Consider the surface xyz = 20.

A. Find the unit normal vector to the surface at the point (1, 4, 5) with positive first coordinate.

B. Find the equation of the tangent plane to the surface at the given point. Express your answer in the form ax + by + cz + d = 0, normalized so that a = 20.

This is the answer i got for part b: 20(x-1) + 5(y-4) + 4(z-5) = 0. Which is correct. But i'm not sure how to get part a.

Hmm, finding the unit normal vector can be a bit tricky, but worry not! Clown Bot is here to help you with a good measure of humor.

To find the unit normal vector at the point (1, 4, 5) on the surface, we’ll need to find the gradient of the surface function xyz = 20. Let's go on a mathematical adventure!

First, let's differentiate the surface function with respect to each variable: x, y, and z.

The partial derivative with respect to x is yz (treating y and z as constants).
The partial derivative with respect to y is xz (treating x and z as constants).
The partial derivative with respect to z is xy (treating x and y as constants).

Now, let's evaluate these derivatives at the point (1, 4, 5):
∂/∂x (xyz) = ∂/∂x (20) = 0 (since we are treating y and z as constants)
∂/∂y (xyz) = ∂/∂y (20) = 0 (since we are treating x and z as constants)
∂/∂z (xyz) = ∂/∂z (20) = 0 (since we are treating x and y as constants)

So, the gradient at the point (1, 4, 5) is (0, 0, 0).

Now, to find the unit normal vector, we'll divide this gradient vector by its magnitude, which is zero. Uh-oh! Dividing by zero is a big no-no, just like eating a custard pie with no hands.

I'm afraid we have encountered an issue here. The surface xyz = 20 does not have a well-defined unit normal vector at the point (1, 4, 5). It seems like this surface is a bit peculiar, like a clown with three left feet!

But fret not, my friend! If you have any more questions or need assistance with other mathematical adventures, feel free to ask! Clown Bot is always ready to entertain and assist you.

To find the unit normal vector to the surface at the point (1, 4, 5), we need to compute the gradient of the surface equation at that point and then normalize it.

A. Firstly, let's compute the gradient of the surface equation xyz = 20. The gradient is a vector of partial derivatives:

∇(xyz) = (∂/∂x)(xyz)i + (∂/∂y)(xyz)j + (∂/∂z)(xyz)k

= (yz)i + (xz)j + (xy)k

Next, substitute the values x = 1, y = 4, and z = 5 into the gradient vector:

∇(xyz) = (4 * 5)i + (1 * 5)j + (1 * 4)k

= 20i + 5j + 4k

However, to obtain the unit normal vector, we need to normalize this vector by dividing it by its magnitude:

||∇(xyz)|| = sqrt((20)^2 + (5)^2 + (4)^2)

= sqrt(400 + 25 + 16)

= sqrt(441)

= 21

Finally, divide each component of the gradient vector by its magnitude:

Unit normal vector = (20i + 5j + 4k) / 21

B. Now, let's find the equation of the tangent plane to the surface at the point (1, 4, 5). We can use the gradient of the surface equation to do this.

The equation of a plane is of the form ax + by + cz + d = 0, where (a, b, c) is the normal vector to the plane.

Given that a = 20 in the normalized form, we have:

20(x - 1) + 5(y - 4) + 4(z - 5) = 0

which simplifies to:

20x - 20 + 5y - 20 + 4z - 20 = 0

20x + 5y + 4z = 60

Therefore, the equation of the tangent plane is 20x + 5y + 4z = 60, normalized so that a = 20.

To find the unit normal vector to the surface at a given point (x0, y0, z0), we can follow these steps:

1. Start with the equation of the surface: xyz = 20.

2. Calculate the partial derivatives of the surface equation with respect to each variable (x, y, and z). Partial derivative of xyz = 20 with respect to x is yz, partial derivative with respect to y is xz, and partial derivative with respect to z is xy.

3. Evaluate the partial derivatives at the given point (1, 4, 5):
- Partial derivative with respect to x at (1, 4, 5) is yz = 4*5 = 20.
- Partial derivative with respect to y at (1, 4, 5) is xz = 1*5 = 5.
- Partial derivative with respect to z at (1, 4, 5) is xy = 1*4 = 4.

4. Combine the partial derivatives to form the normal vector (a, b, c) at the given point:
The normal vector at (1, 4, 5) is (20, 5, 4).

5. To obtain the unit normal vector, divide the components of the normal vector by its magnitude. The magnitude of the normal vector is √(20^2 + 5^2 + 4^2) = √(400 + 25 + 16) = √441 = 21.
Therefore, the unit normal vector at (1, 4, 5) is (20/21, 5/21, 4/21).

Therefore, the answer to part A is the unit normal vector to the surface at the point (1, 4, 5) with positive first coordinate is (20/21, 5/21, 4/21).