find equation of circle of curvature at x,y of curve sqrt(x/a)+sqrt(y/b)=1


To find the equation of the circle of curvature at a given point \((x_0, y_0)\) on the curve \(\sqrt{\frac{x}{a}} + \sqrt{\frac{y}{b}} = 1\), we need to follow these steps:

### 1. **Step 1: Parametrize the curve**
Given the implicit equation:
\[
\sqrt{\frac{x}{a}} + \sqrt{\frac{y}{b}} = 1
\]
You can parametrize the curve as:
\[
x = a \sin^2(\theta), \quad y = b \cos^2(\theta)
\]
This form satisfies the original equation because:
\[
\sqrt{\frac{a \sin^2(\theta)}{a}} + \sqrt{\frac{b \cos^2(\theta)}{b}} = \sin(\theta) + \cos(\theta) = 1
\]
for appropriate values of \(\theta\).

### 2. **Step 2: Calculate the first and second derivatives of the curve**
To calculate the curvature, we need the first and second derivatives \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\).

Implicit differentiation of the original equation gives us:
\[
\frac{1}{2\sqrt{x/a}} \cdot \frac{1}{a} \cdot \frac{dx}{dx} + \frac{1}{2\sqrt{y/b}} \cdot \frac{dy}{dx} = 0
\]
This simplifies to:
\[
\frac{1}{2\sqrt{ax}} + \frac{1}{2\sqrt{by}} \cdot \frac{dy}{dx} = 0
\]
Solving for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = -\frac{\sqrt{b y}}{\sqrt{a x}}
\]

For the second derivative \(\frac{d^2y}{dx^2}\), apply the quotient rule:
\[
\frac{d^2y}{dx^2} = \frac{d}{dx}\left( -\frac{\sqrt{by}}{\sqrt{ax}} \right)
\]
You can compute this by differentiating again with respect to \(x\).

### 3. **Step 3: Curvature formula**
The curvature \(K\) of a curve \(y = f(x)\) at any point is given by:
\[
K = \frac{|y''|}{(1 + y'^2)^{3/2}}
\]
where \(y' = \frac{dy}{dx}\) and \(y'' = \frac{d^2y}{dx^2}\).

### 4. **Step 4: Radius of curvature**
The radius of curvature \(R\) is the reciprocal of the curvature:
\[
R = \frac{1}{K}
\]

### 5. **Step 5: Equation of the circle of curvature**
The equation of the circle of curvature (osculating circle) at a point \((x_0, y_0)\) with curvature \(K\) is:
\[
(x - x_0)^2 + (y - y_0)^2 = R^2
\]
where \(R = \frac{1}{K}\) is the radius of curvature, and \((x_0, y_0)\) is the point on the curve where you are computing the circle.

Would you like to proceed with a specific point on the curve, or do you need further clarification on any of the steps?

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### Related questions:
1. What is the geometric significance of the curvature of a curve?
2. How does the curvature change for different types of curves?
3. What are the key applications of the osculating circle in physics and engineering?
4. How can implicit differentiation be used to solve more complex curves?
5. What is the difference between local curvature and global curvature?

**Tip:** Always check for symmetry when analyzing curves, as it can simplify many computations!

find equation of circle of curvature at x,y of curve sqrt(x/a)+sqrt(y/b)=1

Find the equation of the circle of curvature at a point on the curve sqrt(x/a) + sqrt(y/b) = 1 using parametrization, implicit differentiation, and curvature formulas. Suitable for advanced high school or undergraduate students.

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