Problem 62 - University Calculus: Early Transcendentals

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. $$x y^{3}+\tan (x+y)=1, \quad P\left(\frac{\pi}{4}, 0\right)$$

Step-by-Step Solution

Answer

The derivative at point \( P \) is \(-1\), and the tangent line equation is \( y = -x + \frac{\pi}{4} \).

Step 1: Plot the Implicit Equation

Use a Computer Algebra System (CAS) to plot the implicit equation \( x y^3 + \tan(x+y) = 1 \). Input the equation into the implicit plotter and visually inspect whether the point \( P(\frac{\pi}{4}, 0) \) lies on the curve. Substitute \( x = \frac{\pi}{4} \) and \( y = 0 \) into the equation to verify: \( \frac{\pi}{4} \times 0^3 + \tan(\frac{\pi}{4} + 0) = 1 \). Since \( \tan(\frac{\pi}{4}) = 1 \), the equation simplifies to \( 1 = 1 \), confirming that point \( P \) satisfies the equation.

Step 2: Implicit Differentiation

Differentiate the equation \( x y^3 + \tan(x+y) = 1 \) with respect to \( x \), treating \( y \) as a function of \( x \). The derivative of the first term using the product rule gives \( y^3 + 3xy^2 \frac{dy}{dx} \). The derivative of \( \tan(x+y) \) using the chain rule is \( \sec^2(x+y)(1 + \frac{dy}{dx}) \). Setting the total derivative to zero yields: \[ y^3 + 3xy^2 \frac{dy}{dx} + \sec^2(x+y) + \sec^2(x+y)\frac{dy}{dx} = 0 \] Simplifying for \( \frac{dy}{dx} \), we get: \[ \frac{dy}{dx} = -\frac{y^3 + \sec^2(x+y)}{3xy^2 + \sec^2(x+y)} \].

Step 3: Evaluate the Derivative at Point P

Substitute \( x = \frac{\pi}{4} \) and \( y = 0 \) into the derivative formula found in Step 2: \[ \frac{dy}{dx} = -\frac{0^3 + \sec^2(\frac{\pi}{4} + 0)}{3 \times \frac{\pi}{4} \times 0^2 + \sec^2(\frac{\pi}{4} + 0)} = -\frac{1}{1} = -1 \]. Thus, the slope of the tangent line at point \( P \) is \(-1\).

Step 4: Find the Equation of the Tangent Line

Using the point-slope form of a line equation, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) = \left(\frac{\pi}{4}, 0\right) \), substitute \( m = -1 \) to get: \[ y - 0 = -1\left(x - \frac{\pi}{4}\right) \]. Simplifying, the equation of the tangent line is \( y = -x + \frac{\pi}{4} \).

Step 5: Plot Implicit Curve and Tangent Line

Use the CAS to plot both the implicit curve \( x y^3 + \tan(x+y) = 1 \) and the tangent line \( y = -x + \frac{\pi}{4} \) on the same graph to visualize where the tangent line touches the curve exactly at point \( P \).

Key Concepts

Tangent Line Equation

To find the equation of a tangent line at a particular point on a curve, you need the slope at that point. This slope is essentially the derivative of the curve's equation, evaluated at the point of interest. Let's consider the curve described implicitly by the equation: \[ x y^3 + \tan(x+y) = 1 \]. Here, we want to find the tangent line at the point \( P\left(\frac{\pi}{4}, 0\right) \).
First, after evaluating the derivative and finding that its value at point \( P \) is \(-1\), we use this to determine the slope of the tangent. The point-slope form of a line equation is useful here:

• The general form is: \( y - y_1 = m(x - x_1) \)
• In our exercise, \( m = -1 \), \( (x_1, y_1) = \left(\frac{\pi}{4}, 0\right) \)

Substituting these into the formula gives: \[ y - 0 = -1(x - \frac{\pi}{4}) \] Simplifying, we get: \[ y = -x + \frac{\pi}{4} \]
This equation represents the tangent line, and it helps to visualize how the curve behaves locally at the point \( P \). Adding this tangent to a plot of the original curve will show exactly where and how it touches the curve.

Implicit Plotting

Implicit plotting is a powerful way to visualize relationships defined by equations that are not explicitly solved for one variable. In this case, the equation \[ x y^3 + \tan(x+y) = 1 \] defines a curve implicitly. We don't have \( y \) or \( x \) isolated, so tools like a Computer Algebra System (CAS) are particularly useful.Before delving further into analysis, it is important to confirm that the provided point actually lies on the curve. Substitute the coordinates of point \( P\left(\frac{\pi}{4}, 0\right) \) into the equation:

• Replace \( x = \frac{\pi}{4} \) and \( y = 0 \)
• Check if the equation holds: \( \frac{\pi}{4} \times 0^3 + \tan(\frac{\pi}{4} + 0) = 1 \)

Since \( \tan(\frac{\pi}{4}) = 1 \), the equation simplifies to \( 1 = 1 \), confirming the point is on the curve.
Using implicit plotters, you can observe the full curve and inspect such points visually. This visualization can guide you in understanding the shape and behavior of the curve around points of interest.

Derivative Evaluation

Evaluating derivatives implicitly allows us to differentiate equations not solved for one variable. For our implicit function \( x y^3 + \tan(x+y) = 1 \), we want to find \( \frac{dy}{dx} \). This derivative represents the rate of change of \( y \) with respect to \( x \).Here's the differentiation process:
1. **Apply the Product Rule** to \( x y^3 \): The derivative is \( y^3 + 3xy^2 \frac{dy}{dx} \).
2. **Use the Chain Rule** for \( \tan(x+y) \): The derivative becomes \( \sec^2(x+y) (1 + \frac{dy}{dx}) \).
3. **Integrate these derivatives together** to form the equation: \[ y^3 + 3xy^2 \frac{dy}{dx} + \sec^2(x+y) + \sec^2(x+y) \frac{dy}{dx} = 0 \]Simplifying for \( \frac{dy}{dx} \), we isolate this derivative:
\[ \frac{dy}{dx} = -\frac{y^3 + \sec^2(x+y)}{3xy^2 + \sec^2(x+y)} \]
Substituting \( x = \frac{\pi}{4} \) and \( y = 0 \) gives \( \frac{dy}{dx} = -1 \). This value is crucial for determining the slope of the tangent line at point \( P \), enhancing our understanding of the curve's local behavior.