(x-1)^2 + (y-1)^2 = 13, (3,4) find dy/dx and find the slope of the curve at the given point

We are given the equation of a circle: 

\[
(x - 1)^2 + (y - 1)^2 = 13
\]

We need to find \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\), which will give us the slope of the tangent line to the curve at the given point \((3,4)\).

### Step 1: Differentiate the equation implicitly with respect to \(x\)
Implicit differentiation means differentiating both sides of the equation with respect to \(x\), remembering that \(y\) is a function of \(x\) (i.e., \(y = y(x)\)).

\[
\frac{d}{dx}\left((x - 1)^2 + (y - 1)^2 \right) = \frac{d}{dx}(13)
\]

The right-hand side, being a constant, differentiates to 0:

\[
2(x - 1) \cdot \frac{d}{dx}(x - 1) + 2(y - 1) \cdot \frac{d}{dx}(y - 1) = 0
\]

Now differentiate each term:

- \(\frac{d}{dx}(x - 1) = 1\) (since \(x\) is the independent variable),
- \(\frac{d}{dx}(y - 1) = \frac{dy}{dx}\) (since \(y\) is a function of \(x\)).

Substitute these into the equation:

\[
2(x - 1) \cdot 1 + 2(y - 1) \cdot \frac{dy}{dx} = 0
\]

### Step 2: Solve for \(\frac{dy}{dx}\)
Simplifying the equation:

\[
2(x - 1) + 2(y - 1) \cdot \frac{dy}{dx} = 0
\]

Solve for \(\frac{dy}{dx}\):

\[
2(y - 1) \cdot \frac{dy}{dx} = -2(x - 1)
\]

\[
\frac{dy}{dx} = \frac{-(x - 1)}{(y - 1)}
\]

### Step 3: Find the slope at the given point \((3, 4)\)
Now, substitute \(x = 3\) and \(y = 4\) into the derivative expression:

\[
\frac{dy}{dx} = \frac{-(3 - 1)}{(4 - 1)} = \frac{-2}{3}
\]

Thus, the slope of the curve at the point \((3,4)\) is \(\frac{-2}{3}\).

### Summary
The derivative \(\frac{dy}{dx} = \frac{-(x-1)}{y-1}\), and the slope of the curve at the point \((3,4)\) is \(\frac{-2}{3}\).

Would you like further details or have any questions?

Here are 5 related questions to expand on this:

1. What is the geometric meaning of the slope of a curve at a point?
2. How can we interpret the equation of a circle in terms of its center and radius?
3. How do we apply implicit differentiation to higher-order derivatives?
4. How do we find the equation of the tangent line at a point on the curve?
5. How does implicit differentiation differ from explicit differentiation?

**Tip:** Implicit differentiation is especially useful when \(y\) cannot be easily solved for \(x\). Always treat \(y\) as a function of \(x\) when differentiating implicitly.

Learn how to find the slope of the curve at the point (3,4) for the equation (x-1)^2 + (y-1)^2 = 13 using implicit differentiation. This problem explores key calculus concepts including the equation of a circle and implicit differentiation. Suitable for students in Grades 11-12.

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