Find the angle between the surfaces x2+y2+z2=25 and x2+y2-z-5 at (3,0,4)

To find the angle between two surfaces at a given point, we can use the following approach:

1. **Determine the gradient of each surface**: The gradient vector at a point on a surface is perpendicular (normal) to the surface at that point.

2. **Find the angle between the two gradient vectors**: The angle between the surfaces is the angle between their normal vectors.

### Step 1: Find the gradients of the surfaces

Given surfaces:
1. \( f_1(x, y, z) = x^2 + y^2 + z^2 - 25 = 0 \)
2. \( f_2(x, y, z) = x^2 + y^2 - z - 5 = 0 \)

The gradients (normal vectors) are found by taking the partial derivatives with respect to \(x\), \(y\), and \(z\).

- **Gradient of \( f_1(x, y, z) \):**
  \[
  \nabla f_1 = \left( \frac{\partial f_1}{\partial x}, \frac{\partial f_1}{\partial y}, \frac{\partial f_1}{\partial z} \right) = \left( 2x, 2y, 2z \right)
  \]
  
- **Gradient of \( f_2(x, y, z) \):**
  \[
  \nabla f_2 = \left( \frac{\partial f_2}{\partial x}, \frac{\partial f_2}{\partial y}, \frac{\partial f_2}{\partial z} \right) = \left( 2x, 2y, -1 \right)
  \]

### Step 2: Evaluate the gradients at the point \((3, 0, 4)\)

- **At \( (3, 0, 4) \):**

  \[
  \nabla f_1(3, 0, 4) = \left( 2(3), 2(0), 2(4) \right) = (6, 0, 8)
  \]
  
  \[
  \nabla f_2(3, 0, 4) = \left( 2(3), 2(0), -1 \right) = (6, 0, -1)
  \]

### Step 3: Find the angle between the gradient vectors

The angle \( \theta \) between two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by:
\[
\cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|}
\]

- **Dot product \( \mathbf{A} \cdot \mathbf{B} \):**

  \[
  (6, 0, 8) \cdot (6, 0, -1) = 6(6) + 0(0) + 8(-1) = 36 - 8 = 28
  \]

- **Magnitude of the vectors:**

  \[
  |\mathbf{A}| = \sqrt{6^2 + 0^2 + 8^2} = \sqrt{36 + 0 + 64} = \sqrt{100} = 10
  \]
  
  \[
  |\mathbf{B}| = \sqrt{6^2 + 0^2 + (-1)^2} = \sqrt{36 + 0 + 1} = \sqrt{37}
  \]

- **Cosine of the angle:**

  \[
  \cos \theta = \frac{28}{10 \times \sqrt{37}} = \frac{28}{10\sqrt{37}} = \frac{14}{5\sqrt{37}}
  \]

- **Angle \( \theta \):**

  \[
  \theta = \cos^{-1} \left( \frac{14}{5\sqrt{37}} \right)
  \]

This is the angle between the two surfaces at the point \((3, 0, 4)\).

Would you like to see more details or have any questions?

### Related Questions:
1. How can the angle between two vectors be calculated if their components are given?
2. What is the significance of the gradient vector in surface geometry?
3. How can we find the normal vector of a surface defined parametrically?
4. What does it mean geometrically when the angle between two surfaces is 90 degrees?
5. Can the angle between surfaces be negative, and what would it signify?

**Tip:** Always remember that the gradient of a scalar function gives the direction of the steepest ascent on the surface defined by that function.

Discover how to calculate the angle between the surfaces x^2+y^2+z^2=25 and x^2+y^2-z-5 at the specific point (3,0,4). Learn about gradient vectors, dot products, and vector magnitudes in surface geometry. Suitable for advanced mathematics enthusiasts and students.

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