To find the arc length of the curve $y^3 = x^2$ from $(0, 0)$ to $(1, 1)$, we can use the given formulas for arc length:

Formula (I):

$L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx$

Formula (II):

$L = \int_c^d \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy$

Step 1: Express $y$ as a function of $x$

The given equation is $y^3 = x^2$. To use Formula (I), we first need to find $\frac{dy}{dx}$.

Differentiating $y^3 = x^2$ implicitly with respect to $x$: $3y^2 \frac{dy}{dx} = 2x$ So, $\frac{dy}{dx} = \frac{2x}{3y^2}$

Step 2: Set up Formula (I)

Substitute $\frac{dy}{dx}$ into Formula (I): $L = \int_0^1 \sqrt{1 + \left( \frac{2x}{3y^2} \right)^2} \, dx$ We know $y = x^{2/3}$ from $y^3 = x^2$, so $y^2 = x^{4/3}$. Thus: $L = \int_0^1 \sqrt{1 + \frac{4x^2}{9x^{8/3}}} \, dx$ Simplifying the integrand: $L = \int_0^1 \sqrt{1 + \frac{4}{9x^{2/3}}} \, dx$ This is the improper integral because of the $x^{2/3}$ term, which can cause issues near $x = 0$.

Step 3: Express $x$ as a function of $y$

Next, for Formula (II), we express $x$ in terms of $y$. From $y^3 = x^2$, we solve for $x$: $x = y^{3/2}$ Now, differentiate $x = y^{3/2}$ with respect to $y$: $\frac{dx}{dy} = \frac{3}{2} y^{1/2}$

Step 4: Set up Formula (II)

Substitute $\frac{dx}{dy}$ into Formula (II): $L = \int_0^1 \sqrt{1 + \left( \frac{3}{2} y^{1/2} \right)^2} \, dy$ Simplifying the integrand: $L = \int_0^1 \sqrt{1 + \frac{9}{4} y} \, dy$ This is a standard integral and is not improper.

Final Answer

• Formula (I) (improper integral): $L = \int_0^1 \sqrt{1 + \frac{4}{9x^{2/3}}} \, dx$
• Formula (II): $L = \int_0^1 \sqrt{1 + \frac{9}{4} y} \, dy$

Would you like further details on solving these integrals? Let me know if you need clarification!

Related Questions:

1. How do you handle improper integrals when $x \to 0$?
2. What methods can you use to solve integrals like the one in Formula (II)?
3. How can you evaluate the arc length integral $\int_0^1 \sqrt{1 + \frac{9}{4} y} \, dy$?
4. What other curve equations can be used to illustrate different methods of finding arc length?
5. How can you verify the accuracy of your arc length calculation?

Tip: For improper integrals, check if the integrand has singularities or if limits must be taken carefully at points like 0 or infinity.