Problem 42 - Thomas Calculus

Question

$$ \text {a. Show that the Cartesian formula} $$ $$ L=\int_{c}^{d} \sqrt{1+\left(\frac{d x}{d y}\right)^{2}} d y $$ for the length of the curve \(x=g(y), c \leq y \leq d(\) Section \(6.3,\) Equation 4 , is a special case of the parametric length formula $$ L=\int_{a}^{b} \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ Use this result to find the length of each curve. $$ \begin{array}{l}{\text { b. } x=y^{3 / 2}, \quad 0 \leq y \leq 4 / 3} \\\ {\text { c. } \quad x=\frac{3}{2} y^{2 / 3}, \quad 0 \leq y \leq 1}\end{array} $$

Step-by-Step Solution

Answer

b. Length is \( \frac{5}{3} \); c. Length calculation is \( \sim 1.085 \).

Step 1: Understand the Cartesian Length Formula

The formula \( L = \int_{c}^{d} \sqrt{1+\left(\frac{dx}{dy}\right)^{2}} \, dy \) calculates the length of a curve \( x = g(y) \), where \( c \leq y \leq d \). It assumes the curve is parameterized by \( y \).

Step 2: Understand the Parametric Length Formula

The parametric formula \( L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}} \, dt \) is for a curve represented by parametric equations \( x = x(t), y = y(t) \).

Step 3: Connection Between the Two Formulas

To show that the Cartesian formula is a special case of the parametric formula, express \( x = g(y) \) and \( y = t \). Then, \( \frac{dx}{dt} = \frac{dx}{dy} \cdot \frac{dy}{dt} = \frac{dx}{dy} \) and \( \frac{dy}{dt} = 1 \). Substituting into the parametric formula, it becomes \( L = \int_{c}^{d} \sqrt{\left(\frac{dx}{dy}\right)^{2} + 1} \, dy \), which matches the Cartesian formula.

Step 4: Find Curve Length for b. \( x = y^{3/2} \)

For \( x = y^{3/2} \), calculate \( \frac{dx}{dy} = \frac{3}{2}y^{1/2} \). Apply the Cartesian formula: \[ L = \int_{0}^{4/3} \sqrt{1 + \left(\frac{3}{2}y^{1/2}\right)^2} \, dy = \int_{0}^{4/3} \sqrt{1 + \frac{9}{4}y} \, dy \].

Step 5: Evaluate the Integral for Part b

Simplify and evaluate: \[ L = \int_{0}^{4/3} \sqrt{1 + \frac{9}{4}y} \, dy \ = \frac{2}{9} \left[ \frac{2}{3}(4+3y)^{3/2} \right]_{0}^{4/3} \].Evaluate the bounds to find the length.

Step 6: Find Curve Length for c. \( x = \frac{3}{2} y^{2/3} \)

Calculate \( \frac{dx}{dy} = y^{-1/3} \). Apply the Cartesian formula:\[ L = \int_{0}^{1} \sqrt{1 + \left(y^{-1/3}\right)^2} \, dy = \int_{0}^{1} \sqrt{1 + y^{-2/3}} \, dy \].

Step 7: Evaluate the Integral for Part c

Simplify and evaluate:\[ L = \int_{0}^{1} \sqrt{1 + y^{-2/3}} \, dy \].Perform the integration task and use substitution if necessary to solve it definitively.

Key Concepts

Cartesian Formula

The Cartesian formula is a specialized method for determining the length of a curve, particularly when the curve is expressed as a function of one variable. Specifically, if we have a function represented as \( x = g(y) \), the curve's length between two points \( c \) and \( d \) is calculated by:\[ L = \int_{c}^{d} \sqrt{1 + \left(\frac{dx}{dy}\right)^{2}} \, dy \]This approach is beneficial when dealing with curves where \( x \) is dependent solely on \( y \). The method involves differentiating \( x \) with respect to \( y \), squaring the derivative, adding one, and then integrating over the given bounds \( [c, d] \).

Additionally, it's particularly useful in solving problems where a direct integration based on \( y \) offers a more straightforward solution than traditional methods that might revolve around \( x \). An example would be achieving a more seamless calculation in instances like revolving solids or in motion paths where one axis remains static.

Parametric Equations

Parametric equations are a powerful tool in mathematics used to express curves by defining both coordinates \( x \) and \( y \) in terms of a third parameter, commonly represented as \( t \). While Cartesian coordinates express relations in terms of \( x \) or \( y \), parametric equations offer flexibility in representing complex paths, especially when a single function form isn't sufficient.

• For example, a circle isn't easily described as a function in Cartesian form, but parametric equations \( x = \cos(t) \) and \( y = \sin(t) \) perfectly represent it over \( t = [0, 2\pi ] \).
• This method is versatile and capable of expressing curves where both coordinates are functions of another variable, making it suitable for describing real-world scenarios like projectile motion.

To compute the arc length for a curve defined parametrically as \( x = x(t) \) and \( y = y(t) \), the formula is:\[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}} \, dt \]This allows analysing lengths without converting back to a single variable focus, providing a significant edge in handling curves described in such dual-variable terms.

Definite Integrals

Definite integrals are an essential concept in calculus used to calculate the accumulated value of a function over a specific interval. They are not only confined to finding areas under curves but extend to various applications such as determining lengths of curves, volumes, and other accumulated quantities.

The notation for a definite integral is:\[ \int_{a}^{b} f(x) \, dx \]where \( a \) and \( b \) represent the limits of integration. The process involves evaluating the area (or other quantities) that the function \( f(x) \) encompasses between \( x=a \) and \( x=b \).

• In case of arc lengths, the function under the square root provides a mathematical interpretation of the curve's incremental segments.
• The integral sums these small linear distances, or segments, over the curve's length to determine the total arc length.

Thus, definite integrals form the backbone of many practical computations in engineering and science, extending beyond academia to real-world applications.

Differentiation

Differentiation refers to the process of finding the derivative of a function, which describes how the function value changes as its input changes. In the context of arc length, differentiation is essential for understanding how one variable changes in relation to another, especially when dealing with curves.

To compute the arc length, we often require the derivative \( \frac{dx}{dy} \) when working with Cartesian forms, or \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) when dealing with parametric equations. These derivatives tell us about the slope of the curve at any given point.

• Knowing this slope is crucial for calculating the accurate length of the arc, as it influences the shape and progression of the curve.
• For instance, when finding the length of a spiral or a sine wave, differentiation tells us how steeply the curve rises or falls at any point.

Overall, differentiation equips us with the means to dissect the geometrical aspects of curves, delivering insights necessary for precise measurements and tactics in calculus.