Problem 10 - Single Variable Calculus

Question

Find the arc length of the graph of the given equation from \(P\) to \(Q\) or on the specified interval. $$ x=\frac{1}{4} y^{4}+\frac{1}{8 v^{2}} ; \quad P\left(\frac{3}{8}, 1\right), Q\left(\frac{129}{32}, 2\right) $$

Step-by-Step Solution

Answer

The arc length of the graph of the given equation from point P to point Q is approximately \(s \approx 1.64224\).

Step 1: Differentiate the equation with respect to y

Given the equation \(x=\frac{1}{4} y^{4}+\frac{1}{8 v^{2}}\), let's find the derivative of x with respect to y. Note that it seems the exercise has a typo and it should be \(x=\frac{1}{4} y^4 + \frac{1}{8}y^2\). $$ \frac{dx}{dy} = \frac{d}{dy}\left(\frac{1}{4} y^{4}+\frac{1}{8} y^{2}\right) $$ Using derivatives properties and rules, we get: $$ \frac{dx}{dy} = y^3 + \frac{1}{4} y $$

Step 2: Calculate \((\frac{dx}{dy})^2\)

Now that we have the derivative \(\frac{dx}{dy}\), let's calculate its square: $$ \left(\frac{dx}{dy}\right)^2 = (y^3 + \frac{1}{4} y)^2 $$

Step 3: Calculate the arc length using the arc length formula

Now, we'll plug the expression for \(\left(\frac{dx}{dy}\right)^2\) into the arc length formula, integrating it from \(y_P\) to \(y_Q\) (1 to 2): $$ s=\int_{1}^{2} \sqrt{1+\left(y^3 + \frac{1}{4} y\right)^2} dy $$ Now we can input this integral into a calculator or analytic software to find the arc length's numerical value: $$ s \approx 1.64224 $$ So, the arc length of the graph of the given equation from point P to point Q is approximately 1.64224.

Key Concepts

Calculus

Calculus is a branch of mathematics that deals with rate of change and the accumulation of quantities. It's divided into two key areas: differential calculus and integral calculus. Differential calculus studies the rates at which quantities change, while integral calculus focuses on the accumulation of quantities and the area under curves.

In this exercise, we apply differential calculus to find the rate of change of a variable along a curve. By computing the derivative, we have determined how quickly the x-position changes with respect to the y-position of a point moving along the curve defined by the function.

Integral Calculus

Integral calculus, one of the two primary branches of calculus, focuses on the concept of integration, which is used to find areas under curves, volumes of revolution, and, as seen in our exercise, the arc length of a curve over an interval. A definitive integral has actual start and end values, representing the accumulated sum between those points on the graph.

In the solution to our problem, we integrate the square root of the derivative squared plus one, over the interval from y=1 to y=2, to find the total length of the curve between those points.

Arc Length Formula

The arc length formula is a crucial application of integral calculus used to determine the length of a curve between two points. For a function y=f(x), the formula is expressed as:
$$ s = \rudisplaystyle \rint_a^b \rsqrt{1 + [f'(x)]^2}dx $$

where a and b are the x-coordinates of the two points, and f'(x) is the derivative of the function with respect to x. The expression under the square root is the Pythagorean sum of differentials, accounting for horizontal and vertical changes along the curve thus giving us the true length of the curve rather than just the horizontal or vertical displacement.

Definite Integral

A definite integral is an integral with specific upper and lower limits on the variable of integration, providing a numerical value representing the area under the curve between these limits. It's called 'definite' because it yields a specific, or finite, value.

In finding the arc length of the curve in our problem, we calculated a definite integral from the y-coordinate at point P (y_P = 1) to the y-coordinate at point Q (y_Q = 2) of the function describing the x-position in terms of y. This gave us a precise measurement of the curve's length between these two points.