Finding Tangent Line Slope To Curve √4x+3y + √2xy At (2,8)

In calculus, determining the slope of a tangent line to a curve at a specific point is a fundamental concept with wide-ranging applications. This article delves into the process of finding the slope of a tangent line, providing a step-by-step approach and illustrative examples. We'll specifically address the problem of finding the slope of the tangent line to the curve defined by the equation $\sqrt{4x + 3y} + \sqrt{2xy} = 11.313708498985$ at the point $(2, 8)$.

Understanding Tangent Lines and Slopes

Before we dive into the specific problem, let's establish a solid understanding of the underlying concepts. A tangent line to a curve at a given point is a straight line that touches the curve at that point and has the same instantaneous direction as the curve. The slope of a line, often denoted by m, represents the steepness and direction of the line. It's defined as the change in the vertical coordinate (y) divided by the change in the horizontal coordinate (x), often expressed as m = Δy / Δx.

The slope of a tangent line at a specific point on a curve provides valuable information about the rate of change of the function at that point. In other words, it tells us how much the y-value of the function is changing with respect to the x-value at that particular location. This concept is crucial in various fields, including physics (velocity and acceleration), economics (marginal cost and revenue), and engineering (optimization problems).

Implicit Differentiation: The Key Technique

The equation $\sqrt{4x + 3y} + \sqrt{2xy} = 11.313708498985$ presents a challenge because it's not explicitly solved for y in terms of x. This means we can't directly differentiate the equation with respect to x using standard differentiation rules. Instead, we need to employ a technique called implicit differentiation.

Implicit differentiation allows us to find the derivative dy/dx even when y is not explicitly defined as a function of x. The core idea is to differentiate both sides of the equation with respect to x, treating y as an implicit function of x. This means we'll need to apply the chain rule whenever we differentiate a term involving y. The chain rule states that the derivative of a composite function f(g(x)) is f'(g(x)) * g'(x). In our case, if we have a term like y^2, its derivative with respect to x will be 2y * (dy/dx).

The power rule, which states that the derivative of x^n is nx^(n-1), is another fundamental rule we'll use extensively. For instance, the derivative of x^3 is 3x^2. When combined with the chain rule, the power rule enables us to differentiate expressions involving functions of x raised to a power. Remember, implicit differentiation is a powerful tool that extends our ability to find derivatives beyond explicitly defined functions.

Step-by-Step Solution: Finding the Slope

Let's now apply implicit differentiation to our given equation and find the slope of the tangent line at the point (2, 8).

1. Differentiate both sides of the equation with respect to x.

Given equation: $\sqrt{4x + 3y} + \sqrt{2xy} = 11.313708498985$

Differentiating both sides with respect to x, we get:

$\frac{d}{dx}(\sqrt{4x + 3y}) + \frac{d}{dx}(\sqrt{2xy}) = \frac{d}{dx}(11.313708498985)$

2. Apply the chain rule and power rule.

• For the first term, $\frac{d}{dx}(\sqrt{4x + 3y})$: We can rewrite this as $\frac{d}{dx}((4x + 3y)^{1/2})$. Applying the chain rule, we get $\frac{1}{2}(4x + 3y)^{-1/2} * \frac{d}{dx}(4x + 3y)$. Further differentiating $(4x + 3y)$ with respect to x gives us $4 + 3\frac{dy}{dx}$. So, the derivative of the first term is $\frac{1}{2}(4x + 3y)^{-1/2} * (4 + 3\frac{dy}{dx})$.
• For the second term, $\frac{d}{dx}(\sqrt{2xy})$: We can rewrite this as $\frac{d}{dx}((2xy)^{1/2})$. Applying the chain rule, we get $\frac{1}{2}(2xy)^{-1/2} * \frac{d}{dx}(2xy)$. To differentiate 2xy, we need to use the product rule, which states that the derivative of uv is u'v + uv'. Here, u = 2x and v = y. So, $\frac{d}{dx}(2xy) = 2y + 2x\frac{dy}{dx}$. Therefore, the derivative of the second term is $\frac{1}{2}(2xy)^{-1/2} * (2y + 2x\frac{dy}{dx})$.
• The derivative of the constant $11.313708498985$ is 0.

Putting it all together, we have:

$\frac{1}{2}(4x + 3y)^{-1/2} * (4 + 3\frac{dy}{dx}) + \frac{1}{2}(2xy)^{-1/2} * (2y + 2x\frac{dy}{dx}) = 0$

3. Substitute the point (2, 8) into the equation.

Now, we substitute x = 2 and y = 8 into the equation we obtained in the previous step:

$\frac{1}{2}(4(2) + 3(8))^{-1/2} * (4 + 3\frac{dy}{dx}) + \frac{1}{2}(2(2)(8))^{-1/2} * (2(8) + 2(2)\frac{dy}{dx}) = 0$

Simplifying, we get:

$\frac{1}{2}(32)^{-1/2} * (4 + 3\frac{dy}{dx}) + \frac{1}{2}(32)^{-1/2} * (16 + 4\frac{dy}{dx}) = 0$

4. Solve for dy/dx.

Let's simplify further and solve for $\frac{dy}{dx}$, which represents the slope of the tangent line at the point (2, 8):

$\frac{1}{2\sqrt{32}}(4 + 3\frac{dy}{dx}) + \frac{1}{2\sqrt{32}}(16 + 4\frac{dy}{dx}) = 0$

Multiply both sides by $2\sqrt{32}$ to eliminate the fractions:

$(4 + 3\frac{dy}{dx}) + (16 + 4\frac{dy}{dx}) = 0$

Combine like terms:

$20 + 7\frac{dy}{dx} = 0$

Isolate $\frac{dy}{dx}$:

$7\frac{dy}{dx} = -20$

$\frac{dy}{dx} = -\frac{20}{7}$

Therefore, the slope of the tangent line to the curve at the point (2, 8) is -20/7.

Visualizing the Tangent Line

It's helpful to visualize the tangent line we just calculated. Imagine the curve defined by the given equation. At the point (2, 8), the tangent line is a straight line that touches the curve at that point and has a slope of -20/7. This negative slope indicates that the line is decreasing as we move from left to right. The steeper the slope (in magnitude), the more rapidly the line is decreasing.

Graphing the curve and the tangent line can provide a visual confirmation of our result. You can use graphing calculators or online tools to plot the equation and the tangent line. The visual representation will show how the tangent line closely approximates the curve in the vicinity of the point (2, 8).

Applications of Tangent Lines and Slopes

The concept of tangent lines and their slopes has numerous applications in various fields. Here are a few examples:

• Optimization: In optimization problems, we often need to find the maximum or minimum value of a function. Tangent lines play a crucial role in this process. At a local maximum or minimum, the tangent line to the curve is horizontal, meaning its slope is zero. By finding points where the derivative (which represents the slope of the tangent line) is zero, we can identify potential maximum or minimum points.
• Related Rates: Related rates problems involve finding the rate of change of one quantity in terms of the rate of change of another quantity. Tangent lines and derivatives are essential tools for solving these problems. For example, we might want to find how the volume of a balloon is changing as we inflate it, given the rate at which the radius is increasing.
• Approximation: Tangent lines can be used to approximate the value of a function near a specific point. This is based on the idea that the tangent line closely resembles the curve in a small neighborhood around the point of tangency. This approximation technique is particularly useful when the function is complex and difficult to evaluate directly.

Common Mistakes and How to Avoid Them

When finding the slope of a tangent line using implicit differentiation, there are several common mistakes to watch out for:

• Forgetting the chain rule: When differentiating terms involving y with respect to x, it's crucial to apply the chain rule. Remember to multiply by dy/dx whenever you differentiate a function of y.
• Incorrectly applying the product rule: The product rule is essential for differentiating products of functions. Make sure you apply it correctly, especially when dealing with terms like 2xy.
• Algebraic errors: Simplifying the equation after differentiation can be tricky. Be careful with algebraic manipulations and double-check your work to avoid errors.
• Not substituting the point: After differentiating and before solving for dy/dx, remember to substitute the given point (x, y) into the equation. This step is crucial for finding the slope of the tangent line at that specific point.

By being aware of these common mistakes and taking the time to carefully review your work, you can improve your accuracy and avoid errors.

Conclusion

Finding the slope of a tangent line is a fundamental concept in calculus with wide-ranging applications. Implicit differentiation is a powerful technique that allows us to find derivatives even when functions are not explicitly defined. By following the step-by-step approach outlined in this article, you can confidently tackle problems involving tangent lines and slopes. Remember to practice and visualize the concepts to deepen your understanding and appreciation for the power of calculus.

This article provided a comprehensive guide to finding the slope of the tangent line to the curve $\sqrt{4x + 3y} + \sqrt{2xy} = 11.313708498985$ at the point (2, 8). We explored the underlying concepts, detailed the steps involved in implicit differentiation, and discussed common mistakes to avoid. With practice and a solid understanding of the principles, you'll be well-equipped to solve similar problems and apply these techniques in various contexts.