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Vector Calculus, Partial Derivatives, and Polar Cylindrical and Spherical Coordinates

See the attached file.

Vectors
1. The points A(2; -3, 3), B(3,5;4), C(3;8;-2) and D(4;4;6) are vertices of a tetrahedron. Find the volume of the tetrahedron.

2. There are two vectors: a = (2; -6; -4) and b = (3; -4; 2). Calculate:
(i) a * b
(ii) (2a - 3b) * (a + 2b) ; (a + b)^2 ; (a - b)^2

3. There are two vectors: a = (3; -1; -2) and b = (1; 2; -1). Calculate:
a * b (2a + b) * b and (2a - b) * (2a + b)

4. A body is being moved from the initial point A(0, 0; 5) till the point B(3; -2; 5) (along a straight line) due to the action of the force F = (3; -2; -5). What is the work done by the force?

5. Find the plane that passes through the point M(2; 4; -5) and which is parallel to the vectors: a = (3; 1; -1) and b = (1; -2; 1)

Partial Derivatives
Please see the attached document.

Polar Cylindrical and Spherical Coordinates
Please see the attached document.

Attachments

Solution Preview

Please see the attached MS Word document for the full solutions to the questions posed. Included are diagrams and formulas used to calculate the answers.

1.

The tetrahedron is defined by three vectors, u, v and w:

The volume of the tetrahedron is given by:
(1.1)
Where S is the base area and h is the altitude to the base..
The base is half the area of a parallelepiped defined by the vectors u and v:

The area is:
(1.2)

But if we use the identity:
(1.3)
Thus:
(1.4)
The cross product results in a vector which is perpendicular to S.
The height of the tetrahedron is the projection of w on n:

(1.5)
The angle is the angle between w and n, thus we can use the dot product identity:
(1.6)
Therefore the altitude to the base is:
(1.7)

The volume of the tetrahedron is therefore:
(1.8)
We have to account for the fact that and apply the absolute value ...

Solution Summary

The solution answers the questions on vector calculus, partial derivatives and polar cylindrical and spherical coordinates.

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