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Magazine Chapter 8: Problem 18
Find the exact length of the curve. $$ y=\sqrt{x-x^{2}}+\sin ^{-1}(\sqrt{x}) $$
Short Answer
Expert verified
The exact length of the curve is \( \frac{\pi}{2} \).
Step by step solution
01
Understanding the Function
We need to find the length of the curve defined by the function \[ y = \sqrt{x-x^2} + \sin^{-1}(\sqrt{x}) \]from its domain. The first step is getting a sense of the entire function and its components. The function consists of a square root and an inverse sine.
02
Finding the Derivative
To find the length of the curve, we'll need the derivative. Differentiate:\[ \frac{dy}{dx} = \frac{d}{dx}(\sqrt{x-x^2}) + \frac{d}{dx}(\sin^{-1}(\sqrt{x})) \]Both derivatives require the chain rule. Compute each separately.
03
Differentiating the Square Root
The derivative of \( y_1 = \sqrt{x-x^2} \) is:\[ \frac{dy_1}{dx} = \frac{1}{2\sqrt{x-x^2}}(1-2x) \]This comes from applying the chain rule and the derivative of a square root function.
04
Differentiating the Inverse Sine
The derivative of \( y_2 = \sin^{-1}(\sqrt{x}) \) is:\[ \frac{dy_2}{dx} = \frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{1}{2\sqrt{x}} \]Simplify this to:\[ \frac{1}{2\sqrt{x-x^2}} \]
05
Sum of Derivatives
Add the derivatives of the two parts:\[ \frac{dy}{dx} = \frac{1-2x}{2\sqrt{x-x^2}} + \frac{1}{2\sqrt{x-x^2}} \]Combine terms to simplify:\[ \frac{dy}{dx} = \frac{2-2x}{2\sqrt{x-x^2}} = \frac{1-x}{\sqrt{x-x^2}} \]
06
Curve Length Formula
The formula for the length of a curve from \(a\) to \(b\) is:\[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]Let's prepare to substitute \( \frac{dy}{dx} \) in this formula.
07
Substitute and Simplify
Substitute \( \frac{dy}{dx} = \frac{1-x}{\sqrt{x-x^2}} \) to get:\[ 1 + \left( \frac{1-x}{\sqrt{x-x^2}} \right)^2 = 1 + \frac{(1-x)^2}{x-x^2} \]This simplifies to:\[ \frac{1-x^2}{x-x^2} = \frac{1}{x-x^2} \]. Use trigonometric identities to simplify.
08
Evaluate the Integral
Now evaluate:\[ L = \int_0^1 \frac{1}{\sqrt{x-x^2}} \, dx \]. Using a substitution like \( x = \sin^2(t) \), simplifies the integral to \( \int_0^{\pi/2} \, dt \).The integral evaluates to \( t \) as \( [0,\pi/2] \). Thus, \( L = \pi/2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative of a Function
The derivative of a function gives us the rate at which the function is changing at any given point. When it comes to finding the length of a curve, the derivative is crucial, as it forms part of the curve length formula. To find the derivative of a function like \( y = \sqrt{x-x^2} + \sin^{-1}(\sqrt{x}) \), you need to differentiate it with respect to \( x \). Understanding how different components of the function behave is essential.
- For the square root part, \( y_1 = \sqrt{x-x^2} \), differentiation involves applying the chain rule to handle the composition of the function.
- The derivative of \( y_1 \) is calculated using: \( \frac{1}{2\sqrt{x-x^2}}(1-2x) \).
- For the inverse sine part, \( y_2 = \sin^{-1}(\sqrt{x}) \), the derivative involves both inverse trigonometric differentiation and the chain rule.
- This derivative simplifies to \( \frac{1}{2\sqrt{x-x^2}} \).
Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate compositions of functions. When you have functions nested within each other, like \( \sqrt{x-x^2} \), the chain rule helps break the process into manageable steps.
- Apply the chain rule by differentiating the outer function first, then the inner function.
- For instance, differentiate \( \sqrt{x-x^2} \) by treating the expression inside the square root as a separate entity.
- This results in the derivative \( \frac{1}{2\sqrt{x-x^2}} \) after multiplying by the derivative of the inside, which is \( 1-2x \).
- Similarly, when differentiating \( \sin^{-1}(\sqrt{x}) \), first find the derivative of \( \sin^{-1}(u) \) where \( u = \sqrt{x} \), and then multiply it by the derivative of \( u \).
Integral Calculus
Integral calculus is all about finding the size or length of something, such as the area under a curve. When it comes to a curve's length, the integral provides a systematic way to calculate it across an interval. The formula for the curve length from \( a \) to \( b \) is:\[L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx\]Here鈥檚 what happens at each stage:
- First, substitute your derivative \( \frac{dy}{dx} \) into the formula. For this exercise, that's \( \frac{1-x}{\sqrt{x-x^2}} \).
- Simplify the expression inside the integrand. This might involve algebraic manipulation or even trigonometric identities.
- Finally, solve the integral. This evaluates the total curve length over the specified interval.
Trigonometric Substitution
Trigonometric substitution is a powerful method used to simplify integrals that are otherwise tricky to solve. It relies on replacing variables with trigonometric functions, which can often turn complicated integrals into more manageable ones.
- In this situation, look at replacing expressions like \( x = \sin^2(t) \) when the integral involves \( \sqrt{x-x^2} \).
- This substitution simplifies the expression drastically, as it turns out that \( \sqrt{x-x^2} \) transforms into \( \cos(t) \).
- The integration bounds also change since the angle, \( t \), corresponds to \( x \) values from 0 to 1, mapping onto \( t \) values from 0 to \( \pi/2 \).
- Finally, integrate with respect to \( t \) rather than \( x \), resulting in a simplified calculation.
