Problem 9 - Single Variable Calculus

Question

Find the arc length of the graph of the given equation from \(P\) to \(Q\) or on the specified interval. $$ y=2(x-1)^{3 / 2} ; \quad P(1,0), Q(5,16) $$

Step-by-Step Solution

Answer

The arc length of the graph of the given equation from P(1, 0) to Q(5, 16) is approximately \(22.70\).

Step 1: Find the derivative of the function

First, we need to calculate the derivative of the function, \(\frac{dy}{dx}\): $$ y = 2(x-1)^{3/2} $$ Using the chain rule, we have: $$ \frac{dy}{dx} = 3 \cdot 2(x-1)^{1/2} \cdot \frac{d}{dx}(x-1) = 6(x-1)^{1/2} $$

Step 2: Plug the derivative into the arc length formula

Now that we have the derivative, we can plug it into the arc length formula: $$ L = \int_{1}^{5} \sqrt{1 + \left(6(x-1)^{1/2}\right)^2} dx $$ Simplify the expression: $$ L = \int_{1}^{5} \sqrt{1 + 36(x-1)} dx $$

Step 3: Evaluate the integral

To solve the integral, we use substitution method. Let: $$ u = 1 + 36(x-1) $$ Differentiate with respect to x: $$ \frac{du}{dx} = 36 $$ Then, the integral becomes: $$ L = \int_{u(1)}^{u(5)} \sqrt{u} \frac{du}{36} $$ Now, we can evaluate the integral: $$ L = \frac{1}{36} \cdot \frac{2}{3} \left[u^{3/2} \right]_{u(1)}^{u(5)} $$ Plug back the value of u: $$ L = \frac{1}{54} \left[(1 + 36(5-1))^{3/2} - (1 + 36(1-1))^{3/2}\right] $$

Step 4: Simplify and compute the arc length

Simplify and compute the arc length: $$ L = \frac{1}{54} \left[(1 + 36 \cdot 4)^{3/2} - (1)^{3/2}\right] $$ $$ L = \frac{1}{54} \left[(145)^{3/2} - 1\right] $$ $$ L \approx 22.70 $$ The arc length of the graph of the given equation from P(1, 0) to Q(5, 16) is approximately 22.70.

Key Concepts

Derivative of a Function

Understanding derivatives is crucial for many calculus applications, including calculating arc lengths. A derivative represents the rate at which a function is changing at any given point. It's like capturing a 'snapshot' of the slope of the function's graph at a single point.

For the function in our example, we considered the function \(y = 2(x-1)^{3/2}\). Its derivative \(dy/dx\) represents the slope of the function at any point \(x\). To find this derivative, we applied the chain rule, a fundamental technique in calculus that provides a way to differentiate composite functions.

Arc Length Formula

The arc length formula is a beautiful tool in calculus used to determine the length of a curve between two points. You can imagine stretching out a curve until it becomes a straight line and then measuring that line's length. That's the arc length.

The formula is \(L = \int \sqrt{1 + (dy/dx)^2} dx\), where \(L\) is the arc length, \(dy/dx\) is the derivative of the function, and the integral runs from point \(P\) to point \(Q\). In our example, we plugged the derivative of \(y\) into this formula.

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. When we have a function nested inside another, the chain rule guides us on how to take the derivative. We treat the inner function as a separate variable, find the derivatives of the outer and inner functions separately, and then multiply them together.

In our exercise, we applied the chain rule to differentiate \(y = 2(x-1)^{3/2}\), resulting in \(dy/dx = 6(x-1)^{1/2}\).

Integral Evaluation

When it comes to understanding integral evaluation, one can think of it as the process of finding the total size or value, such as area under a curve, from a continuous rate of change (the derivative). It's the 'accumulation' of all those small changes.

In the arc length calculation, after substituting the derivative into the arc length formula, the next step is to evaluate the definite integral which gives us the arc length between two points on the curve.

Substitution Method

The substitution method is a powerful technique for simplifying integrals. By choosing a new variable to replace a more complex expression in the integral, the integral often becomes easier to evaluate.

For instance, in our example, the substitution \(u = 1 + 36(x-1)\) is used to transform the original integral into one in terms of \(u\) that is less complicated to solve. After finding the antiderivative in terms of \(u\) and simplifying, we can then substitute back in terms of \(x\) to find the actual length of the arc.