View the curve (y−x)2+2=xy−3 as a contour of f(x,y).

(a) Use ∇f(2,3) to find a vector normal to the curve at (2,3).


(b) Use your answer to part (a) to find an implicit equation for the tangent line to the curve at (2,3).



We are given the equation \((y - x)^2 + 2 = xy - 3\) and asked to treat it as a **contour** of the function \(f(x, y)\). 

To solve the problem, we need to:  
1. Express \(f(x, y)\) and compute the gradient \(\nabla f(x, y)\).  
2. Use the gradient \(\nabla f(2, 3)\) to find a normal vector to the curve at the point \((2, 3)\).  
3. Use this normal vector to derive the equation of the tangent line at that point.

---

### **Part (a): Find the Gradient and the Normal Vector**  
First, rewrite the given contour equation:  
\[
(y - x)^2 + 2 = xy - 3.
\]
This suggests we define the function:
\[
f(x, y) = (y - x)^2 + 2 - (xy - 3).
\]
Simplify:
\[
f(x, y) = (y - x)^2 + 2 - xy + 3 = (y - x)^2 - xy + 5.
\]

The **gradient of \(f(x, y)\)** is:
\[
\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right).
\]

1. **Partial derivative with respect to \(x\)**:
   \[
   \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( (y - x)^2 - xy + 5 \right).
   \]
   Using the chain rule:
   \[
   \frac{\partial}{\partial x} (y - x)^2 = 2(y - x) \cdot (-1) = -2(y - x),
   \]
   and:
   \[
   \frac{\partial}{\partial x} (-xy) = -y.
   \]
   So:
   \[
   \frac{\partial f}{\partial x} = -2(y - x) - y.
   \]

2. **Partial derivative with respect to \(y\)**:
   \[
   \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( (y - x)^2 - xy + 5 \right).
   \]
   Again, applying the chain rule:
   \[
   \frac{\partial}{\partial y} (y - x)^2 = 2(y - x),
   \]
   and:
   \[
   \frac{\partial}{\partial y} (-xy) = -x.
   \]
   So:
   \[
   \frac{\partial f}{\partial y} = 2(y - x) - x.
   \]

Thus, the gradient is:
\[
\nabla f(x, y) = \left( -2(y - x) - y, \; 2(y - x) - x \right).
\]

**Evaluate the gradient at \((2, 3)\)**:  
At the point \((2, 3)\), we have \(y - x = 3 - 2 = 1\). Substituting into the gradient:
\[
\nabla f(2, 3) = \left( -2(1) - 3, \; 2(1) - 2 \right) = (-5, 0).
\]

So, the normal vector to the curve at \((2, 3)\) is:
\[
\mathbf{n} = (-5, 0).
\]

---

### **Part (b): Find the Equation of the Tangent Line at \((2, 3)\)**  
A normal vector to the curve at \((2, 3)\) is \(\mathbf{n} = (-5, 0)\), which is perpendicular to the tangent line. Recall that the general form of a line with a given normal vector \((a, b)\) passing through a point \((x_0, y_0)\) is:
\[
a(x - x_0) + b(y - y_0) = 0.
\]

Using \((a, b) = (-5, 0)\) and \((x_0, y_0) = (2, 3)\), the equation becomes:
\[
-5(x - 2) + 0(y - 3) = 0.
\]
Simplify:
\[
-5x + 10 = 0 \quad \Longrightarrow \quad 5x = 10 \quad \Longrightarrow \quad x = 2.
\]

Thus, the equation of the **tangent line** at the point \((2, 3)\) is:
\[
x = 2.
\]

---

### **Final Answers**  
(a) The vector normal to the curve at \((2, 3)\) is \(\mathbf{n} = (-5, 0)\).  
(b) The equation of the tangent line at \((2, 3)\) is \(x = 2\).

---

Do you have any questions or need further clarification?  

Here are five related questions you might find useful:  
1. What is the geometric meaning of the gradient vector?  
2. How do we generalize the concept of contour lines to 3D surfaces?  
3. How can we find tangent planes instead of tangent lines for functions of several variables?  
4. How do implicit differentiation and gradients relate to each other?  
5. How would the tangent line equation change if the normal vector were different?

**Tip:** The gradient vector \(\nabla f(x, y)\) always points in the direction of the steepest ascent of the function \(f(x, y)\).

View the curve (y − x)^2 + 2 = xy − 3 as a contour of f(x, y).

(a) Use ∇f(2,3) to find a vector normal to the curve at (2,3).

(b) Use your answer to part (a) to find an implicit equation for the tangent line to the curve at (2,3).

Learn how to find the normal vector and the equation of the tangent line to the curve (y − x)^2 + 2 = xy − 3 by treating it as a contour of a function f(x, y). This problem involves using gradient calculus and implicit differentiation. Suitable for college-level calculus students.

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