Problem 48 - Essential Calculus

Question

Find the arc length of the curve on the given interval. $$ x=\arcsin t, \quad y=\ln \sqrt{1-t^{2}} \quad 0 \leq t \leq \frac{1}{2} $$

Step-by-Step Solution

Answer

The arc length of the curve on the given interval is \( 1/2 \).

Step 1: Find the Derivatives

First we need to find the first derivatives of \( x(t) \) and \( y(t) \) with respect to \( t \). Let's calculate these derivatives:\n The derivative of \( x(t) \) with respect to \( t \) is: \( dx/dt = 1/ \sqrt{1 - t^2} \).\n The derivative of \( y(t) \) with respect to \( t \) is: \( dy/dt = -t/ \sqrt{1 - t^2} \).

Step 2: Length of the curve

To find the length of the curve, we will use the formula \[ L= \int_a^b \sqrt{[dx/dt]^2+[dy/dt]^2} dt \], a being 0 and b being 1/2. Substituting the values in, we get \( L = \int_0^{1/2} \sqrt{1^2 + (-t/ \sqrt{1 - t^2})^2} dt \). This simplifies to \( L = \int_0^{1/2} dt \), which is simply \( t|_0^{1/2} \).

Step 3: Evaluate the integral over the interval

The last step is to evaluate the integral over the interval from 0 to \( 1/2 \). The integral \( \int_0^{1/2} dt \) evaluates to \( t|_0^{1/2} \), which equals \( 1/2 \) when \( t \) equals \( 1/2 \) and equals 0 when \( t \) equals 0. The difference is \( 1/2 - 0 = 1/2 \). Therefore, the arc length of the curve on the given interval is \( 1/2 \).

Key Concepts

Derivative of Parametric Functions

Understanding the derivative of parametric functions is essential when dealing with arc length in calculus. Parametric functions express a set of related quantities as explicit functions of an independent variable, often time. In the case of arc length problems, you're usually given two functions, one for each coordinate, namely, x(t) and y(t), which are parameterized by a common variable t.

The rate of change of these functions with respect to t is key to finding the arc length because it gives us the slope of the tangent and thereby the infinitesimal changes along the curve. To find this derivative, you would use differentiation rules such as chain rule, product rule, quotient rule, and sometimes trigonometric derivatives.

Integral Calculus

Integral calculus is a part of calculus that deals with accumulation or aggregation of quantities, such as area under a curve, or in this situation, the total length along a curve. The fundamental theorem of calculus connects differentiation with integration, making them inverse processes. For the arc length problem, integration is used to sum up an infinite number of infinitesimal lengths along the curve obtained from the derivatives of the parametric functions.

The formula for arc length involves the integral of the square root of the sum of the squares of these derivatives. The exercise demonstrates this where the length L is found using the definite integral from point a to b on the function describing the curve's infinitesimal lengths.

Trigonometric Antiderivatives

Trigonometric antiderivatives are important when tackling integrals which involve trigonometric functions. In an arc length problem involving trigonometric expressions, identifying these antiderivatives is crucial to evaluate the integrals correctly.

For instance, when integrating functions involving sine, cosine, or their inverses, one must often use substitution methods or trigonometric identities to simplify the integrand into a form where the antiderivative is more obvious. The antiderivative is simply the function that, when differentiated, returns the integrand we started with. Familiarity with basic trigonometric antiderivatives and integration techniques aid greatly in solving calculus problems related to arc lengths and areas beneath trigonometric curves.