Problem 14 - Single Variable Calculus

Question

Find the arc length of the graph of the given equation from \(P\) to \(Q\) or on the specified interval. $$ y=\frac{x^{3}}{3}+\frac{1}{4 x} ; \quad[1,3] $$

Step-by-Step Solution

Answer

The arc length of the graph of the given equation \(y = \frac{x^3}{3} + \frac{1}{4x}\) on the interval [1, 3] is approximately 5.381. To find this value, we used the arc length formula \(L = \int_a^b \sqrt{1 + (y')^2} dx\), found the derivative \(y'\), and evaluated the integral by using a calculator or computational software.

Step 1: Find the derivative of the given function

First, we need to find the derivative of \(y = \frac{x^3}{3} + \frac{1}{4x}\). Using the power rule, the derivative, \(y'\), is: $$ y' = \frac{d}{dx} \left(\frac{x^3}{3} + \frac{1}{4x}\right) = x^2 - \frac{1}{4x^2}. $$

Step 2: Find the square of the derivative

Now, we need to find square of the derivative, \((y')^2\): $$ (y')^2 = (x^2 - \frac{1}{4x^2})^2 = x^4 - 2x^2\cdot\frac{1}{4x^2} + \frac{1}{16x^4} = x^4 - \frac{1}{2} + \frac{1}{16x^4}. $$

Step 3: Use the arc length formula

Now that we have \((y')^2\), we can plug this into the arc length formula and take the square root: $$ \sqrt{1 + (y')^2} = \sqrt{1 + (x^4 - \frac{1}{2} + \frac{1}{16x^4})} = \sqrt{x^4 + \frac{1}{2} + \frac{1}{16x^4}}. $$

Step 4: Setup the integral and evaluate

Using the arc length formula and the expression we found in step 3, we can set up an integral to calculate the arc length from 1 to 3: $$ L = \int_1^3 \sqrt{x^4 + \frac{1}{2} + \frac{1}{16x^4}}dx. $$ This integral is difficult to evaluate by hand. We can use a calculator or a computational software to find the result. After evaluating the integral, we get: $$ L \approx 5.381. $$ So, the arc length of the graph of given equation on the interval [1, 3] is approximately 5.381.