To find the unit normal to the surface $x^2 + y^2 = z$ at the point $(1, -2, 5)$, follow these steps:

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Step 1: Represent the surface as a level surface

The given surface can be written as: $F(x, y, z) = x^2 + y^2 - z = 0$ This representation allows us to use the gradient of $F(x, y, z)$ to find a vector normal to the surface.

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Step 2: Compute the gradient of $F(x, y, z)$

The gradient of $F(x, y, z)$ is: $\nabla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) = (2x, 2y, -1)$

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Step 3: Evaluate the gradient at the point $(1, -2, 5)$

Substitute $x = 1$, $y = -2$, and $z = 5$ into $\nabla F$: $\nabla F(1, -2, 5) = (2(1), 2(-2), -1) = (2, -4, -1)$ This gives the normal vector to the surface at the given point.

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Step 4: Normalize the normal vector

To find the unit normal vector, divide the normal vector by its magnitude: $\text{Magnitude of } (2, -4, -1) = \sqrt{2^2 + (-4)^2 + (-1)^2} = \sqrt{4 + 16 + 1} = \sqrt{21}$ $\text{Unit normal vector} = \frac{1}{\sqrt{21}}(2, -4, -1) = \left(\frac{2}{\sqrt{21}}, \frac{-4}{\sqrt{21}}, \frac{-1}{\sqrt{21}}\right)$

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Final Answer:

The unit normal vector to the surface $x^2 + y^2 = z$ at the point $(1, -2, 5)$ is: $\boxed{\left(\frac{2}{\sqrt{21}}, \frac{-4}{\sqrt{21}}, \frac{-1}{\sqrt{21}}\right)}$

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Would you like me to expand any steps or clarify further?

Related Questions:

1. How is the gradient of a scalar field used to find normals?
2. What is the geometric significance of normal vectors?
3. How can this method be applied to other surfaces like spheres or cylinders?
4. Why is normalizing the vector important for unit normals?
5. How would the result change if the point did not satisfy the surface equation?

Tip:

When finding the unit normal, always verify that the point lies on the surface by substituting it into the given equation.