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Understanding the Derivative dΘ/dx for x rcosΘ
Understanding the Derivative dΘ/dx for x rcosΘ
Introduction: When dealing with trigonometric functions in the context of polar coordinates, we often need to understand how the angle Θ changes relative to the Cartesian coordinate x. Specifically, in the context of the relationship x rcosΘ, finding dΘ/dx is crucial for numerous applications, including physics, engineering, and calculus.
Deriving dΘ/dx
Given the equation x rcosΘ, we can express Θ in terms of x and r. This requires a series of steps involving implicit differentiation.
Step 1: Express Θ in Terms of x and r
From the given equation, we have:
cosΘ x/r
Therefore, we can write:
Θ arccos(x/r)
Step 2: Implicit Differentiation
To find dΘ/dx, we need to differentiate both sides of the equation cosΘ x/r implicitly with respect to x.
-sinΘ * dΘ/dx 1/r
Solving for dΘ/dx, we get:
dΘ/dx -1/(r*sinΘ)
Step 3: Simplification
Since sinΘ √(1 - cos^2Θ), we can substitute cosΘ x/r into the equation:
sinΘ √(1 - (x/r)^2) √(r^2 - x^2)/r
Substituting this back into our derivative, we obtain:
dΘ/dx -1/(r*√(1 - (x/r)^2)) -1/√(r^2 - x^2)
Example and Verification
Let's verify our results through a step-by-step example using the relationship x rcosΘ and the derived expression for dΘ/dx.
Example Calculation
Suppose x 5 and r 10. We need to find dΘ/dx at this point.
1. Using Θ arccos(x/r), we find Θ arccos(5/10) arccos(0.5).
2. Similarly, we calculate sinΘ √(1 - (0.5)^2) √0.75 0.866.
3. Using the derived formula, we find:
dΘ/dx -1/√(10^2 - 5^2) -1/√75 ≈ -0.1155
Conclusion
To summarize, understanding the derivative dΘ/dx is essential in polar coordinate transformations. By applying trigonometric identities and implicit differentiation, we have derived the expression:
dΘ/dx -1/√(r^2 - x^2)
This can be extremely useful in various fields, including but not limited to, physics and engineering problems involving polar coordinates.