Problem 75
Question
Find a point on the curve $$ y=x^{3}+2 x+2 $$ whose tangent line is parallel to the line \(3 x-y=2\). Is there more than one such point? If so, find all other points with this property.
Step-by-Step Solution
Verified Answer
There are two points: \(\left(\frac{1}{\sqrt{3}}, \frac{7}{3\sqrt{3}} + 2\right)\) and \(\left(-\frac{1}{\sqrt{3}}, -\frac{7}{3\sqrt{3}} + 2\right)\). Both have tangents parallel to the line.
1Step 1: Understand the Problem
We need to find a point on the curve \(y = x^3 + 2x + 2\) where the tangent is parallel to a given line \(3x - y = 2\). This condition means the slope of the tangent line of the curve at this point must equal the slope of the given line.
2Step 2: Determine the Slope of the Given Line
To find the slope of the line \(3x - y = 2\), rewrite it in the slope-intercept form \(y = mx + b\). Rearranging gives \(y = 3x - 2\), which shows the slope \(m\) is 3.
3Step 3: Find the Slope of the Tangent to the Curve
The slope of the tangent to the curve \(y = x^3 + 2x + 2\) at any point is given by the derivative. Thus, calculate \(\frac{dy}{dx} = 3x^2 + 2\).
4Step 4: Set the Slope of the Tangent Equal to the Line's Slope
We need the slope of the tangent, \(3x^2 + 2\), to be equal to 3, the slope from Step 2. Set up the equation: \[ 3x^2 + 2 = 3 \]
5Step 5: Solve for x
By solving the equation \(3x^2 + 2 = 3\), subtract 2 from both sides to get: \[ 3x^2 = 1 \]Divide by 3:\[ x^2 = \frac{1}{3} \]Therefore, \(x = \pm \frac{1}{\sqrt{3}}\).
6Step 6: Find the Corresponding y-values
For each \(x\) value found in the previous step, substitute into the original curve equation \(y = x^3 + 2x + 2\) to find the \(y\)-coordinates.- When \(x = \frac{1}{\sqrt{3}}\), substitute in to get: \[ y = \left(\frac{1}{\sqrt{3}}\right)^3 + 2\left(\frac{1}{\sqrt{3}}\right) + 2 = \frac{1}{3\sqrt{3}} + \frac{2}{\sqrt{3}} + 2\]- When \(x = -\frac{1}{\sqrt{3}}\), substitute in to get: \[ y = \left(-\frac{1}{\sqrt{3}}\right)^3 + 2\left(-\frac{1}{\sqrt{3}}\right) + 2 = -\frac{1}{3\sqrt{3}} - \frac{2}{\sqrt{3}} + 2\]
7Step 7: Simplify the Results
For \(x = \frac{1}{\sqrt{3}}\), calculate further:\[ y = \frac{1 + 6}{3\sqrt{3}} + 2 = \frac{7}{3\sqrt{3}} + 2\]For \(x = -\frac{1}{\sqrt{3}}\), calculate further:\[ y = -\frac{1 + 6}{3\sqrt{3}} + 2 = -\frac{7}{3\sqrt{3}} + 2\]These define two points on the curve.
Key Concepts
DerivativesTangent lineParabolaSlopeEquation solving
Derivatives
Derivatives are a key concept in calculus and are used to describe how a function changes as its input changes. They provide us with the slope of a function at any given point, which is essential in understanding the rate of change.
To find the derivative of a function, we use rules like the power rule, product rule, and chain rule. In the context of this exercise, the derivative of the function \( y = x^3 + 2x + 2 \) with respect to \( x \) gives us \( \frac{dy}{dx} = 3x^2 + 2 \).
This derivative tells us the slope of the tangent line to the curve at any point \( x \). Understanding derivatives is crucial for determining how the function behaves and for solving problems involving tangents and rates of change.
To find the derivative of a function, we use rules like the power rule, product rule, and chain rule. In the context of this exercise, the derivative of the function \( y = x^3 + 2x + 2 \) with respect to \( x \) gives us \( \frac{dy}{dx} = 3x^2 + 2 \).
This derivative tells us the slope of the tangent line to the curve at any point \( x \). Understanding derivatives is crucial for determining how the function behaves and for solving problems involving tangents and rates of change.
Tangent line
A tangent line is a straight line that touches a curve at a single point. It represents the instantaneous direction of the curve at that point and is the best linear approximation of the curve near that point.
The slope of the tangent line is given by the derivative of the function at the point of tangency. In this exercise, we're interested in finding where the tangent line to our parabola, \( y = x^3 + 2x + 2 \), is parallel to the given line \( 3x - y = 2 \).
Parallel lines have the same slope. Therefore, we set the derivative of the curve equal to the slope of the given line, which enables us to find the points where the tangents are parallel.
The slope of the tangent line is given by the derivative of the function at the point of tangency. In this exercise, we're interested in finding where the tangent line to our parabola, \( y = x^3 + 2x + 2 \), is parallel to the given line \( 3x - y = 2 \).
Parallel lines have the same slope. Therefore, we set the derivative of the curve equal to the slope of the given line, which enables us to find the points where the tangents are parallel.
Parabola
A parabola is a specific type of curve on a graph. It is often represented by quadratic or cubic equations. In this problem, our function \( y = x^3 + 2x + 2 \) forms a curve that looks like a parabolic shape.
Although generally associated with quadratic functions like \( y = ax^2 + bx + c \), our cubic function exhibits similar traits at points where it behaves like a parabola. Understanding the shape helps us predict and verify where tangent lines can be drawn parallel to other lines.
Although generally associated with quadratic functions like \( y = ax^2 + bx + c \), our cubic function exhibits similar traits at points where it behaves like a parabola. Understanding the shape helps us predict and verify where tangent lines can be drawn parallel to other lines.
Slope
The slope is a measure of the steepness or inclination of a straight line. Mathematically, it is defined as the ratio of the vertical change to the horizontal change between two points on a line, often expressed as \( m \).
For a given line, converting its equation into the form \( y = mx + b \) reveals its slope \( m \). In the equation \( y = 3x - 2 \), the slope is 3. This constant slope tells us how much the line rises or falls for every unit it moves horizontally.
In the context of derivatives, the slope of a tangent line at a point on a curve is given by the value of the derivative at that point.
For a given line, converting its equation into the form \( y = mx + b \) reveals its slope \( m \). In the equation \( y = 3x - 2 \), the slope is 3. This constant slope tells us how much the line rises or falls for every unit it moves horizontally.
In the context of derivatives, the slope of a tangent line at a point on a curve is given by the value of the derivative at that point.
Equation solving
Equation solving involves finding the values that satisfy a given equation or system of equations. In our problem, we must solve the equation \( 3x^2 + 2 = 3 \) to find the points where the tangent line's slope matches the slope of the given line.
Rearranging and solving, we find \[ 3x^2 = 1 \] and thus \[ x^2 = \frac{1}{3} \]. The solutions are \( x = \pm \frac{1}{\sqrt{3}} \). This method involves solving a quadratic equation and helps identify where our conditions are met for tangent parallelism.
After finding the \( x \)-values, substitute them back into the original function to find the corresponding \( y \)-values, completing the solution process.
Rearranging and solving, we find \[ 3x^2 = 1 \] and thus \[ x^2 = \frac{1}{3} \]. The solutions are \( x = \pm \frac{1}{\sqrt{3}} \). This method involves solving a quadratic equation and helps identify where our conditions are met for tangent parallelism.
After finding the \( x \)-values, substitute them back into the original function to find the corresponding \( y \)-values, completing the solution process.
Other exercises in this chapter
Problem 75
Use logarithmic differentiation to find the first derivative of the given functions. $$ y=x^{\cos x} $$
View solution Problem 75
Differentiate $$ f(x)=\frac{a x}{3+x} $$ with respect to \(x\). Assume that \(a\) is a positive constant.
View solution Problem 76
Use logarithmic differentiation to find the first derivative of the given functions. $$ y=(\cos x)^{x} $$
View solution Problem 76
Differentiate $$ f(x)=\frac{a x}{k+x} $$ with respect to \(x\). Assume that \(a\) and \(k\) are positive constants.
View solution