Prove: If √(1-x²) + √(1-y²) = A(x-y), Then Dy/dx

by SLV Team 49 views

Hey guys! Today, we're diving into a super interesting math problem. Our mission, should we choose to accept it (and we totally do!), is to prove that if √(1-x²) + √(1-y²) = a(x-y), then dy/dx = √(1-y²)/√(1-x²). Sounds like a mouthful, right? But don't worry, we'll break it down step by step, making it as clear as crystal. So grab your favorite beverage, maybe a notepad, and let's get started!

Understanding the Problem

Before we jump into the solution, let's make sure we're all on the same page. The problem presents us with an equation relating x and y: √(1-x²) + √(1-y²) = a(x-y). Our goal is to find the derivative of y with respect to x (dy/dx) and show that it equals √(1-y²)/√(1-x²). This involves implicit differentiation, which is a fancy way of saying we'll differentiate both sides of the equation with respect to x, treating y as a function of x. Remember, when we differentiate y terms, we'll need to use the chain rule. Think of it like peeling an onion – we go layer by layer until we get to the core! This problem combines algebraic manipulation with calculus, making it a fantastic exercise for honing our skills.

So, understanding what we're trying to prove is half the battle. We know the given equation and the target derivative. Now, let's strategize our attack. We'll use trigonometric substitution to simplify the initial equation, making it easier to work with. Then, we will differentiate both sides with respect to x, carefully applying the chain rule wherever necessary. Finally, we'll rearrange the terms to isolate dy/dx and show that it matches the expression we're aiming for. This might seem like a lot of steps, but each one is manageable, and by following them systematically, we'll conquer this problem. Remember, math isn't about memorizing formulas; it's about understanding the underlying concepts and applying them creatively.

Let’s look at the key concepts we'll be using! First, implicit differentiation is super important because we don't have y explicitly defined as a function of x. Second, the chain rule will pop up whenever we differentiate a term involving y. Third, trigonometric substitution helps simplify the square roots. By understanding these three concepts we are well on our way.

Solution

Here's where the magic happens! We'll walk through the solution step-by-step, explaining each move we make.

Step 1: Trigonometric Substitution

To simplify the given equation, we'll use trigonometric substitution. Let's set:

  • x = sin(θ)
  • y = sin(φ)

This substitution is motivated by the identity sin²(x) + cos²(x) = 1. It will help us get rid of the square roots.

Substituting these into the given equation, we get:

√(1 - sin²(θ)) + √(1 - sin²(φ)) = a(sin(θ) - sin(φ))

Which simplifies to:

cos(θ) + cos(φ) = a(sin(θ) - sin(φ))

Step 2: Applying Trigonometric Identities

Now, we'll use the sum-to-product trigonometric identities to further simplify the equation. These identities are:

  • cos(θ) + cos(φ) = 2 cos((θ + φ)/2) cos((θ - φ)/2)
  • sin(θ) - sin(φ) = 2 cos((θ + φ)/2) sin((θ - φ)/2)

Substituting these identities into our equation, we get:

2 cos((θ + φ)/2) cos((θ - φ)/2) = a * 2 cos((θ + φ)/2) sin((θ - φ)/2)

Step 3: Simplifying the Equation

We can now cancel out the common factor of 2 cos((θ + φ)/2) from both sides (assuming cos((θ + φ)/2) ≠ 0):

cos((θ - φ)/2) = a sin((θ - φ)/2)

Dividing both sides by cos((θ - φ)/2), we get:

1 = a tan((θ - φ)/2)

Which implies:

tan((θ - φ)/2) = 1/a

Therefore:

(θ - φ)/2 = arctan(1/a)

And:

θ - φ = 2 arctan(1/a)

Let's call 2 arctan(1/a) = k, where k is a constant.

So, we have:

θ - φ = k

Step 4: Implicit Differentiation

Now, we'll differentiate both sides of the equation θ - φ = k with respect to x. Remember that θ = arcsin(x) and φ = arcsin(y).

d/dx (arcsin(x) - arcsin(y)) = d/dx (k)

The derivative of arcsin(x) is 1/√(1 - x²), and the derivative of arcsin(y) is 1/√(1 - y²) * dy/dx (using the chain rule).

So, we have:

1/√(1 - x²) - 1/√(1 - y²) * dy/dx = 0

Step 5: Isolating dy/dx

Now, we'll isolate dy/dx:

1/√(1 - y²) * dy/dx = 1/√(1 - x²)

Multiplying both sides by √(1 - y²), we get:

dy/dx = √(1 - y²)/√(1 - x²)

Which is exactly what we wanted to prove!

Conclusion

And there you have it, folks! We've successfully proven that if √(1-x²) + √(1-y²) = a(x-y), then dy/dx = √(1-y²)/√(1-x²). This problem demonstrates the power of trigonometric substitution and implicit differentiation. It might have seemed daunting at first, but by breaking it down into smaller, manageable steps, we were able to conquer it. Keep practicing, keep exploring, and most importantly, keep having fun with math! Remember, every problem you solve makes you a little bit stronger and a little bit wiser. So, keep challenging yourself and never stop learning!

This was a wonderful exploration of calculus and trigonometric identities. Hopefully, this step-by-step solution helps you understand the concepts better. Keep practicing and exploring, and you'll become a math whiz in no time!

Key takeaways:

  • Trigonometric substitution can simplify equations with square roots.
  • Implicit differentiation is essential when y is not explicitly defined as a function of x.
  • The chain rule is crucial when differentiating terms involving y with respect to x.

Now go forth and conquer more math problems! You got this! Let me know in the comments if you have any questions or want to see more problems like this. Happy calculating!