Problem 62 - Calculus for Biology and Medicine

Question

Find the length of the curve $$ 3 y^{2}=4 x^{3} $$ from \(x=0\) to \(x=1\).

Step-by-Step Solution

Answer

Express x in terms of y, differentiate, and integrate to find the length.

Step 1: Understanding the formula for curve length

The length of a curve described by a function \( y = f(x) \) from \( x = a \) to \( x = b \) is calculated using the formula: \[ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx \]. Our task is to identify \( f(x) \) or \( x \) as a function of \( y \) if more appropriate.

Step 2: Expressing x as a function of y

Given the equation \( 3y^2 = 4x^3 \), we can express \( x \) in terms of \( y \) as follows: \( x = \left(\frac{3}{4}y^2\right)^{1/3} \). Differentiating this with respect to \( y \) will allow us to use the adaptation of the formula for parametric forms.

Step 3: Finding the derivative

Differentiate \( x = \left(\frac{3}{4}y^2\right)^{1/3} \) with respect to \( y \) to find \( \frac{dx}{dy} \). Using the power rule and chain rule, we get: \( \frac{dx}{dy} = 2 \cdot \frac{1}{3}\left(\frac{3}{4}y^2\right)^{-2/3} \cdot \frac{3}{4}y = \frac{1}{2} \left(\frac{3}{4}y^2\right)^{-2/3} \cdot y \).

Key Concepts

Integral Calculus

Integral calculus is all about finding the total sum or accumulation of quantities. It helps in calculating areas under curves, amongst many other applications. In our exercise, we're interested in finding the length of a curve. This involves integrating a function over a specific interval.

When calculating the length of a curve given by the function \(y = f(x)\), we use the integral:

• \( L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx \)

This formula comes from approximating the curve using a series of small, straight-line segments, then summing their lengths using a limit process as the segments become infinitely small.

In this problem, however, the curve is described in terms of \(y\) rather than \(x\), which leads us to use a version of the formula adapted for parametric equations.

Parametric Curves

Parametric curves come into play when you define a curve using parameters rather than a single function. Instead of \(y\) being expressed solely as \(f(x)\), both \(x\) and \(y\) are expressed in terms of another variable, often \(t\) or, in this case, each other.

For the equation \(3y^2 = 4x^3\), it's easier to express \(x\) as a function of \(y\), transforming our approach. We represent the curve in parametric form as:

• \(x = \left(\frac{3}{4}y^2\right)^{1/3}\)

Using this representation, we can find derivatives that help us calculate lengths and other integral calculus problems on the curve. Parametric curves allow us to deal with situations where the traditional \(y = f(x)\) approach doesn't work, providing flexibility in analyzing curves with complex relationships.

Differentiation

Differentiation is a key concept in calculus. It's used to find rates of change and slope of curves, which can be applied to find the lengths of curves.

In our scenario, differentiation provides the tool for transforming a curve into an integrable form for calculating its length. After expressing \(x\) in terms of \(y\), we take the derivative to help us set up our integral for the length of the curve:

• Differentiate the function \(x = \left(\frac{3}{4}y^2\right)^{1/3}\) with respect to \(y\).
• Apply the power rule and chain rule to find \(\frac{dx}{dy} = \frac{1}{2} \left(\frac{3}{4}y^2\right)^{-2/3} y\).

This derivative becomes part of the integral setup for finding the curve length, showing how differentiation is intertwined with integral calculus to solve geometric problems.