Question

1. $y = 2x^{3/2}$, $left\frac{1}{3}, \frac{5}{3}\

ight$

Explanation:

Step1: Find the derivative of \( y \)

The function is \( y = 2x^{3/2} \). Using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \), the derivative \( y' \) is:
\( y'=2\times\frac{3}{2}x^{\frac{3}{2}-1}=3x^{1/2} \)

Step2: Calculate the arc length formula

The arc length \( L \) of a function \( y = f(x) \) from \( x = a \) to \( x = b \) is given by \( L=\int_{a}^{b}\sqrt{1+(y')^{2}}dx \).
Substitute \( y' = 3x^{1/2} \) into the formula:
\( (y')^{2}=(3x^{1/2})^{2}=9x \)
So, \( L=\int_{\frac{1}{3}}^{\frac{5}{3}}\sqrt{1 + 9x}dx \)

Step3: Integrate the function

Let \( u = 1+9x \), then \( du=9dx \) or \( dx=\frac{1}{9}du \).
When \( x=\frac{1}{3} \), \( u = 1 + 9\times\frac{1}{3}=1 + 3 = 4 \)
When \( x=\frac{5}{3} \), \( u = 1+9\times\frac{5}{3}=1 + 15 = 16 \)
The integral becomes \( \frac{1}{9}\int_{4}^{16}\sqrt{u}du \)
\( \sqrt{u}=u^{1/2} \), so integrating \( u^{1/2} \) gives \( \frac{2}{3}u^{3/2} \)
Thus, \( \frac{1}{9}\times\frac{2}{3}[u^{3/2}]_{4}^{16}=\frac{2}{27}(16^{3/2}-4^{3/2}) \)

Step4: Evaluate the definite integral

Calculate \( 16^{3/2}=(16^{1/2})^{3}=4^{3}=64 \) and \( 4^{3/2}=(4^{1/2})^{3}=2^{3}=8 \)
Substitute these values:
\( \frac{2}{27}(64 - 8)=\frac{2}{27}\times56=\frac{112}{27} \)

Answer:

The arc length of the curve \( y = 2x^{3/2} \) from \( x=\frac{1}{3} \) to \( x=\frac{5}{3} \) is \( \frac{112}{27} \)