Question

In Exercises \(3-14,\) find the arc length of the graph of the function over the indicated interval. $$ y=\frac{2}{3} x^{3 / 2}+1, \quad[0,1] $$

Short answer

The arc length of the graph of the function \( y = \frac{2}{3} x^{3 / 2} + 1 \) on the interval \([0, 1]\) is \( L = \frac{2}{3}(2\sqrt{2} - 1) \).

Step-by-step solution

Compute the derivative

Find the derivative \( f'(x) \) of the function \( y = f(x) = \frac{2}{3} x^{3 / 2} + 1\). The derivative is \( f'(x) = (2/3) \cdot (3/2)x^{1/2} = x^{1/2}\).

Plug the derivative into the arc length formula

Substitute \( f'(x)\) into the integral that represents the arc length formula which is \( \int_a^b \sqrt{1+[f'(x)]^2} dx\). This gives us: \( L = \int_0^1 \sqrt{1 + [(x^{1/2})]^2} dx = \int_0^1 \sqrt{1 + x} dx\).

Evaluate the integral

This integral can be evaluated by the method of substitution. Let \( u = x +1 \), then \( dx = du \) and when \( x = 0, u = 1 \), and when \( x = 1, u = 2 \). So the integral becomes \( L = \int_1^2 \sqrt{u} du \). Integrating, we get \( L = [\frac{2}{3}u^{3/2}]_1^2 = \frac{2}{3}(2^{3/2} - 1) = \frac{2}{3}(2\sqrt{2} - 1)\).