Analyzing The Equation X³ + (y-1)² = 1 Finding Derivatives And Tangent Lines
Hey guys! Today, we're diving deep into the fascinating world of calculus and analytical geometry with a twist. We're going to explore the equation x³ + (y - 1)² = 1, a seemingly simple equation that holds a lot of interesting secrets. We'll tackle finding the derivative y', pinpoint where the tangent lines go horizontal, and even where they stand tall and vertical. Buckle up, because this is going to be a fun ride!
(a) Finding y': Implicit Differentiation in Action
The heart of calculus often lies in its ability to handle implicit functions, and that's precisely what we have here. Our equation, x³ + (y - 1)² = 1, doesn't explicitly express y in terms of x, so we need to employ the powerful technique of implicit differentiation. This means we'll differentiate both sides of the equation with respect to x, keeping in mind that y is a function of x, meaning we'll need to use the chain rule whenever we differentiate a term involving y. Let's get started!
First, let's rewrite the equation to make the differentiation process clearer: x³ + (y - 1)² = 1. Now, we differentiate both sides with respect to x. Differentiating x³ with respect to x is straightforward using the power rule, which gives us 3x². For the second term, (y - 1)², we'll need the chain rule. The derivative of the outer function, something squared, is two times that something. So, we get 2(y - 1). But remember, y is a function of x, so we need to multiply by the derivative of the inner function, (y - 1), with respect to x. The derivative of y with respect to x is simply y', and the derivative of the constant -1 is 0. Thus, the derivative of (y - 1)² with respect to x is 2(y - 1)y'. Finally, the derivative of the constant 1 on the right side of the equation is 0.
Putting it all together, the derivative of the equation x³ + (y - 1)² = 1 with respect to x is:
3x² + 2(y - 1)y' = 0
Now, our goal is to isolate y', so we can express it in terms of x and y. Let's subtract 3x² from both sides:
2(y - 1)y' = -3x²
Next, we divide both sides by 2(y - 1) to get y' by itself:
y' = -3x² / [2(y - 1)]
And there we have it! We've successfully found y', the derivative of y with respect to x, using implicit differentiation. This expression tells us the slope of the tangent line to the curve at any point (x, y) that satisfies the original equation. But our journey doesn't end here. We're now ready to explore where these tangent lines are horizontal and vertical, which will give us valuable insights into the shape and behavior of the curve.
In summary, we found the derivative y' by applying the principles of implicit differentiation and the chain rule. This is a fundamental technique in calculus, and mastering it opens the door to analyzing a wide range of functions and curves. Remember, the key is to treat y as a function of x and carefully apply the chain rule whenever differentiating a term involving y. Now, let's move on to the next exciting part: finding the horizontal and vertical tangents!
(b) Finding Horizontal and Vertical Tangents
Now that we have the derivative, y' = -3x² / [2(y - 1)], we can use it to find the points on the curve where the tangent line is horizontal or vertical. This is where the fun really begins, as we start to visualize the shape of the curve and how the tangent lines behave at different points. Remember, the derivative represents the slope of the tangent line, so horizontal tangents correspond to where the slope is zero, and vertical tangents correspond to where the slope is undefined.
(i) Horizontal Tangents: When the Slope is Zero
A horizontal tangent line has a slope of zero. This means we need to find the points (x, y) on the curve where y' = 0. Looking at our expression for y', y' = -3x² / [2(y - 1)], we see that a fraction is zero only when its numerator is zero. So, we need to solve the equation:
-3x² = 0
Dividing both sides by -3 gives us:
x² = 0
Taking the square root of both sides, we get:
x = 0
So, the tangent line is horizontal when x = 0. But we're not done yet! We need to find the corresponding y-coordinate(s) by plugging x = 0 back into the original equation, x³ + (y - 1)² = 1:
0³ + (y - 1)² = 1
This simplifies to:
(y - 1)² = 1
Taking the square root of both sides, we get:
y - 1 = ±1
Adding 1 to both sides, we have two possible values for y:
y = 1 + 1 = 2
y = 1 - 1 = 0
Therefore, there are two points on the curve where the tangent line is horizontal: (0, 2) and (0, 0). These points are critical points where the curve changes direction, and the horizontal tangents provide valuable information about the curve's shape.
(ii) Vertical Tangents: When the Slope is Undefined
A vertical tangent line has an undefined slope. This happens when the denominator of our expression for y' is zero, because division by zero is undefined. So, we need to find the points (x, y) on the curve where the denominator, 2(y - 1), is zero:
2(y - 1) = 0
Dividing both sides by 2, we get:
y - 1 = 0
Adding 1 to both sides, we find:
y = 1
So, the tangent line is vertical when y = 1. Now, we need to find the corresponding x-coordinate(s) by plugging y = 1 back into the original equation, x³ + (y - 1)² = 1:
x³ + (1 - 1)² = 1
This simplifies to:
x³ = 1
Taking the cube root of both sides, we get:
x = 1
Therefore, there is one point on the curve where the tangent line is vertical: (1, 1). At this point, the curve has a sharp turn, and the vertical tangent indicates a change in the curve's horizontal direction.
In summary, we found the horizontal tangents by setting the numerator of y' to zero and solving for x, then finding the corresponding y values. We found the vertical tangents by setting the denominator of y' to zero and solving for y, then finding the corresponding x values. These horizontal and vertical tangents give us a clear picture of the curve's key features, such as its turning points and points of inflection. By combining this information with the derivative, we can gain a deep understanding of the curve's behavior.
Conclusion: Putting It All Together
Wow, guys, we've really taken a deep dive into the equation x³ + (y - 1)² = 1! We started by finding the derivative y' using implicit differentiation, which gave us a powerful tool for analyzing the curve's behavior. Then, we used y' to find the points where the tangent lines are horizontal and vertical, revealing crucial information about the curve's shape and direction.
We discovered that the tangent line is horizontal at the points (0, 2) and (0, 0), indicating turning points where the curve changes its vertical direction. We also found that the tangent line is vertical at the point (1, 1), suggesting a sharp turn or a point of inflection where the curve changes its horizontal direction. By analyzing these tangent lines, we've gained a comprehensive understanding of the curve's characteristics.
This exercise highlights the power of calculus in analyzing curves and functions. Implicit differentiation allows us to handle equations where y is not explicitly defined in terms of x, and the derivative provides invaluable information about the slope of the tangent line and the curve's behavior. Finding horizontal and vertical tangents is a key technique for sketching curves and understanding their properties.
So, the next time you encounter a seemingly simple equation, remember that there might be a whole world of mathematical beauty hidden beneath the surface. By applying the principles of calculus, we can unlock these secrets and gain a deeper appreciation for the elegance and power of mathematics. Keep exploring, keep questioning, and keep having fun with math!
