Problem 77 - Thomas Calculus

Question

Use a CAS to perform the following steps in Exercises \(77-84\) . 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. $$ x^{3}-x y+y^{3}=7, \quad P(2,1) $$

Step-by-Step Solution

Answer

The derivative at point P is -11, and the tangent line equation is \(y = -11x + 23\).

Step 1: Plot the Equation Using CAS

Use a Computer Algebra System (CAS) to plot the equation \(x^3 - xy + y^3 = 7\). Verify that the point \(P(2,1)\) lies on this curve by substituting \(x = 2\) and \(y = 1\) into the equation. Check if both sides of the equation are equal to confirm the point.

Step 2: Verify Point P on the Curve

Substitute \(x = 2\) and \(y = 1\) into the equation: \(2^3 - 2 \cdot 1 + 1^3 = 7\). Calculate: \(8 - 2 + 1 = 7\). Since both sides are equal, the point \(P(2, 1)\) satisfies the equation.

Step 3: Implicit Differentiation to Find \(\frac{dy}{dx}\)

Differentiate the equation \(x^3 - xy + y^3 = 7\) implicitly with respect to \(x\). The derivative will be \(3x^2 - y - x \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 0\). Solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{y - 3x^2}{3y^2 - x} \]

Step 4: Evaluate the Derivative at Point P

Substitute \(x = 2\) and \(y = 1\) into the derivative formula: \[ \frac{dy}{dx} = \frac{1 - 3(2)^2}{3(1)^2 - 2} = \frac{1 - 12}{3 - 2} = \frac{-11}{1} = -11 \]

Step 5: Find Equation of the Tangent Line

Use the point-slope form of the equation of a line: \[ y - y_1 = m(x - x_1) \] where \(m = -11\), \((x_1, y_1) = (2, 1)\). Thus, the equation becomes \[ y - 1 = -11(x - 2) \]. Simplify to \[ y = -11x + 22 + 1 \] \[ y = -11x + 23 \].

Step 6: Plot Implicit Curve and Tangent Line

Using the CAS, plot the implicit curve \(x^3 - xy + y^3 = 7\) and the tangent line \(y = -11x + 23\) together. Ensure that the tangent line touches the curve at the point \(P(2, 1)\).

Key Concepts

Tangent Line

The tangent line to a curve at a given point is a straight line that just touches the curve at that point. It has the same slope as the curve at the point of tangency.
In this problem, the slope was found using the derivative of the curve at point \( P(2,1) \). The derivative represents how steep the curve is at a specific point.
With the slope known as \(-11\), we can use the point-slope form of a line to write the equation of the tangent line. This line equation is \( y = -11x + 23 \), which shows how the tangent gently kisses the original curve without actually crossing it.
The tangent line can give a lot of insight into the behavior and curvature of the graph at the specific point.

Implicit Plotting

Implicit plotting is used to graph equations where \( y \) cannot be easily isolated. In other words, in implicit plotting, an expression involving both \( x \) and \( y \) is set equal to a constant, leading to a curve.
Using a Computer Algebra System (CAS), you can visually understand the curve of the equation \( x^3 - xy + y^3 = 7 \). It helps us see the shape without explicitly solving for \( y \) in terms of \( x \), which might be difficult or even impossible.
You can also verify that specific points, like \( P(2,1) \), lie on the curve by plotting it and checking manually. It's a powerful tool for visual learners and those dealing with complex multivariable functions.

Equation of a Curve

An equation of a curve like \( x^3 - xy + y^3 = 7 \) represents all the points \((x, y)\) that satisfy the given relation. Unlike regular functions, curves can loop, vertical points, and are not bound by the same vertical line test that functions are.
You can determine whether points lie on a curve by substituting them into the equation and seeing if they satisfy it. This process, shown when point \( P(2,1) \) was verified, is crucial for confirming positions on the graph.
Understanding the equation of a curve is essential for grasping the geometrical representation of mathematical relations and determining properties like intercepts and areas enclosed.

Derivative Evaluation

Finding the derivative, \( \frac{dy}{dx} \), is an essential skill in calculus implying how one quantity changes with respect to another. When dealing with implicit functions, we apply implicit differentiation.
By differentiating each term of the equation implicitly and solving for \( \frac{dy}{dx} \), we derived the slope formula \( \frac{y - 3x^2}{3y^2 - x} \).
To evaluate the derivative at a specific point, like \( P(2,1) \), we substitute the respective \( x \) and \( y \) values, resulting in \( \frac{dy}{dx} = -11 \).
This number tells us how steep the original curve is at the given point, hence allowing us to draw the tangent line accurately. Derivative evaluation is about understanding how curves behave at particular spots.