Hey guys! Today, we're diving into a fascinating problem involving implicit differentiation. We'll explore the equation x⁴ + y⁴ = 82, tackle finding dy/dx, and then zoom in on the tangent line at a specific point. So, buckle up and let's get started!
1. Cracking Implicit Differentiation to Find dy/dx
Let's dive straight into finding dy/dx for the equation x⁴ + y⁴ = 82 using implicit differentiation. Now, what exactly is implicit differentiation? Well, it's our trusty tool when we can't easily isolate y in terms of x. Our equation here is a perfect example. We're not dealing with a simple y = f(x) situation; instead, we have a relationship tying x and y together. Think of it like this: y is implicitly defined as a function of x, even though we don't have an explicit formula for it.
So, how does implicit differentiation work its magic? The core idea is to differentiate both sides of the equation with respect to x, keeping in mind that y is a function of x. This is where the chain rule becomes our best friend. Remember the chain rule? It tells us how to differentiate composite functions. When we encounter a term involving y, we'll treat it as a function within a function, applying the chain rule to peel away the layers. Let's break it down step by step:
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Differentiate both sides: We start by applying the differentiation operator d/dx to both sides of our equation: d/dx (x⁴ + y⁴) = d/dx (82)
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Apply the power rule and chain rule: On the left side, we have the sum of two terms. We'll differentiate each term separately. For x⁴, the power rule (d/dx xⁿ = nxⁿ⁻¹) gives us 4x³. For y⁴, we use the power rule and the chain rule. We treat y⁴ as a composite function, where the outer function is the power function and the inner function is y(x). Applying the power rule gives us 4y³, and then the chain rule tacks on dy/dx: 4x³ + 4y³ (dy/dx) = d/dx (82)
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Differentiate the constant: On the right side, we have the constant 82. The derivative of any constant is always 0: 4x³ + 4y³ (dy/dx) = 0
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Isolate dy/dx: Now comes the algebraic maneuvering! Our goal is to isolate dy/dx on one side of the equation. Let's subtract 4x³ from both sides: 4y³ (dy/dx) = -4x³
And finally, divide both sides by 4y³: dy/dx = (-4x³) / (4y³)
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Simplify: We can simplify our expression by canceling out the common factor of 4: dy/dx = -x³ / y³
And there we have it! We've successfully found dy/dx using implicit differentiation. This expression tells us the slope of the tangent line to the curve x⁴ + y⁴ = 82 at any point (x, y) on the curve. It's like having a slope-finding machine for this specific curve!
This formula, dy/dx = -x³/y³, is super important because it gives us the slope of the tangent line at any point (x, y) on the curve defined by x⁴ + y⁴ = 82. Think of it as a slope calculator that works for every single point on this curve. To get the actual equation of a tangent line, we'll need a specific point, which leads us perfectly into the next part of our adventure.
2. Finding the Tangent Line Equation at (1, -3)
Now that we've conquered finding dy/dx, let's put it to work! Our mission: to find the equation of the tangent line to the curve x⁴ + y⁴ = 82 at the point (1, -3). This is where things get really tangible. We're not just dealing with abstract derivatives anymore; we're finding the equation of a line that actually touches the curve at a specific spot.
To find the equation of a line, we need two key ingredients: its slope and a point on the line. Luckily, we've already got both! We know the point (1, -3), and we have a formula for the slope, dy/dx = -x³/y³. All that's left is to plug in the coordinates of our point into the dy/dx formula to find the slope at that specific location.
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Calculate the slope: We'll substitute x = 1 and y = -3 into our formula for dy/dx: dy/dx = -(1)³ / (-3)³ = -1 / -27 = 1/27
So, the slope of the tangent line at the point (1, -3) is 1/27. That's a pretty gentle slope, meaning the tangent line is fairly flat at this point.
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Use the point-slope form: Now that we have the slope (m = 1/27) and a point (x₁, y₁) = (1, -3), we can use the point-slope form of a line to write the equation. The point-slope form is: y - y₁ = m(x - x₁)
Plugging in our values, we get: y - (-3) = (1/27)(x - 1)
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Simplify to slope-intercept form (optional): While the point-slope form is perfectly valid, we can also simplify the equation to the more familiar slope-intercept form (y = mx + b). Let's do that: y + 3 = (1/27)x - 1/27
Subtract 3 from both sides: y = (1/27)x - 1/27 - 3
To combine the constants, we need a common denominator. We can rewrite 3 as 81/27: y = (1/27)x - 1/27 - 81/27
And finally, combine the fractions: y = (1/27)x - 82/27
So, the equation of the tangent line to the curve x⁴ + y⁴ = 82 at the point (1, -3) is y = (1/27)x - 82/27. Woohoo! We've found the line that just kisses the curve at that specific spot. This equation gives us a complete picture of the tangent line, telling us its slope (1/27) and its y-intercept (-82/27).
This result is pretty cool, guys! It means we've not just found some abstract mathematical formula, but we've actually pinned down the equation of a line that has a very specific geometric relationship with our original curve. It touches the curve at (1, -3) and has a slope of 1/27 there. Think about visualizing this – it really brings the math to life!
Final Thoughts
There we have it, folks! We've successfully navigated the world of implicit differentiation, found dy/dx, and even calculated the equation of a tangent line. This journey highlights the power of implicit differentiation in handling equations where y isn't explicitly defined as a function of x. It's a versatile tool that unlocks a whole new level of problem-solving in calculus.
Remember, the key to implicit differentiation is to treat y as a function of x and diligently apply the chain rule. Once you've mastered that, the rest is algebraic maneuvering. And as we saw in finding the tangent line, these abstract derivatives can have very concrete geometric interpretations.
So, keep practicing, keep exploring, and keep the calculus fires burning! You've got this!