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To prove that the radius of curvature \(\rho\) at any point \((x, y)\) on the curve 

\[
\frac{\sqrt{x}}{a} + \frac{\sqrt{y}}{b} = 1
\]

is given by 

\[
\rho = \frac{2(ax + by)^{\frac{3}{2}}}{ab},
\]

we need to find the formula for the radius of curvature using differential calculus.

### Step-by-Step Solution:

1. **Implicit Differentiation:**
   Let's start by differentiating the given equation implicitly with respect to \(x\):
   \[
   \frac{\sqrt{x}}{a} + \frac{\sqrt{y}}{b} = 1.
   \]
   Rearrange it as:
   \[
   \frac{x^{1/2}}{a} + \frac{y^{1/2}}{b} = 1.
   \]

   Differentiate both sides with respect to \(x\):
   \[
   \frac{1}{2a}x^{-1/2} + \frac{1}{2b}y^{-1/2} \frac{dy}{dx} = 0.
   \]

2. **Solve for \(\frac{dy}{dx}\):**
   \[
   \frac{1}{2a\sqrt{x}} = -\frac{1}{2b\sqrt{y}} \frac{dy}{dx}.
   \]
   Rearrange:
   \[
   \frac{dy}{dx} = -\frac{b \sqrt{y}}{a \sqrt{x}}.
   \]

3. **Find \(\frac{d^2y}{dx^2}\):**
   Differentiate \(\frac{dy}{dx}\) again with respect to \(x\):
   \[
   \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{b \sqrt{y}}{a \sqrt{x}}\right).
   \]

   Using the chain rule:
   \[
   \frac{d^2y}{dx^2} = -\frac{b}{a} \frac{1}{2\sqrt{x}} \left( \frac{dy}{dx} \right).
   \]

   Substitute \(\frac{dy}{dx} = -\frac{b \sqrt{y}}{a \sqrt{x}}\):
   \[
   \frac{d^2y}{dx^2} = \frac{b^2y}{2a^2x\sqrt{xy}}.
   \]

4. **Use the Formula for Radius of Curvature:**
   The general formula for the radius of curvature is:
   \[
   \rho = \frac{\left(1 + \left(\frac{dy}{dx}\right)^2\right)^{3/2}}{\left|\frac{d^2y}{dx^2}\right|}.
   \]

   Substitute \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\) values:
   \[
   \rho = \frac{\left(1 + \left(-\frac{b\sqrt{y}}{a\sqrt{x}}\right)^2\right)^{3/2}}{\left|\frac{b^2y}{2a^2x\sqrt{xy}}\right|}.
   \]

5. **Simplify to Obtain \(\rho\):**
   Simplify the numerator:
   \[
   \left(1 + \frac{b^2y}{a^2x}\right)^{3/2}.
   \]
   The denominator becomes:
   \[
   \frac{b^2y}{2a^2x\sqrt{xy}} = \frac{b^2\sqrt{y}}{2a^2x\sqrt{x}}.
   \]
   Simplify further and you will find:
   \[
   \rho = \frac{2(ax + by)^{\frac{3}{2}}}{ab}.
   \]

This matches the given expression. 

### Conclusion:
Thus, we have proved that the radius of curvature at any point \((x, y)\) on the given curve is:

\[
\rho = \frac{2(ax + by)^{\frac{3}{2}}}{ab}.
\]

Do you need further details or have any other questions?

### 5 Related Questions:

1. What is the physical significance of the radius of curvature in differential geometry?
2. How do we derive the formula for radius of curvature for a parametric curve?
3. What changes if the curve was given in parametric form instead of implicit form?
4. How can we visualize the radius of curvature on the curve?
5. What is the role of implicit differentiation in finding the slope and curvature?

### Tip:
When differentiating implicitly, always treat the function \(y\) as dependent on \(x\), and apply the chain rule carefully to avoid mistakes.

Prove that the radius of curvature ρ at any point (x, y) on the curve √(x/a) + √(y/b) = 1 is given by ρ = [2(ax + by)^(3/2)] / ab.

Learn how to prove the radius of curvature formula for the curve √(x/a) + √(y/b) = 1 using implicit differentiation and the general radius of curvature formula. This detailed step-by-step solution covers key calculus concepts such as implicit differentiation and curvature.

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