By substituting these values into the formula, you efficiently solve for the equation: \(w = -\frac{1}{2}z + \frac{7}{2}\). The whole process showcases the elegance and simplicity of using the point-slope formula.

Derivative Evaluation

Evaluating a derivative at a particular point provides specific information about the function's behavior at that point. The derivative represents the rate of change of the function, essentially the slope of the tangent line.In the given exercise, the derivative \(\frac{dw}{dz}\) was calculated using the chain rule as \(-\frac{1}{2\sqrt{4-z}}\). Evaluating this derivative at \(z = 3\), you substitute the value in to get:\[ -\frac{1}{2\sqrt{4-3}} = -\frac{1}{2}\times 1 = -\frac{1}{2} \]This resulting value provides the slope of the tangent line, indicating how steep the function graph is at \(z = 3\). Such evaluation is crucial for a nuanced understanding of the function's instantaneous rate of change, filling out the essential characteristics captured by a tangent line.