Finding The Slope Of Tangent Line To -2x^2 + 2xy - 2y^3 = 58 At (-1, -3)

Finding the slope of a tangent line to a curve at a specific point is a fundamental problem in calculus. This problem combines concepts from algebra and calculus, requiring a solid understanding of implicit differentiation and the geometric interpretation of derivatives. In this comprehensive guide, we will delve into the step-by-step process of finding the slope of the tangent line to the curve defined by the equation $-2x^2 + 2xy - 2y^3 = 58$ at the point $(-1, -3)$. This exploration will not only provide a solution to the given problem but also enhance your understanding of related calculus techniques. Understanding how to find the slope of a tangent line is crucial for various applications, including optimization problems, curve sketching, and analyzing rates of change. Mastering these techniques provides a strong foundation for advanced topics in mathematics and its applications in science and engineering. Let's embark on this journey to unravel the intricacies of this problem and solidify your understanding of calculus. We'll break down each step, ensuring clarity and a deeper grasp of the underlying principles. This will enable you to tackle similar problems with confidence and precision. By the end of this discussion, you will have a clear methodology for finding the slope of tangent lines to implicitly defined curves.

Understanding Implicit Differentiation

Implicit differentiation is a powerful technique used to find the derivative of a function that is not explicitly defined in the form $y = f(x)$. In our case, the equation $-2x^2 + 2xy - 2y^3 = 58$ implicitly defines a relationship between $x$ and $y$. To differentiate such an equation, we apply the chain rule and differentiate each term with respect to $x$, treating $y$ as a function of $x$. The concept of implicit differentiation is essential when dealing with equations where isolating $y$ is difficult or impossible. It allows us to find the derivative $\frac{dy}{dx}$ without explicitly solving for $y$. Implicit differentiation involves differentiating both sides of the equation with respect to $x$, remembering to apply the chain rule whenever differentiating a term involving $y$. This technique is particularly useful in scenarios where the relationship between variables is complex and cannot be easily expressed in a straightforward manner. For example, equations involving trigonometric, exponential, or logarithmic functions combined with algebraic terms often require implicit differentiation. Mastery of implicit differentiation is crucial for solving a wide range of calculus problems, including finding tangent lines, related rates, and optimization problems. It provides a flexible and powerful tool for analyzing curves and their properties. The key to successful implicit differentiation is careful application of the chain rule and algebraic manipulation to isolate the desired derivative, $\frac{dy}{dx}$. This process requires attention to detail and a solid understanding of differentiation rules.

Step-by-Step Solution

1. Implicitly Differentiate the Equation

To begin, we differentiate both sides of the equation $-2x^2 + 2xy - 2y^3 = 58$ with respect to $x$. Remember to apply the product rule to the term $2xy$ and the chain rule to the term $-2y^3$. Differentiating $-2x^2$ with respect to $x$ gives us $-4x$. For the term $2xy$, we use the product rule, which states that $(uv)' = u'v + uv'$. Here, $u = 2x$ and $v = y$, so we have: $\frac{d}{dx}(2xy) = 2y + 2x\frac{dy}{dx}$. Differentiating $-2y^3$ with respect to $x$ requires the chain rule: $\frac{d}{dx}(-2y^3) = -6y^2\frac{dy}{dx}$. The derivative of the constant 58 is 0. Combining these results, we get: $-4x + 2y + 2x\frac{dy}{dx} - 6y^2\frac{dy}{dx} = 0$. This equation represents the derivative of the original equation with respect to $x$, and it implicitly defines the slope of the curve at any point $(x, y)$. The next step involves isolating $\frac{dy}{dx}$ to find an expression for the slope of the tangent line. This requires algebraic manipulation and careful attention to detail. By correctly applying the differentiation rules and the chain rule, we have successfully found the implicit derivative of the given equation. This is a crucial step in determining the slope of the tangent line at a specific point.

2. Isolate $\frac{dy}{dx}$

Next, we need to isolate $\frac{dy}{dx}$ in the equation $-4x + 2y + 2x\frac{dy}{dx} - 6y^2\frac{dy}{dx} = 0$. First, group the terms containing $\frac{dy}{dx}$: $2x\frac{dy}{dx} - 6y^2\frac{dy}{dx} = 4x - 2y$. Now, factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(2x - 6y^2) = 4x - 2y$. Finally, divide both sides by $(2x - 6y^2)$ to solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{4x - 2y}{2x - 6y^2}$. We can simplify this expression by dividing both the numerator and the denominator by 2: $\frac{dy}{dx} = \frac{2x - y}{x - 3y^2}$. This simplified expression gives us the slope of the tangent line at any point $(x, y)$ on the curve. The process of isolating $\frac{dy}{dx}$ involves careful algebraic manipulation and attention to detail. Each step is crucial to ensure the correct expression for the derivative. This expression will be used in the next step to find the slope of the tangent line at the specific point $(-1, -3)$. Mastering the technique of isolating the derivative is essential for solving implicit differentiation problems and understanding the behavior of curves.

3. Substitute the Point (-1, -3)

Now, we substitute the coordinates of the given point $(-1, -3)$ into the expression for $\frac{dy}{dx}$ to find the slope of the tangent line at that point. We have $\frac{dy}{dx} = \frac{2x - y}{x - 3y^2}$. Plugging in $x = -1$ and $y = -3$, we get: $\frac{dy}{dx} = \frac{2(-1) - (-3)}{(-1) - 3(-3)^2}$. Simplifying the numerator: $2(-1) - (-3) = -2 + 3 = 1$. Simplifying the denominator: $(-1) - 3(-3)^2 = -1 - 3(9) = -1 - 27 = -28$. Therefore, the slope of the tangent line at the point $(-1, -3)$ is: $\frac{dy}{dx} = \frac{1}{-28} = -\frac{1}{28}$. This calculation gives us the precise slope of the tangent line at the specified point on the curve. The substitution and simplification process requires careful arithmetic to avoid errors. The resulting slope provides valuable information about the direction and steepness of the curve at the point $(-1, -3)$. Understanding how to substitute and evaluate the derivative at a specific point is crucial for many applications of calculus, including curve sketching and optimization problems. This step completes the process of finding the slope of the tangent line to the given curve at the point $(-1, -3)$.

Conclusion

In conclusion, we have successfully found the slope of the tangent line to the curve $-2x^2 + 2xy - 2y^3 = 58$ at the point $(-1, -3)$. The slope is $-\frac{1}{28}$. This process involved implicit differentiation, isolating $\frac{dy}{dx}$, and substituting the given point into the resulting expression. This problem demonstrates the power of calculus in analyzing curves and finding their properties at specific points. By mastering implicit differentiation and related techniques, you can tackle a wide range of problems in mathematics and its applications. The ability to find the slope of a tangent line is a fundamental skill in calculus, with applications in optimization, curve sketching, and understanding rates of change. Each step in the solution process is crucial, from correctly applying differentiation rules to careful algebraic manipulation and substitution. By understanding these steps, you can confidently approach similar problems and gain a deeper appreciation for the elegance and utility of calculus. This example serves as a valuable learning experience, reinforcing the importance of precision and attention to detail in mathematical problem-solving. The techniques discussed here are applicable to a variety of curves and provide a solid foundation for further study in calculus and related fields.