Problem 14 - Essential Calculus

Question

In Exercises \(3-14,\) find the arc length of the graph of the function over the indicated interval. $$ x=\frac{1}{3} \sqrt{y}(y-3), \quad 1 \leq y \leq 4 $$

Step-by-Step Solution

Answer

The arc length of the graph of the function over the indicated interval [1, 4] can be found by evaluating the integral \(\int_{1}^{4}\sqrt{1+\left(\frac{1}{3}[1/2\sqrt{y}(y-3)+\sqrt{y}]\right)^2}dy\)

Step 1: Compute the derivative

We first compute the derivative \( dx/dy \) of the given function. Let's differentiate \( x=\frac{1}{3}\sqrt{y}(y-3) \) with respect to y. Using the product rule gives \( dx/dy=\frac{1}{3}[(1/2\sqrt{y}(y-3)+\sqrt{y}(1)] \).

Step 2: Square the derivative

Now square the derivative computed in the previous step to get \((dx/dy)^2 = [\frac{1}{3}(1/2\sqrt{y}(y-3)+\sqrt{y})]^2\)

Step 3: Insert into the arc length formula

The formula for the arc length includes the square root of 1 plus the derivative squared. Inserting, it leads to the integral \(\int_{1}^{4}\sqrt{1+\left(\frac{1}{3}[1/2\sqrt{y}(y-3)+\sqrt{y}]\right)^2}dy\)

Step 4: Evaluate the integral

Now that we've formulated the integral, evaluate it from the limits 1 to 4 to get the arc length of the graph of the function over the indicated interval.

Key Concepts

Integral Calculus

Integral calculus is a fundamental concept in calculus that helps us find the total value of functions, especially when dealing with curves and areas under curves. It allows us to calculate quantities that are defined in terms of a continuous change, like position or in this instance, the arc length of a curve. In this exercise, we are tasked with calculating the arc length of a function.
To achieve this, we set up a definite integral, which allows us to sum an infinite number of infinitesimally small distances along the curve. Calculating these requires understanding the function’s rate of change or its derivative at each point in the interval. Integral calculus, when applied to arc length, helps us understand how long the path or the curve is between two points.

Derivative

The derivative is a tool in calculus that measures how a function changes as its input changes. Simply put, it tells us the rate of change or the slope of the function at any given point. In the context of arc length calculation, understanding the behavior of the derivative is essential.
In our problem, we first computed the derivative \( \frac{dx}{dy} \) for the given curve \( x=\frac{1}{3}\sqrt{y}(y-3) \). Using the product rule, we are able to break down the curve's expression and differentiate it. This gives us insight into how the function’s x-coordinate changes with respect to y within the specified interval. Getting the correct derivative is crucial because the arc length formula involves this squared derivative.

Definite Integral

A definite integral is a specific use of integral calculus that finds the total accumulation of a quantity over a given interval. It gives us the net area under a curve, but in the context of arc length, it helps us compute the length of the curve over a certain interval.
When we set up the definite integral for the arc length, \( \int_{1}^{4}\sqrt{1+\left(\frac{1}{3}\left[\frac{1}{2}\sqrt{y}(y-3)+\sqrt{y}\right]\right)^2}dy \), we are essentially finding the total distance traveled along the curve from \( y=1 \) to \( y=4 \). Evaluating this integral provides the arc length, which represents the physical length of the curve in space.
It's a powerful technique that lets us calculate complex lengths that would be difficult, if not impossible, to measure directly.

Arc Length Formula

The arc length formula is a specific application of the integral calculus used to find the length of a curve. In essence, this formula translates the idea of summing smaller line segments into the language of calculus. For a curve parametrized by a variable, like the function given in our problem, the formula involves integrating the square root of \( 1 \) plus the square of the derivative over the specified interval.
The general formula for arc length \( L \) is given by

• \( L = \int_{a}^{b}\sqrt{1+(f'(y))^2}dy \) for functions of y, where \( f'(y) \) is the derivative of the function with respect to y.

This formula captures the essential idea of adding up infinitely many infinitesimal distances to get the total length of the curve. Computing the definite integral of this expression across the interval \([1, 4]\) provides the precise arc length of the function curve.
It's impressive how calculus leverages concepts like derivatives and definite integrals to solve real-world geometric problems efficiently.