Problem 8

Question

Find an equation of the tangent line to the curve at the given point. Graph the curve and the tangent line. $$y=2 x-x^{3} \quad \text { at }(1,1)$$

Step-by-Step Solution

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Answer

The equation of the tangent line is $ y = -x + 2 $.

1Step 1: Differentiate the function

To find the equation of the tangent line to the curve at a given point, we first need to differentiate the function to find the slope of the tangent line. The given function is $ y = 2x - x^3 $. Differentiate this with respect to $ x $:\[ \frac{dy}{dx} = 2 - 3x^2 \].

2Step 2: Evaluate the derivative at the point

The slope of the tangent line at the point $(1, 1)$ is given by evaluating the derivative $ \frac{dy}{dx} $ at $ x = 1 $. Substitute $ x = 1 $ into the derivative:\[ \frac{dy}{dx} \bigg|_{x=1} = 2 - 3(1)^2 = 2 - 3 = -1 \].So, the slope of the tangent line is $-1$.

3Step 3: Use the point-slope form to find the equation of the tangent line

Using the point-slope form of a line, which is $ y - y_1 = m(x - x_1) $, where $ m $ is the slope, and $(x_1, y_1)$ is the given point $(1, 1)$, substitute the values:\[ y - 1 = -1(x - 1) \].Simplify this to get the equation of the tangent line:\[ y = -x + 2 \].

4Step 4: Graph the curve and the tangent line

Graph the original function $ y = 2x - x^3 $ and the tangent line $ y = -x + 2 $ on the same set of axes. The tangent line should touch the curve at the point $(1, 1)$ and have the same slope as the curve at that point.

Key Concepts

DifferentiationPoint-Slope FormDerivative of Polynomial Functions

Differentiation

Differentiation is a fundamental concept in calculus that helps us understand how a function changes at any given point. In simple terms, when we differentiate a function, we find its derivative, which tells us the rate of change or slope of the function at any particular point.
Think of it as peeling back a layer of the function to see how it behaves.To differentiate a function, we use a limit process that gives us the instantaneous rate of change. For example, with the function given in the exercise, $ y = 2x - x^3 $, differentiation involves calculating the derivative with respect to $ x $. The derivative here is found to be $ \frac{dy}{dx} = 2 - 3x^2 $.
This tells us, at any $ x $, the slope of the tangent to the function. This step is key in finding the tangent line, which represents a straight line that just touches the curve at a point, perfectly mimicking its instantaneous direction of travel.

Point-Slope Form

The point-slope form is a straightforward method used to write the equation of a line when we know a point on the line and its slope. This is particularly useful in situations like ours, where we need to find the equation of a tangent line.

The point-slope form is expressed as: $ y - y_1 = m(x - x_1) $.
"$ m $" represents the slope of the line. In this problem, the slope of the tangent line is $-1$, which we found by evaluating the derivative at $ x = 1 $.
"$(x_1, y_1)$" is the specific point on the line, which is $(1, 1)$ here from the exercise.

By substituting $ m = -1 $, $ x_1 = 1 $, and $ y_1 = 1 $ into the formula, it results in $ y - 1 = -1(x - 1) $. Upon simplifying, we find the equation of the tangent line as $ y = -x + 2 $. This efficient method allows us to quickly reach the equation with minimal fuss.

Derivative of Polynomial Functions

Polynomial functions like $ y = 2x - x^3 $ are commonly encountered in calculus. Differentiating these functions follows specific rules that make the process straightforward and systematic. Since polynomials consist of terms like $ ax^n $, we apply the power rule for differentiation.
The power rule states that for any term $ ax^n $, its derivative is $ anx^{n-1} $. Let's break it down further:

For the term $ 2x $, differentiation gives $ 2 \times 1x^{1-1} = 2 $.
For the term $ -x^3 $, it yields $ -3 \times 1x^{3-1} = -3x^2 $.

This systematic approach transforms the entire polynomial into its derivative, $ 2 - 3x^2 $, providing us the slope function. Understanding these differentiation rules, such as the power rule, allows students to handle any polynomial terms with confidence and ease. This ability is essential for more complex applications involving calculus and offers a solid foundation for tackling other types of functions beyond polynomials.

Problem 8

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