Can anybody help me with this?

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.
= 0.

For part b i got 20(x-1) + 5(y-4) + 4(z-5) which is correct. And i got (x-1),(y-4)/4,(z-5)/5 for a normal line. But i'm not sure how to get part A...

User Icon for Explain Bot Explain Bot answered
1 year ago

To find the unit normal vector to the surface at the point (1, 4, 5), we can use the gradient vector.

The gradient vector is a vector that points in the direction of the maximum rate of increase of a function. In this case, we can think of the equation xyz = 20 as a function f(x, y, z) = xyz - 20.

The gradient vector of a function is a vector whose components are the partial derivatives of the function with respect to each variable. So, to find the gradient vector of f(x, y, z) = xyz - 20, we need to find the partial derivatives.

∂f/∂x = yz
∂f/∂y = xz
∂f/∂z = xy

Evaluate these partial derivatives at the point (1, 4, 5):
∂f/∂x = (4)(5) = 20
∂f/∂y = (1)(5) = 5
∂f/∂z = (1)(4) = 4

The gradient vector at the point (1, 4, 5) is then:
∇f(1, 4, 5) = (20, 5, 4)

To find the unit normal vector, we need to normalize this vector by dividing each component by its magnitude.

Magnitude of ∇f(1, 4, 5) = sqrt((20)^2 + (5)^2 + (4)^2) = sqrt(541) ≈ 23.258
Unit normal vector = (∇f(1, 4, 5))/|∇f(1, 4, 5)| = (20/23.258, 5/23.258, 4/23.258)
≈ (0.86, 0.22, 0.17)

Therefore, the unit normal vector to the surface at the point (1, 4, 5) with positive first coordinate is (0.86, 0.22, 0.17).