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Magazine Chapter 1: Problem 46
There are two points on the graph of \(y=x^{3}\) where the tangent lines are parallel to \(y=x .\) Find these points.
Short Answer
Expert verified
The points are \(\left( \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{9} \right)\) and \(\left( -\frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{9} \right)\).
Step by step solution
01
- Understand the Problem
We need to find the points on the curve of the function \[y = x^3\] where the tangent lines are parallel to the line \[y = x\]. This implies that the slopes of the tangent lines to the curve equal the slope of the line \[y = x\].
02
- Derive the Slope of the Tangent Line
The slope of the line \[y = x\] is 1. We need to find the derivative of the function \[y = x^3\] to get the slope of the tangent line at any point \(x\).
03
- Calculate the Derivative
The derivative of \[y = x^3\] with respect to \(x\) is \[\frac{dy}{dx} = 3x^2\].
04
- Set the Derivative Equal to 1
Since the tangent lines are parallel to \[y = x\], we set the derivative equal to 1:\[3x^2 = 1\].
05
- Solve for x
Solve the equation \[3x^2 = 1\]:\[x^2 = \frac{1}{3}\]\[x = \pm\sqrt{\frac{1}{3}}\]\[x = \pm\frac{1}{\sqrt{3}}\]\[x = \pm\frac{\sqrt{3}}{3}\].
06
- Find the Corresponding y-Values
Substitute \[x = \pm\frac{\sqrt{3}}{3}\] back into the original function \[y = x^3\] to find the corresponding \(y\) values:For \(x = \frac{\sqrt{3}}{3}\):\[y = \left(\frac{\sqrt{3}}{3}\right)^3 = \frac{(\sqrt{3})^3}{27} = \frac{3\sqrt{3}}{27} = \frac{\sqrt{3}}{9}\]For \(x = -\frac{\sqrt{3}}{3}\):\[y = \left( -\frac{\sqrt{3}}{3} \right)^3 = -\frac{(\sqrt{3})^3}{27} = -\frac{3\sqrt{3}}{27} = -\frac{\sqrt{3}}{9}\].
07
- State the Points
The points on the graph \[y = x^3\] where the tangent lines are parallel to \[y = x\] are \(\left( \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{9} \right)\) and \(\left( -\frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{9} \right)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
derivatives
In this problem, understanding the concept of derivatives is crucial. A derivative represents the rate of change of a function with respect to a variable. For a given function \(f(x)\), its derivative \(f'(x)\) indicates how \(f(x)\) changes as \(x\) changes.
Step 3 of the solution involves finding the derivative of \(y = x^3\). This tells us how the function \(y\) changes with \(x\). The formula for the derivative of \(x^n\) is \(nx^{n-1}\).
Applying this formula, the derivative of \(x^3\) is \(3x^2\). This derivative gives us the slope of the tangent line to the curve \(y = x^3\) at any point \(x\), which is essential for solving the problem.
Step 3 of the solution involves finding the derivative of \(y = x^3\). This tells us how the function \(y\) changes with \(x\). The formula for the derivative of \(x^n\) is \(nx^{n-1}\).
Applying this formula, the derivative of \(x^3\) is \(3x^2\). This derivative gives us the slope of the tangent line to the curve \(y = x^3\) at any point \(x\), which is essential for solving the problem.
slope of tangent line
The slope of a tangent line to a curve at a point tells us the steepness of the curve at that point. For the line \(y = x\), the slope is 1 because it rises one unit for each unit it moves horizontally.
We want to find points on the curve \(y = x^3\) where the tangent lines are parallel to the line \(y = x\). This means setting the derivative \(3x^2\) equal to 1 to identify the points where the slopes are equal.
By solving \(3x^2 = 1\), we determine the points on the curve where the tangent has a slope of 1. These points are where the curve \(y = x^3\) and the line \(y = x\) have tangents with the same steepness.
We want to find points on the curve \(y = x^3\) where the tangent lines are parallel to the line \(y = x\). This means setting the derivative \(3x^2\) equal to 1 to identify the points where the slopes are equal.
By solving \(3x^2 = 1\), we determine the points on the curve where the tangent has a slope of 1. These points are where the curve \(y = x^3\) and the line \(y = x\) have tangents with the same steepness.
solving equations
To find the points on the curve where the tangent lines are parallel to \(y = x\), we need to solve the equation \(3x^2 = 1\). Solving an equation involves finding the value(s) of \(x\) that satisfy it.
\begin{itemize}First, we divide both sides by 3: \(x^2 = \frac{1}{3}\) Next, we take the square root of both sides to isolate \(x\): \(x = \pm \frac{1}{\root 3}\) \begin{itemize}This results in two potential \(x\) values: \(x = \frac{\root 3}{3}\) and \(x = -\frac{\root 3}{3}\) We then substitute these values back into the original function \(y = x^3\) to find the corresponding y-values, giving us the points where the tangent lines are parallel to \(y = x\).
\begin{itemize}
function graphing
Graphing functions helps us visually understand their behavior. In this problem, graphing \(y = x^3\) and the lines parallel to \(y = x\) enhances our comprehension.
The curve \(y = x^3\) is a cubic function, characterized by its symmetric shape around the origin. Its slope changes at different points, determined by its derivative \(3x^2\).
By finding where the derivative equals 1, we locate points with tangents parallel to \(y = x\). Plotting these points on the graph of \(y = x^3\) with their tangent lines makes it clearer how they relate to the line \(y = x\). This visualization supports the solution and deepens our understanding of the problem.
The curve \(y = x^3\) is a cubic function, characterized by its symmetric shape around the origin. Its slope changes at different points, determined by its derivative \(3x^2\).
By finding where the derivative equals 1, we locate points with tangents parallel to \(y = x\). Plotting these points on the graph of \(y = x^3\) with their tangent lines makes it clearer how they relate to the line \(y = x\). This visualization supports the solution and deepens our understanding of the problem.
