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...
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).