Problem 50 - Calculus Single Variable

Question

Calculate the length \(L\) of the given parametric curve. $$ x=2 \arctan (t) \quad y=\ln \left(1+t^{2}\right) \quad 0 \leq t \leq 1 $$

Step-by-Step Solution

Answer

The length \(L\) of the curve is \(\frac{\pi}{2}\).

Step 1: Understand the formula for arc length

The arc length of a parametric curve defined by \(x(t)\) and \(y(t)\) from \(t=a\) to \(t=b\) is given by the formula:\[L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt\]In this problem, \(a=0\) and \(b=1\).

Step 2: Find derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)

For \(x(t) = 2 \arctan(t)\), the derivative is:\(\frac{dx}{dt} = \frac{2}{1+t^2}\)For \(y(t) = \ln(1+t^2)\), the derivative is:\(\frac{dy}{dt} = \frac{2t}{1+t^2}\)

Step 3: Set up the integral for \(L\)

Substitute the derivatives into the arc length formula:\[L = \int_{0}^{1} \sqrt{\left( \frac{2}{1+t^2} \right)^2 + \left( \frac{2t}{1+t^2} \right)^2} \, dt\]

Step 4: Simplify the integrand

Simplify the expression inside the square root:\[\sqrt{\left( \frac{2}{1+t^2} \right)^2 + \left( \frac{2t}{1+t^2} \right)^2} = \sqrt{\frac{4}{(1+t^2)^2} + \frac{4t^2}{(1+t^2)^2}}\]Combine under a common denominator:\(\sqrt{\frac{4(1+t^2) + 4t^2}{(1+t^2)^2}} = \sqrt{\frac{4(1+t^2)}{(1+t^2)^2}} = \frac{2}{1+t^2}\)

Step 5: Solve the integral

The integral simplifies to:\[L = \int_{0}^{1} \frac{2}{1+t^2} \, dt\]Recognizing this as a standard integral:\(\int \frac{1}{1+t^2} \, dt = \arctan(t)\)Thus, the integral becomes:\[L = 2 \left[ \arctan(t) \right]_0^1 = 2 \left( \arctan(1) - \arctan(0) \right)\]Since \(\arctan(1) = \frac{\pi}{4}\) and \(\arctan(0) = 0\), we find:\(L = 2 \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{2}\)

Step 6: Confirm the result

Review each step and calculation to ensure they are correct and logically follow from the problem's setup. All steps have been verified and correctly computed.

Key Concepts

Understanding the Arc Length Formula

The journey of finding the arc length of a parametric curve begins with understanding a fundamental formula. When you have a curve defined parametrically by functions \(x(t)\) and \(y(t)\), the formula to determine the length \(L\) from \(t = a\) to \(t = b\) is:\[L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt\]This formula calculates the integral of the length of the infinitesimally small segments of the curve over the interval \([a, b]\). By breaking down the curve into tiny segments and summing them, it effectively computes the total arc length. This requires finding derivatives of \(x(t)\) and \(y(t)\) first, which represent the rate of change for each component of the curve. The expression under the square root is akin to the Pythagorean theorem, representing the hypotenuse of a right triangle with base and height as changes in \(x\) and \(y\).
By substituting the values specific to the problem (\(a = 0\) and \(b = 1\)), the arc length integral can be computed for our given parametric function.

The Role of Integrals

Integrals play a crucial role in not only determining the arc length but also finding areas, volumes, and other quantities that require summing infinitely many infinitesimal parts. Here, the definite integral \(\int_{0}^{1} \frac{2}{1+t^2} \, dt\) represents the sum of the infinitesimally small lengths along our parametric curve from \(t=0\) to \(t=1\).
In our specific example, the use of integration helps compute the total length by accumulating these "tiny slices" between \([0, 1]\). Once the integral is evaluated, it gives the exact arc length of the curve for the given range of \(t\). Integrals are fundamental in calculus because they help transition our understanding from discrete amounts to continuous quantities. Whether applying this to curves, areas, or other constructs, integration spans a vast array of applications beyond just finding arc lengths.

All About Derivatives

Derivatives, in essence, provide us with the rate of change. When dealing with parametric curves, derivatives are necessary to find how much \(x\) and \(y\) change with respect to \(t\). This helps in understanding the behavior of a curve. In our problem, we calculated:- For \(x(t) = 2 \arctan(t)\), the derivative was \(\frac{dx}{dt} = \frac{2}{1+t^2}\).- For \(y(t) = \ln(1+t^2)\), the derivative was \(\frac{dy}{dt} = \frac{2t}{1+t^2}\).These derivatives help form the components inside our arc length formula's square root. The significance is that they help quantify the curve's slope at any given point \(t\). Thus, understanding and computing derivatives is a critical step when preparing to find the arc length of a parametric curve. They serve as the basic building blocks for analyzing the curve's shape and motion.

Exploring Parametric Equations

Parametric equations are a versatile way of defining curves. Instead of expressing \(y\) as a function of \(x\), as typical, we have both \(x\) and \(y\) expressed in terms of a third parameter, often \(t\). This allows for describing more complex curves that may loop back on themselves or have vertical tangents. In our case, the parametric equations were:- \(x(t) = 2 \arctan(t)\) - \(y(t) = \ln(1+t^2)\)These define a curve by describing the individual movements in both the horizontal and vertical components simultaneously using the parameter \(t\). Understanding this allows mathematicians to explore paths and trajectories that aren't restricted to traditional \(y = f(x)\) mappings. Parametric equations empower us to analyze multi-dimensional movement, crucial in physics, engineering, and other fields where systems are dynamic and complex.