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Chapter 3: Problem 53

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. $$x^{3}-x y+y^{3}=7, \quad P(2,1)$$

Short Answer

Expert verified
The tangent line at \(P(2,1)\) is \(y = -11x + 23\).

Step by step solution

01

Plot the Equation

To begin, we use a CAS (Computer Algebra System) to plot the implicit equation \(x^3 - xy + y^3 = 7\). We check if the point \(P(2,1)\) satisfies the equation by substituting \(x = 2\) and \(y = 1\) into the equation:\[2^3 - 2 \cdot 1 + 1^3 = 8 - 2 + 1 = 7\]Since the left side equals 7, the point \(P(2,1)\) satisfies the equation, confirming it lies on the curve.
02

Implicit Differentiation

Next, we perform implicit differentiation on the equation \(x^3 - xy + y^3 = 7\) with respect to \(x\). Applying the derivative rules:- \(\frac{d}{dx}(x^3) = 3x^2\)- \(\frac{d}{dx}(-xy) = -(y + x\frac{dy}{dx})\)- \(\frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx}\)The derivative of the equation is:\[3x^2 - (y + x\frac{dy}{dx}) + 3y^2\frac{dy}{dx} = 0\]Simplifying, we obtain:\[3x^2 - y - x\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0\]Rearranging for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{3x^2 - y}{x - 3y^2}\]
03

Evaluate Derivative at P(2,1)

To find the slope at \(P(2,1)\), we substitute \(x = 2\) and \(y = 1\) into the derivative formula:\[\frac{dy}{dx} = \frac{3(2)^2 - 1}{2 - 3(1)^2} = \frac{12 - 1}{2 - 3} = \frac{11}{-1} = -11\]Thus, the slope of the curve at \(P(2,1)\) is \(-11\).
04

Find Equation of Tangent Line

Using the slope \(m = -11\) and the point \(P(2,1)\), we use the point-slope form to determine the equation of the tangent line:\[y - 1 = -11(x - 2)\]Simplifying, we find:\[y = -11x + 22 + 1\]\[y = -11x + 23\]The equation of the tangent line at \(P(2,1)\) is \(y = -11x + 23\).
05

Plot Curve and Tangent Line

Finally, in the CAS, plot the original implicit curve \(x^3 - xy + y^3 = 7\) and the tangent line \(y = -11x + 23\) on the same graph to visually verify the tangent line at \(P(2,1)\). The tangent should just touch the curve at the point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangent Line Equation
One of the beautiful concepts in calculus is the tangent line to a curve. A tangent line is the straight line that just "touches" a curve at a particular point without crossing it at that point. This line has the same slope as the curve at the touchpoint, making it a very useful tool in calculus. To find the tangent line, we use the point-slope formula of a line.
  • The formula is: \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the point on the curve, and \(m\) is the slope of the line.
  • In our exercise, after finding the slope at point \(P(2,1)\) as \(-11\), we substitute into the formula: \(y - 1 = -11(x - 2)\).
  • Simplifying this, we derive the equation: \(y = -11x + 23\).
Ultimately, this equation represents the tangent line to the curve at point \(P(2,1)\), giving us a linear approximation of the curve's behavior near that point.
It's critical because it helps us understand the curve's local behavior and can be used for further approximations.
Implicit Equations
Implicit equations are those where the relationship between variables isn't obviously solved for one specific variable. Unlike explicit functions where \(y\) might be given clearly as a function of \(x\), implicit equations often involve multiple variables combined in a single equation.
  • An example of an implicit equation is \(x^3 - xy + y^3 = 7\).
  • Solving an implicit equation often requires techniques like implicit differentiation to analyze relationships between the variables, because expressing one variable solely in terms of the other can be challenging.
  • These equations naturally arise from real-world situations where direct functional relationships aren鈥檛 clearly defined.
The beauty of implicit equations is in how they allow us to study curves and surfaces that aren't easily handled by standard functions.
By learning to work with them, we open the door to understanding more complex mathematical scenarios.
Derivative Evaluation
Evaluating the derivative of a function tells us the rate at which one quantity changes with respect to another. When dealing with implicit equations, however, derivative evaluation can take a different path using the method of implicit differentiation.
  • Implicit differentiation involves taking the derivative of both sides of an equation with respect to \(x\), while treating \(y\) as an implicit function of \(x\).
  • In our case, for the equation \(x^3 - xy + y^3 = 7\), we differentiate each term to obtain the expression \(3x^2 - (y + x\frac{dy}{dx}) + 3y^2\frac{dy}{dx} = 0\).
  • Rearranging terms helps solve for \(\frac{dy}{dx}\), giving us: \(\frac{dy}{dx} = \frac{3x^2 - y}{x - 3y^2}\).
Evaluating this at specific points, like \((2,1)\), helps us determine the slope of the tangent line at that point.
This evaluation is integral for drawing tangent lines and understanding the behavior of the curve near those points.

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Most popular questions from this chapter

When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity \(g\). The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in \(g.\) By keeping track of \(\Delta T,\) we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L\). a. With \(L\) held constant and \(g\) as the independent variable, calculate \(d T\) and use it to answer parts (b) and (c). b. If \(g\) increases, will \(T\) increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a \(100-\mathrm{cm}\) pendulum is moved from a location where \(g=980 \mathrm{cm} / \mathrm{s}^{2}\) to a new location. This increases the period by \(d T=0.001\) s. Find \(d g\) and estimate the value of \(g\) at the new location.

A highway patrol plane flies 3 km above a level, straight road at a steady \(120 \mathrm{km} / \mathrm{h}\). The pilot sees an oncoming car and with radar determines that at the instant the line-of-sight distance from plane to car is \(5 \mathrm{km},\) the line-of-sight distance is decreasing at the rate of \(160 \mathrm{km} / \mathrm{h}\). Find the car's speed along the highway.

Suppose that a piston is moving straight up and down and that its position at time \(t\) s is $$s=A \cos (2 \pi b t)$$ with \(A\) and \(b\) positive. The value of \(A\) is the amplitude of the motion, and \(b\) is the frequency (number of times the piston moves up and down each second). What effect does doubling the frequency have on the piston's velocity, acceleration, and jerk? (Once you find out, you will know why some machinery breaks when you run it too fast.)

Suppose that functions \(f\) and \(g\) and their derivatives with respect to \(x\) have the following values at \(x=2\) and \(x=3\) $$\begin{array}{ccccc}\hline x & f(x) & g(x) & f^{\prime}(x) & g^{\prime}(x) \\\\\hline 2 & 8 & 2 & 1 / 3 & -3 \\ 3 & 3 & -4 & 2 \pi & 5 \\\\\hline\end{array}$$ Find the derivatives with respect to \(x\) of the following combinations at the given value of \(x\) a. \(2 f(x), \quad x=2\) b. \(f(x)+g(x), \quad x=3\) c. \(f(x) \cdot g(x), \quad x=3\) d. \(f(x) / g(x), \quad x=2\) e. \(f(g(x)), \quad x=2\) f. \(\sqrt{f(x)}, \quad x=2\) g. \(1 / g^{2}(x), \quad x=3\) h. \(\sqrt{f^{2}(x)+g^{2}(x)}, \quad x=2\)

Using the Chain Rule, show that the Power Rule \((d / d x) x^{n}=n x^{n-1}\) holds for the functions \(x^{n}\). $$x^{3 / 4}=\sqrt{x \sqrt{x}}$$
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