Problem 64
Question
Use a CAS to perform the following steps a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\) c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$x \sqrt{1+2 y}+y=x^{2}, \quad P(1,0)$$
Step-by-Step Solution
Verified Answer
The tangent line equation at point \(P(1,0)\) is \(y = x - 1\).
1Step 1: Verify Point on Curve
Use a Computer Algebra System (CAS) to plot the implicit equation \(x \sqrt{1+2y} + y = x^2\) and visually verify that the point \(P(1,0)\) lies on the graph. Substitute \(x = 1\) and \(y = 0\) into the equation to check if it satisfies the equation: \(1 \cdot \sqrt{1+2 \cdot 0} + 0 = 1^2\). The equation simplifies to \(1 = 1\), confirming that the point satisfies the equation.
2Step 2: Perform Implicit Differentiation
Differentiate both sides of the equation \(x \sqrt{1+2y} + y = x^2\) with respect to \(x\). The differentiation steps are:- Differentiate \(x \sqrt{1+2y}\): Apply the product rule to get \(\sqrt{1+2y} + x \cdot \frac{d}{dx}(\sqrt{1+2y})\).- Differentiate \(y\): Apply the chain rule to get \(\frac{dy}{dx}\).- Differentiate \(x^2\): Result is \(2x\).Now, solve the equation \(\sqrt{1+2y} + \frac{x}{2\sqrt{1+2y}} \cdot 2 \cdot \frac{dy}{dx} + \frac{dy}{dx} = 2x\) for \(\frac{dy}{dx}\). Simplify to get the derivative formula.
3Step 3: Evaluate Derivative at Given Point
Substitute \(x=1\) and \(y=0\) into the derivative formula derived in Step 2 to find \(\frac{dy}{dx}\) at the point \(P(1,0)\). Calculate the slope of the tangent, which is \(\frac{dy}{dx} = 2 - 1 = 1\).
4Step 4: Find Equation of Tangent Line
Use the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = (1, 0)\) and \(m = 1\), to write the equation of the tangent line: \(y - 0 = 1(x - 1)\). Simplify to get \(y = x - 1\).
5Step 5: Plot Curve and Tangent Line
Plot the original implicit curve \(x \sqrt{1+2y} + y = x^2\) and the tangent line \(y = x - 1\) on the same graph using the CAS. Verify that the tangent line touches the curve at the given point \(P(1,0)\) and represents the slope accurately.
Key Concepts
Tangent LineDerivative of Implicit FunctionsImplicit Plotting
Tangent Line
A tangent line to a curve at a particular point is a straight line that just "touches" the curve at that point without crossing it. Imagine you have a curve on a graph, and you want to place a ruler so that it barely touches the curve at one point. That's your tangent line. To find the equation of the tangent line, you need to know the slope of the line at the point of tangency and the point's coordinates.
In this context, the exercise provides us with a point \(P=(1,0)\) on a curve given by the equation \(x \sqrt{1+2y} + y = x^2\). By calculating the slope of the curve at this specific point, we can use the point-slope form \(y-y_1=m(x-x_1)\) to derive the equation for the tangent line. Here, \((x_1, y_1) = (1, 0)\), and the slope \(m\) is the derivative \(\frac{dy}{dx}\) evaluated at \(P\).
The final equation for the tangent line explains how the straight line behaves precisely at that specific point on the curve, such as whether it's rising or falling.
In this context, the exercise provides us with a point \(P=(1,0)\) on a curve given by the equation \(x \sqrt{1+2y} + y = x^2\). By calculating the slope of the curve at this specific point, we can use the point-slope form \(y-y_1=m(x-x_1)\) to derive the equation for the tangent line. Here, \((x_1, y_1) = (1, 0)\), and the slope \(m\) is the derivative \(\frac{dy}{dx}\) evaluated at \(P\).
The final equation for the tangent line explains how the straight line behaves precisely at that specific point on the curve, such as whether it's rising or falling.
Derivative of Implicit Functions
The concept of differentiating implicit functions is a remarkable tool in calculus. Unlike explicit functions where \(y\) is expressed solely in terms of \(x\), implicit functions involve both variables intertwined together.
In our exercise, the equation \(x \sqrt{1+2y} + y = x^2\) doesn't present \(y\) explicitly in terms of \(x\). Thus, we perform implicit differentiation to find \(\frac{dy}{dx}\), which denotes how \(y\) changes with respect to \(x\).
To achieve this, you apply differentiation rules such as the product rule and chain rule. For instance, when differentiating the term \(x \sqrt{1+2y}\), the product rule comes into play, and you apply the chain rule to differentiate \(\sqrt{1+2y}\). After performing implicit differentiation across all terms, you'll rearrange the resulting equation to isolate \(\frac{dy}{dx}\).
This isolated \(\frac{dy}{dx}\) gives you the derivative at any point \((x, y)\) on the curve. It is essential for determining the slope of the tangent line to the curve at any desired point.
In our exercise, the equation \(x \sqrt{1+2y} + y = x^2\) doesn't present \(y\) explicitly in terms of \(x\). Thus, we perform implicit differentiation to find \(\frac{dy}{dx}\), which denotes how \(y\) changes with respect to \(x\).
To achieve this, you apply differentiation rules such as the product rule and chain rule. For instance, when differentiating the term \(x \sqrt{1+2y}\), the product rule comes into play, and you apply the chain rule to differentiate \(\sqrt{1+2y}\). After performing implicit differentiation across all terms, you'll rearrange the resulting equation to isolate \(\frac{dy}{dx}\).
This isolated \(\frac{dy}{dx}\) gives you the derivative at any point \((x, y)\) on the curve. It is essential for determining the slope of the tangent line to the curve at any desired point.
Implicit Plotting
Implicit plotting allows us to graph mathematical relationships that aren’t easily expressed as \(y\) being a function of \(x\), or vice versa.
In this exercise, the given equation \(x \sqrt{1+2y} + y = x^2\) is plotted using an implicit plotter provided by a Computer Algebra System (CAS). This software assists in creating a visual representation of our curve, even when it cannot be easily solved for one variable.
Through implicit plotting, we can visually verify that the point \(P=(1,0)\) indeed lies on the curve, as it fits the equation when substituted back. This graphically confirmed point becomes vital when deriving other related properties like the tangent line.
Moreover, plotting both the implicit equation and the tangent line on the same graph illustrates how the curve behaves at and around the point of interest. The visual examination helps in understanding the nature of solutions better and provides a clear picture of the slope and direction of the tangent line with respect to the curve.
In this exercise, the given equation \(x \sqrt{1+2y} + y = x^2\) is plotted using an implicit plotter provided by a Computer Algebra System (CAS). This software assists in creating a visual representation of our curve, even when it cannot be easily solved for one variable.
Through implicit plotting, we can visually verify that the point \(P=(1,0)\) indeed lies on the curve, as it fits the equation when substituted back. This graphically confirmed point becomes vital when deriving other related properties like the tangent line.
Moreover, plotting both the implicit equation and the tangent line on the same graph illustrates how the curve behaves at and around the point of interest. The visual examination helps in understanding the nature of solutions better and provides a clear picture of the slope and direction of the tangent line with respect to the curve.
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