Chapter 12: Problem 8
Find an equation of the tangent line to the curve at the given point. Graph the curve and the tangent line. $$y=2 x-x^{3} \quad \text { at }(1,1)$$
Short Answer
Expert verified
The equation of the tangent line is \( y = -x + 2 \).
Step by step solution
01
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 \( x \):\[ \frac{dy}{dx} = 2 - 3x^2 \].
02
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 \(-1\).
03
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:\[ y = -x + 2 \].
04
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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:
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 \( -3 \times 1x^{3-1} = -3x^2 \).