Let
. The tangent plane to the surface at (0, 0, 8) is

The gradient is

so the tangent plane's equation is

The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by
, then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation

or
,
, and
.
(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)
Answer:
1580.0
Step-by-step explanation:
- Add all numbers and divide them by 10 (the amount of numbers).
- Mean is when you add all the numbers and divide them by how many their are.
Answer:
14.037
Step-by-step explanation:
0.037+5+9
14.037
Answer:
(36π -72) cm²
Step-by-step explanation:
The area of a segment that subtends arc α (in radians) is given by ...
A = (1/2)r²·(α - sin(α))
Here, you have r = 12 cm and α = π/2 radians, so the area of the segment is ...
A = (1/2)(12 cm)²·(π/2 -1) = (36π -72) cm²