If f(x,y)=x2+y2−4x , find the equation of the tangent line to the level curve f(x,y)=33 at the point (1,6) .


**A**. −2x+12y=70
**B**. −2x+12y=72
**C**. x−6y=−40
**D**. Not enough information

To find the equation of the tangent line to the level curve \( f(x, y) = 33 \) at the point \( (1, 6) \), follow these steps:

### Step 1: Compute the Gradient of \( f(x, y) \)

The gradient vector of \( f(x, y) \), denoted by \( \nabla f(x, y) \), points in the direction perpendicular to the level curve. The gradient is given by:

\[
\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)
\]

The function is \( f(x, y) = x^2 + y^2 - 4x \). So, we compute the partial derivatives:

- \( \frac{\partial f}{\partial x} = 2x - 4 \)
- \( \frac{\partial f}{\partial y} = 2y \)

Thus, the gradient is:

\[
\nabla f(x, y) = (2x - 4, 2y)
\]

### Step 2: Evaluate the Gradient at \( (1, 6) \)

Now, evaluate the gradient at the point \( (1, 6) \):

\[
\nabla f(1, 6) = (2(1) - 4, 2(6)) = (-2, 12)
\]

### Step 3: Equation of the Tangent Line

The gradient \( (-2, 12) \) is perpendicular to the tangent line. Therefore, the direction of the tangent line is perpendicular to this vector.

The general form of a line is \( A(x - x_0) + B(y - y_0) = 0 \), where \( (A, B) \) is a vector perpendicular to the line, and \( (x_0, y_0) \) is a point on the line.

Using \( (-2, 12) \) as the perpendicular vector and \( (1, 6) \) as the point on the line, we get:

\[
-2(x - 1) + 12(y - 6) = 0
\]

Simplify:

\[
-2x + 2 + 12y - 72 = 0
\]
\[
-2x + 12y - 70 = 0
\]

Thus, the equation of the tangent line is:

\[
-2x + 12y = 70
\]

### Final Answer:

The correct answer is **A**. \( -2x + 12y = 70 \).

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Would you like more details or have any questions?

Here are five related questions you might find useful:
1. What is the geometric interpretation of the gradient in multivariable calculus?
2. How do you find the equation of a tangent line to an implicit curve?
3. Why is the gradient perpendicular to level curves in general?
4. How do partial derivatives relate to the slope of the tangent line?
5. Can you find the tangent line for a different level curve of this function?

**Tip:** In problems involving level curves, the gradient vector always points in the direction of steepest ascent and is perpendicular to the curve.

If f(x,y)=x^2+y^2−4x, find the equation of the tangent line to the level curve f(x,y)=33 at the point (1,6).

Learn how to find the equation of the tangent line to the level curve f(x, y) = x^2 + y^2 - 4x = 33 at the point (1,6). This detailed solution covers the use of gradient vectors, partial derivatives, and the geometric interpretation of tangent lines. Suitable for college students studying multivariable calculus.

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