Problem 20 - The Calculus with Analytic Geometry

Question

Find an equation of the tangent line to the curve \(y=(6-2 x)^{1 / 3}\) at each of the following points: \((-1,2),(1, \sqrt[3]{4})\), \((3,0),(5,-\sqrt[3]{4}),(7,-2)\). Draw a sketch of the graph and segments of the tangent lines at the given points.

Step-by-Step Solution

Answer

Tangent lines: At (-1,2): \( y - 2 = -\frac{1}{6}(x + 1) \); At (1, \sqrt[3]{4}): \( y - \sqrt[3]{4} = -\frac{2}{3}(4)^{-2/3}(x - 1) \); At (3,0): \( x = 3 \); At (5, -\sqrt[3]{4}): \( y + \sqrt[3]{4} = -\frac{2}{3}(-4)^{-2/3}(x - 5) \); At (7,-2): \( y + 2 = -\frac{1}{6}(x - 7) \).

Step 1: Understand the Problem

The goal is to find the equations of the tangent lines to the curve given by the function \( y=(6-2x)^{1/3} \) at the specified points.

Step 2: Find the Derivative of the Function

The derivative represents the slope of the tangent line at any point on the curve. Use the chain rule to differentiate \( y=(6-2x)^{1/3} \). The derivative \( y' \) will be: \[ \frac{dy}{dx} = \frac{d}{dx} \big((6-2x)^{1/3}\big) = \frac{1}{3}(6-2x)^{-2/3} \times (-2) \rightarrow y' = -\frac{2}{3}(6-2x)^{-2/3} \text{So, } y' = -\frac{2}{3}(6-2x)^{-2/3} \]

Step 3: Calculate the Slope at Each Point

Evaluate the derivative at each given point to find the slope of the tangent line at that point.For (-1, 2): \[ y'(-1) = -\frac{2}{3}(8)^{-2/3} = -\frac{2}{3} \times \frac{1}{4} = -\frac{1}{6} \] For (1, \sqrt[3]{4}): \[ y'(1) = -\frac{2}{3}(4)^{-2/3} = -\frac{2}{3} \times \frac{1}{\root{3}{4^2}} \] For (3, 0): \[ y'(3) = -\frac{2}{3}(0)^{-2/3} = undefined \] For (5, -\sqrt[3]{4}): \[ y'(5) = - \frac{2}{3}(–4)^{-2/3} = -\frac{2}{3} \times \frac{1}{\root{3}{16}} \] For (7, -2): \[ y'(7) = -\frac{2}{3}(–8)^{-2/3} = -\frac{2}{3} \times \frac{1}{4} = -\frac{1}{6} \]

Step 4: Write the Equations of the Tangent Lines

Use the point-slope form of a line equation: \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a given point and \( m \) is the slope.At (-1, 2): \( y - 2 = -\frac{1}{6}(x + 1) \) At (1, \sqrt[3]{4}): \( y - \root{3}{4} = -\frac{2}{3}(4)^{-2/3}(x - 1) \) At (3, 0): The slope is undefined, indicating a vertical tangent line, so the equation is \( x = 3 \).At (5, -\sqrt[3]{4}): \( y + \root{3}{4} = -\frac{2}{3}(–4)^{-2/3}(x - 5) \) At (7, -2): \( y + 2 = -\frac{1}{6}(x - 7) \)

Step 5: Sketch the Graph

Graph the curve \( y = (6-2x)^{1/3} \) and the tangent lines at the given points. Remember to plot the points (-1,2), (1, \sqrt[3]{4}), (3,0), (5, -\sqrt[3]{4}), and (7,-2). Draw the corresponding linear segments representing the tangent lines.

Key Concepts

derivative

A derivative is a measure of how a function changes as its input changes. It represents the slope of the function at any given point and is a key element in finding the equation of a tangent line. To find the derivative of the function in the exercise, we need to apply differentiation rules.
In this case, we start with the function \(y = (6-2x)^{1/3}\). The derivative of this function requires the chain rule, which is a method used for differentiating compositions of functions. We ultimately end up with:

• \(y' = -\frac{2}{3}(6-2x)^{-2/3}\)

This derivative formula provides a way to calculate the slope of the tangent line at any given point on the curve.

chain rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. When we have a function inside another function, we use the chain rule to find the derivative. This involves taking the derivative of the outer function and multiplying it by the derivative of the inner function.
For our function \(y = (6-2x)^{1/3}\), we need to differentiate \( (6-2x)^{1/3} \).
By applying the chain rule, we get:

• \(y' = \frac{d}{dx}((6-2x)^{1/3}) = \frac{1}{3}(6-2x)^{-2/3} \times (-2)\)
• Simplifying gives us: \( y' = -\frac{2}{3}(6-2x)^{-2/3}\)

This step-by-step differentiation helps us find the slope of the tangent line at any point on the curve.

point-slope form

The point-slope form is a linear equation used to find the equation of a line given a point on the line and its slope. The formula is:

• \(y - y_1 = m(x - x_1)\)

where \((x_1, y_1)\) is a specific point on the line and \(m\) is the slope.
Once we have the slope from the derivative, we can plug it, along with the given point, into the point-slope formula. For example, for the point \((-1, 2)\) with the slope \(-\frac{1}{6}\):

• \(y - 2 = -\frac{1}{6}(x + 1)\)

This gives us the equation of the tangent line at \((-1, 2)\).

graphing functions

Graphing functions involves plotting a function's curve on a coordinate plane. This helps visualize the behavior of the function, and in our case, the positions of tangents. The exercise requires sketching the graph of \(y = (6-2x)^{1/3}\) and the tangent lines at specific points.
Steps to graph:

• Plot the given points: \((-1, 2)\), \((1, \sqrt[3]{4})\), \((3, 0)\), \((5, -\sqrt[3]{4})\), and \((7, -2)\)
• Draw the curve of the function based on its equation
• At each point, use the derived slope to draw the tangent lines

Sketching helps to see where each tangent touches the curve and can aid in understanding the relationship between the curve and its tangent lines.