Civil Engineering Board Exams Problems Philippines - Polar Coordinates and Equations, Lecture notes of Analytical Geometry

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CIVIL ENGINEERING BOARD EXAMS PROBLEMS PHILIPPINES November 6, 2020
POLAR COORDINATES AND EQUATIONS
To form a polar coordinate system, start with a fixed point and call it the pole or origin. From this point draw a half line, or ray (usually horizontal and to the
right) and call this line the polar axis. In this system, the location of a point is expressed by its distance r from a fixed point and its angle from a fixed line.
Sign convention
1. θ is positive for counterclockwise and negative for clockwise.
2. r is positive for laid offs at terminal side and negative for laid offs at prologation through O from the terminal side θ .
DISTANCE BETWEEN TWO POINTS
CONVERSION OF RECTANGULAR TO POLAR COORDINATES AND VICE VERSA.
POLAR CURVES AND RECTANGULAR CURVES
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CIVIL ENGINEERING BOARD EXAMS PROBLEMS PHILIPPINES – November 6, 2020

POLAR COORDINATES AND EQUATIONS

To form a polar coordinate system, start with a fixed point and call it the pole or origin. From this point draw a half line, or ray (usually horizontal and to the

right) and call this line the polar axis. In this system, the location of a point is expressed by its distance r from a fixed point and its angle from a fixed line.

Sign convention

  1. θ is positive for counterclockwise and negative for clockwise.
  2. r is positive for laid offs at terminal side and negative for laid offs at prologation through O from the terminal side θ.

DISTANCE BETWEEN TWO POINTS

CONVERSION OF RECTANGULAR TO POLAR COORDINATES AND VICE VERSA.

POLAR CURVES AND RECTANGULAR CURVES

EXAMPLES:

  1. Convert (-4 , 1.077) to rectangular coordinates. (1.077 radians)

SOLUTION:

  1. Rewrite to rectangular form: r = 3/(1 – 2cos θ)

SOLUTION:

SOLID ANALYTIC GEOMETRY

SPACE COORDINATE SYSTEM – There three coordinate system used in solid analytic geometry:

  1. RECTANGULAR COORDINATES – a point P(x , y , z) in space is fixed by its three distance x, y, and z from the coordinate planes.
  2. CYLINDRICAL COORDINATES – A point P in space may be imagined as being on the surface of a cylinder perpendicular to the XY- plane. P(r, θ

,z) is fixed by its distance z from the xy- plane and by the polar coordinates (r , θ) of the projection of P on the XY-plane.

  1. SPHERICAL COORDINATES – A point P in space may be imagined as being on the surface of a sphere with center at the origin O and radius r.

P(r, ϕ , θ) is fixed by its distance r from O, the angle ϕ between OP and z-axis , and the angle θ which is the angle between the x axis and the

projection OP on the xy- plane.

DISTANCE BETWEEN TWO POINTS

CONVERSION FACTORS : FROM RECTANGULAR AND VICE VERSA.

TO CYLINDRICAL : TO SPHERICAL:

DISTANCE FROM A POINT TO A PLANE: GENERAL EQUATION OF THE PLANE

EQUATION OF THE PLANE (INTERCEPT FORM) ANGLE BETWEEN TWO PLANES

PERPENDICULAR DISTANCE BETWEEN TWO PLANES

POINT OF INTERSECTION OF THE MEDIANS

  1. What is the distance between the point P(1,2,3) and the plane 2x + 2y – 3y + 3 = 0?

SOLUTION:

  1. Convert to rectangular coordinates: (4 , 2π/3 , - 2)

SOLUTION:

  1. Determine the angles of the radius vector of the point (3,-2,5) that forms with the coordinate axes.

SOLUTION:

  1. Find the direction cosines of a point having a coordinates of (2, 3, - 6).

SOLUTION:

  1. Convert the point (-1 , 1, - √2) to spherical coordinates.

SOLUTION:

EXERCISES – Answer the following questions.

  1. Convert from cylindrical to spherical coordinates: (1, π/2 , 1) (Hint: Convert first to rectangular form) Ans. (√2 , π/2 , π/4)
  2. Find the distance from a point (4, - 4, 3) to the plane 2x – 2y + 5z + 8 = 0. Ans. 6.
  3. If 1/2 , 1/√ 2 , cos γ are the direction cosines of the vector, find γ. Ans. No solution
  4. A given sphere has the equation x

2

  • y

2

  • z

2

  • 4x – 6y – 10 z + 13 = 0. Find the radius. Ans. 5
  1. Determine the direction cosines of the normal to the plane x + y + z = 1. Ans. cos α = cos β = cos γ = 1/√ 3
  2. Calculate the distance of the planes 2x – y – 2z + 5 = 0 and 4x – 2y – 2z + 15 = 0. Ans. 0.
  3. Find the direction cosines of the line passing through points ( - 2, 4, - 5) and (1, 2, 3). Ans. cos α = 3/√77 , cos β = - 2/√77 , cos γ = 8/√ 77
  4. The vertices of a triangle are (1,1,0) , (1, 0 , 1) and ( 0, 1, 1). Find the point of intersection of the medians of the triangle. Ans. (2/3 , 2/3, 2/3)
  5. Find the midpoint of the points (5, 12, 10) and (3, 0 , - 1). Ans. (4,6,4.5)
  6. Find the angle between two planes x – 2y +z = 0 and 2x + 3y – 2z = 0. Ans. 126.448º
  7. Find the distance between the given plane 2x + 4y – 4x – 6 = 0 and P(0,3,6). Ans. 3
  8. Convert (1, - 1 , - √2) to spherical form. Ans. (2, 7 π/4 , 3π/4)
  9. Find the equation of the sphere x

2

  • y

2

  • z

2

= 4 in cylindrical form. Ans. r

2

  • z

2

  1. Determine the cos γ for the point (2 , 3 , 4). Ans. 4/√ 29
  2. Calculate the distance between points (6, 11, 3) and (4, 6, 12). Ans. 10. 5

END OF ANALYTIC GEOMETRY

Next Topics on November 9, 2020 – Solid Mensuration

  1. Squares and Rectangles
  2. Parallelograms , Trapezoids and Other Quadrilaterals