Faculty of Science and Information Technology
Department of Natural Sciences
Coordinate geometry is the branch of mathematics in which geometry is studied with the help of algebra. The great France mathematician and philosopher Rene Descartes (1596-1650) first applied algebraic formulae in geometry. A system of geometry where the position of points on the plane is described using two numbers called an ordered pair of numbers or coordinates. The first element of the ordered pair represents the distance of that point on x-axis called abscissa and second element on y-axis called ordinate. This abscissa and ordinate makes coordinates of that point.
The method of describing the location of points in this way was proposed by the French mathematician René Descartes (1596 - 1650). (Pronounced "day CART"). He proposed further that curves and lines could be described by equations using this technique, thus being the first to link algebra and geometry. In honor of his work, the coordinates of a point are often referred to as its Cartesian coordinates and the coordinate plane as the Cartesian Coordinate Plane and coordinate geometry sometimes called Cartesian geometry.
1 Cartesian/Rectangular Coordinates System:
Every plane is two dimensional so to locate the position of a point in a plane there is needed two coordinates. The Mathematician Rene Descartes first considered two perpendicular intersecting fixed straight lines in a plane as axes of coordinates. These two straight lines are named as rectangular axes and intersecting point as the origin denotes by the symbol O and the symbol O comes from the first letter of the word origin. The Cartesian coordinate system is named after the inventor name Rene Descartes. The Cartesian coordinate system is also known as Rectangular coordinates system as the axes are in right angle (Rec means right).In Cartesian coordinate system position of a point measured by the distance on both axes. First one is on x-axis called abscissa or x-coordinate denoted by the symbol x and second one is on y-axis called ordinate or y-coordinate denoted by the symbol y.We express the coordinate of a point P in Cartesian plane by the ordered pair P(x , y) or P(abscissa, ordinate).
The horizontal line XOX’ is called x-axis Y and the vertical line YOY’ is called y-axis ( + , + ) ( - , + ) .Both axes divide the whole plane into four P ( ) parts called Quadrants .Four Quadrants x , y XOY, X’OY, X’OY’ and XOY’ are called y st nd rd th anti -clock-wisely 1 ,2 ,3 and 4 X O X ’ x quadrant respectively. The coordinate of Origin the origin is O(0,0) because all distances measured considering origin as starting ( - , - ) ( + , - ) point. Y
’
Polar coordinates System:
In similar manner of Cartesian system for fixing or locating the point P in a plane we take a fixed point O called the pole and a fixed straight line OX called the initial line. Joining line of the points P and O is called radius vector and length of radius vector OP = r and the positive angle XOP is called vectorial angle. It is sometimes convenient to locate the position of a point P in terms of its distances from a fixed point and its direction from a fixed line through this point. So the coordinates of locating points in this system is called Polar coordinates system. The coordinates of point in this system are called Polar coordinates. The polar coordinates of the point P are expressed as Pr(,) . In expressing the polar coordinates of the point P the radius Y vector is always written as the first coordinate. It is P(r,) considered positive if measured from the pole along the line r bounding the vectorial angle otherwise negative. In a polar system the same point has an infinite number of representations and it is the demerits of polar coordinate X O(0,0) X system to Cartesian system. For example: The point P has the coordinates 푟, 휃 , −푟, 휃 + 휋 , −푟, 휃 − 휋 , 푟, 휃 − 2휋 etc. Y
2 Relation between Cartesian and Polar Coordinate System:
Suppose that the coordinates of the point P in Cartesian system is P(,) x y and in Polar system is Pr(,) .Our target here to establish the relation between two coordinates systems. From the triangle with the help of trigonometry we can find the relation between the Cartesian system & the polar system.
Y From the pictorial triangle P(x,y) We get, P(r,) 푥 cos 휃 = ⇒ 푥 = 푟 cos 휃 푟 푦 And , sin 휃 = ⇒ 푦 = 푟 sin 휃 r y 푟 X’ X Again O Applying Pythagorean Theorem from x geometry we have a relation, 푟2 = 푥2 + 푦2 ⇒ 푟 = 푥2 + 푦2 푦 And, tan 휃 = 푥 푦 휃 = tan−1 {Principal Argument} Y’ 푥
Therefore the relations are:
22 xr cos r x y y tan yr sin x
Distance between two Points:
If the coordinates of two points in Cartesian system are A x11, y and B x22, y , then the distance between
22 two points is AB x2 x 1 y 2 y 1 .
Y Again,
If the coordinates of two points in Cartesian A system are 퐴 푟 , 휃 푎푛푑 퐵 푟 , 휃 then the 1 1 2 2 r2 B distance between two points is, r1 2 ퟐ ퟐ 1 X 푨푩 = 풓ퟏ + 풓ퟐ − ퟐ풓ퟏ풓ퟐ 퐜퐨퐬 휽ퟐ − 휽ퟏ O X ’
Y’
3 Area of a triangle:
If the coordinates of the vertices of the triangle in Cartesian system are respectively A x11, y , B x22, y and
C x33, y , then the area of the triangle is xy1 1 11 ABC x y 1 Sq. Units 2 22 xy331
= − + − + −
Area by Sarrus Diagram Method:
1 x x x x △ABC 1 2 3 1 2 y1 y 2 y 3 y 1 1 {(xy xy xy ) ( yx yx yx )} 2 12 23 31 12 23 31
An alternative representation of Sarrus diagram Method:
xy11 1 xy △ABC 22 2 xy33
xy11 1 {(xy12 xy 23 xy 31 ) ( yx 12 yx 23 yx 31 )} 2 Note: 1. In the determinant point must be choose in anti-clock-wise direction. 2. In Sarrus Diagram method first point is repeated. 3. Applying Sarrus Diagram Method we find the area of a polygon.
4 Again,
If the coordinates of the vertices of the triangle in Cartesian system are respectively Ar 11, , Br 22, and
Cr 33, , then the area of the triangle is rrcos sin 1 1 1 1 1 1 ABC rcos r sin 1 Sq. Units 2 2 2 2 2 rr3cos 3 3 sin 3 1 1 ABC rr12sin 2123 r r sin 3213 rr sin 31 2
Mathematical Problem
Problem 01:- Determine the polar coordinates of the point 3, 1 . Solution: We have given xy, 3, 1 . Therefore x 3 and y 1 We know that, r x22 y
2 2 r 31 A.C.W. For calculating angle r 3 1 4 2 C.W. 1’st quadrant 1’st quadrant And 2’nd quadrant 1’st quadrant 1 tan 휋 − 휃 or, −2휋 + 휃 휃 3
1 휋 + 휃 or, −휋 − 휃 2휋 − 휃 or, −휋 + 휃 tan 3 3’rd quadrant 4’th quadrant C.W. A.C.W. C.W. tan tan A.C.W. 6 tan tan 2 6 12 tan tan 6 11 tan tan 6 11 6 11 Therefore the polar form of the given point is r, 2, orr, 2, . 6 6
5 5 Problem 02:- Determine the Cartesian coordinates of the point 2 2, . 4 Solution: 5 We have given r, 2 2, . 4 5 Therefore r 22 and 4 We know that, 51 xrcos 2 2 cos 2 2 cos 2 2 cos 2 2 2 4 4 4 2 And 51 yrsin 2 2 sin 2 2 sin 2 2 sin 2 2 2 4 4 4 2 Therefore the Cartesian form of the given point is xy, 2, 2 .
H.W:
1. Convert the following points to the polar form: 3 3 3 i) ii) iii) 1, 1 iv) 1, 3 2 3, 2 , 22 5 2 5 2 v) vi) vii) aa,3 , 22
2. Convert the following points to the Cartesian form: 2 2 i) 3, ii) 5, iii) 2,a iv) 2, v) 1, 6 4 3 3 6 11 5 vi) 2, vii) 3, viii) 2, ix) 4, x) 2, 3 2 6 6 4
Problem 03:- Transform the equation x33 y3 axy to Polar equation.
Solution:
Given Cartesian Equation is . We have xr cos and yr sin Now replacing x and y from the above equation by its values given equation reduces to the following form r3cos 3 r 3 sin 3 3 a . r cos . r sin r3cos 3 sin 3 3 a . r 2 cos sin racos33 sin 3 cos sin 3 racos33 sin 2cos sin 2
6 3 racos33 sin cos2 2 This required polar equation.
2 Problem 04:- If x, y be related by means of the equation x2 y 2 a 2 x 2 y 2 , find the corresponding relation between r and .
Solution:
Given Cartesian Equation is . We have xr cos and yr sin Putting and the above relation is transformed into the following form 2 r2cos 2 r 2 sin 2 a 2 r 2 cos 2 r 2 sin 2 2 r4cos 2 sin 2 a 2 r 2 cos 2 sin 2 ra212 2 cos 2 sin 2 ra22 cos2 This required relation.
H.W:
Convert the followings to the polar form: 2 i) x2 x 2 y 2 a 2 x 2 y 2 ii) xy3 yx 3 a 2 iii) x2 y 2 y 2(2 x 1) 2 x 3 0 iv) 4x3 y 3 3( x y ) x 2 y 2 5 kxy v) x32 y(2 a x ) vi) x4 x 2 y ( x y ) 2 0
Problem 05:- Transform the equation 2ar sin2 cos 0 to Cartesian equation.
Solution:
Given Polar Equation is . We have , y rsin and x2 y 2 r 2 Putting , the above relation is transformed into the following form,
2ar22 sin r cos 0 r 2 2ar sin 2 r cos 0 r 2 2ay2 x 0 xy22
7 20a y2 x x 2 y 2 2a y2 x 3 xy 2 x32 a y 2 xy 2 x32 y2 a x (As desired) This required Cartesian equation.
H.W:
Convert the followings to the Cartesian form: i) ra2cos 2 2 cos2 ii) r42 2 a cos ec 2 iii) r cos 2 2sin2 2 iv) rk(cos3 sin3 ) 5 sin cos v) 2ar sin2 cos vi) r 1 tan
Broad Questions: 1. Find the distances among the points 1, , 2, and3, . 6 3 2
2. Find the area of the triangle formed by the points , and .
3. If three points 1,2 , 2, 1 andh,3 are collinear then show that h 2 . 4. Find the area of a polygon whose vertices are given 1,3 ,4,1 ,5,3 , 3,2 and 2,4 . 11 5. Show that the three points1, 1 , , and 1,2 form a right angled triangle. 22 6. Find locus of the parametric coordinates point abcos , sin where is a parameter.
8 Change of Axes Transformation of coordinates: The process of changing the coordinates of point or the equation of the curves is called transformation of coordinates. Transformation of coordinates is of three types such as follows
1. Translation of axes: In this process the position of the origin is changed but the direction of coordinate axes is being parallel to the old system. Y Y
When origin 0,0 shifted to the new point ℎ, 푘 and keeping the direction of coordinate axes fixed then the pair of equations h,k X
represents the relation between new system 푋′, 푌′ and old X 0,0 system 푋, 푌 and is called the translation of axes. Fig. Translation of axes
2. Rotation of axes : In this process the position of the origin is not changed but the direction of coordinate axes is being changed through a fixed angle with the x-axis.
When the position of the origin is not changed and the
direction of coordinate axes is being changed through a fixed angle with the X-axis then then this is called rotation of axes and the relation between new 푋′, 푌′ system and old system
푋, 푌 are given below.
Fig. Rotation of axes
3. Translation-Rotation:
In this process the position of the origin is changed and the direction of coordinate axes is being changed through a fixed angle with the x-
axis. The relation between new system and old system are given below.
9 Mathematical problem
Mathematical problem on Translation of Axis
Problem 01: Determine the equation of the curve 2x22 3 y 8 x 6 y 7 0when the origin is transferred to the point2, 1 . Solution: Given Equation of the curve is, 2x22 3 y 8 x 6 y 7 0 ( i ) Origin is transferred to the point hk, 2, 1 so as the transformed relations are x x h x 2 and y y k y 1. Using the above transformation given equation ()i becomes 2x 222 3 y 1 8 x 2 6 y 1 7 0 2x22 443 x y 218 y x 26 y 170 2x22 883 x y 638166670 y x y 2x22 8 x 83 y 6 y 38166 x y 670 2xy22 3 18 0 2xy22 3 18 Removing suffices from the above equation we get the transformed equation of the given curve. 2xy22 3 18 Thich is the required equation that represents an ellipse.
Problem 02: What does the equation x22 y 4 x 6 y 6 0becomes when the origin is transferred to the point2,3and the direction of axes remain unaltered. Solution: Given Equation of the curve is, x22 y 4 x 6 y 6 0 ( i ) Origin is transferred to the point hk, 2,3so as the transformed relations are and y y k y 3. Using the above transformation given equation reduces to x224 x 4 y 6 y 9 4 x 2 6 y 3 6 0 x2244 x y 694861860 y x y x224 x 4 y 6 y 94861860 x y xy22 17 0
Removing suffices from the above equation we get the transformed equation of the given curve. xy2217 Which is the required equations that represents a circle.
10 H.W.
1. Transform to parallel axes through the point 3,5 the equation x22 y 6 x 10 y 2 0 . 2. Transform x222 y 6 x 7 0 to parallel axes through the point3,1 . 3. Transform the equation − 2 + = to parallel axes through − ,2 . 4. Transform the equation x223 y 4 x 6 y 0 by transferring the origin to the point 2,1, coordinate axes remaining parallel. 5. Transform the equation 3x22 14 xy 24 y 22 x 110 y 121 0 shifting the origin to the point 1,2 and keeping the direction of axes fixed.
Mathematical problem on Rotation of Axis
Problem 03: Transform the equation 2 + 2 − = 0 to axes turned through .
Solution: Given that, 2 + 2 − = 0 i Since the axes rotated are an angle and origin be unchanged.
So, = cos − sin and = sin + cos = ′cos − ′sin , = sin + cos 1 1 1 1 = ′ − ′ , = ′ + ′ √2 √2 √2 √2 Using this value in equation i , we get,
1 1 1 1 ′ − ′ 2 + ′ + ′ 2 − = 0 √2 √2 √2 √2 1 1 Or, ′ − ′ 2 + ′ + ′ 2 − = 0 √2 √2 3 5 2 Or, ( ′2 − 2 ′ ′ + ′2) + ′ + 2 ′ ′ + ′2 − = 0 2 2 Or, ′2 − ′ ′ + ′2 + ′2 + 0 ′ ′ + ′2 − = 0 ′2 + ′ ′ + ′2 − = 0 Now removing suffixes, we can write,
2 + 2 + 2 − = 0 This is the required equation.
Problem 04: If the axes be turned through an angle tan1 2 , what does the equation 43xy x22 a become? Solution: Given Equation of the curve is, 4xy 3 x22 a ( i ) The coordinate axes turned through an angle tan1 2 that implies tan 2. Now 2 sin 5 1 2 cos 5 1
11 Considering the new coordinate of the point is xy, and rotating the axes through an angle tan1 2 and origin be unchanged as the transformed equations are as follows
x xcos y sin xy2 1 xy2 55 5
And, yxy sin cos 2xy 1 2xy 55 5
Putting the value of x and y the above equation ()i becomes 2 1 1 1 2 4x 2 y 2 x y 3 x 2 y a 5 5 5
132 Or, 4.x 2 y 2 x y x 2 y a2 55 Or, 4x 2 y 2 x y 3 x 2 y 2 5 a2 Or, 4 2xxyxyy2 4 2 2 3 x 2 4 xyy 4 2 5 a 2 Or, 4 2x2 3 x y 2 y 2 3 x 2 4 x y 4 y 2 5 a 2 Or, 8x2 12 x y 8 y 2 3 x 2 12 x y 12 y 2 5 a 2
Or, 8x2 12 x y 8 y 2 3 x 2 12 x y 12 y 2 5 a 2
Or, 5x2 20 y 2 5 a 2
Or, x24 y 2 a 2 Removing suffices from the above equation we get the transformed equation of the given curve. 2 2 2 x4 y a
H.W:
1. Transformed the equation 2 − 2 + 2 + = 0 to the axes turned through an angle tan−1 . 1 2. Transformed the equation 2 − 2 + 2 + = 0 to the axes turned through an angle tan−1 . 2 3. Determine the equation of the parabola x222 xy y 2 x 4 y 3 0 after rotating of axes through . 4
12 Mathematical problem on Translation-Rotation
Problem 05: Determine the transform equation of − 2 + = 0 when the origin is transferred to the point −2, − and the axes turned through an angle .
Solution: Given Equation is, − 2 + = 0 ………………….(i) Origin is transferred to the point ℎ, = −2, − so as the transformed relations are = ′ − 2 and = ′ − Using the above transformation given equation ()i becomes ′ − 2 − 2 ′ − + = 0 ⇒ ′ − − 2 ′ + 2 + = 0 ⇒ ′ − 2 ′ + = 0 Now removing suffixes, we can write,
− 2 + = 0 ……………………(ii) Again the axes rotated are an angle
So, = cos − sin and = sin + cos = ′cos − ′sin , = sin + cos 1 1 1 1 = ′ − ′ , = ′ + ′ √2 √2 √2 √2 Using this value in equation ii , we get,
1 1 1 1 ′ − ′ − 2 ′ + ′ + = 0 √2 √2 √2 √2
⇒ ′ − ′ − 2 ′ + ′ + √2 = 0
⇒ ′ − ′ − 2 ′ − 2 ′ + √2 = 0
⇒ ′ − ′ + √2 = 0 Now removing suffixes, we can write,
− + √2 = 0 This is the required transform equation.
H.W:
1. Find transform equation of 2 + 2 + 2 − − 22 + 0 = 0 when the origin is transferred to the point 2, and the axes turned through an angle . 2. Find transform equation of x222 xy y 2 x 4 y 3 0 when the origin is transferred to the point −2, and the axes turned through an angle 0 . 3. Find transform equation of 2 + − 2 − + 2 + = 0 when the origin is transferred to the point − , −2 and the axes turned through an angle .
13 Equation & its Geometry ◙ Equation in 2 variables: 1st degree General equation: + + =
It always represents a straight line in plane, provided at least a 0 or, b 0
Homogeneous equation: An equation in which degree of each term in it is equal is called Homogeneous equation. Such as ax2220 hxy by is a homogeneous equation of degree or order 2 because degree of its each term is two. It is noted that homogeneous equation always represents straight lines passing through the origin.
2nd degree homogenous equation:
+ + = ------(*)
In plane, it always represents two straight lines passing through the origin (0, 0).
Lines be perpendicular if
Lines be parallel/coincident if Since two lines passes through the point , so lines must be coincident.
Lines be real and different if .
Lines be imaginary if . But passes through the point .
Non-homogeneous equation: An equation in which degree of each term in it is not equal is called Non-homogeneous equation. Such as ax222 hxy by 2 gx 2 fy c 0 is a non- homogeneous equation of degree or order 2.
2nd degree General equation :
+ + + + + = ------(**)
Descartes found that the graphs of second-degree equations in two variables in plane always fall into one of seven categories: [1] single point, [2] pair of straight lines, [3] circle, [4] parabola, [5] ellipse, [6] hyperbola, and [7] no graph at all.
14
Note: , represents pair of straight lines if
(1) Two parallel lines if = 0, ℎ2 = ab (2) Two perpendicular lines if = 0, a + b = 0
If = ퟎ then Otherwise, (**) will represent: (**) will represent 1. A circle if 0, 푎 = 푏, ℎ = 0 pair of straight line 2. A parabola if 0, ℎ2 = ab and if then (**) 3. An ellipse if 0, ℎ2 − ab < 0 ퟎ 2 will represent curve. 4. A hyperbola if 0, ℎ − ab > 0 5. A rectangular hyperbola if 푎 + 푏 = 0, ℎ2 − 푎푏 > 0, 0
If none of the above conditions are satisfied then (**) represents [1] or [7].
Also,
if c = 0 , then (**) always passes through the origin(0,0). term containing xy can be transformed by a suitably chosen rotation into a form which does not contain xy term. The standard equation is easily obtained. − 2 the rotation angle ‘ ’ that will eliminate the ‘xy’ term is given by cot 2 = or tan 2 = 2 −
Determination of intersecting point (,) of the straight lines represented by the equation ax222 hxy by 2 gx 2 fy c 0 :
Let f( x , y ) ax22 2 hxy by 2 gx 2 fy c f Then set 2ax 2 hy 2 g 0 ax hy g 0 ( i ) x f 2hx 2 by 2 f 0 hx by f 0 ( ii ) y Now calculate the intersecting point of the equation ()i and ()ii we get the intersecting point of the straight lines represented by the equation .
Angle ‘Ө’ between 2 (real) straight lines represented by (*) or (**):
We know that 2 straight lines always cut at an angle (real or imaginary). If ‘Ө’ be that angle, we have to use following formula to find ‘Ө’.
2√ℎ2 − tan = +
15
Case 1 : If 풉ퟐ < 풂풃, then angle ‘Ө’ is imaginary and we can’t view it.
Case 2 : When 풉ퟐ ≮ 풂풃, then angle ‘Ө’ is not imaginary.
Some condition for angle ‘Ө’:- If Ө = 00 , we say, the straight lines are either parallel or, coincident (i.e. same ). If a + b = 0, then Ө = 900 , and we say, the straight lines are perpendicular. If Ө is +ve, we accept it & say that the angle is acute. If Ө is -ve, we add 1800 , and then get an obtuse angle.
Separation of equation of each line from (*) or (**): To determine the equation of each line separately, we need to solve (*) or (**).i.e. we write the equation as Ax2 + B x + C = 0 & then solve it.
Some figure and standard form:-
Circle Ellipse Y
X (g, f )
Std.form. + + + + = Std.form. + =
Parabola Hyparabola
Std.form. 푥2 − 푦2 = 2 2 Std.form. 푦2 = 푎푥 x y 1 Re c tan gular Hyparabula a2 b2
Invariant:
By the rotation of the rectangular coordinate axes about the origin the following equation ax222 hxy by 2 gx 2 fy c 0 changes to the equation ax222 hxyby 2 gx 2 fyc 0 in which the following expression remain unchanged
16 1. a b a b 2. h22 a b h ab a h g a h g 3. where h b f and h b f g f c g f c
2 The unchanged quantities ab , h ab and are called invariants in coordinate transformation.
Mathematical problem Solution
Problem 01:- If the equation is3x22 16 xy 5 y 0 , then find, (a) angle between the lines (b) equation of line.
Solution: (a) Given homogeneous equation is as follows
Comparing the given equation with the general homogeneous equation ax2220 hxy by we have a3, h 8 and b 5. Let an angle between the lines is . 2 h2 ab Then we have tan ab 2 ( 8)2 3.5 Or, tan 35 2 64 15 Or, tan 8 2 49 2.7 14 Or, tan 8 8 8
1014 tan 60.26 8 Therefore the angle between the lines is 60.260 .
(b) Given homogeneous equation is as follows
We expressed the given equation as
Or, 3x22 16 y . x 5 y 0 16y ( 16 y )22 4.3.5 y Or, x 2.3
17 16y 256 y22 60 y Or, x 6 16yy 196 2 Or, x 6 16yy 14 Or, x 6 16y 14 y 30 y Taking positive sign we get xy 5 66 Therefore x5 y x 5 y 0 16y 14 y 2 y y And taking negative sign we get x 6 6 3 y Therefore, x 3 x y 3 x y 0 3 Therefore xy50 and 30xy These are the straight lines passing through the origin.
Problem 02:- Show that 6x22 5 xy 6 y 14 x 5 y 4 0 represents pair of straight lines.
Solution: Given equation is, 6x22 5 xy 6 y 14 x 5 y 4 0 ( i ) Comparing this above equation with the standard equation ax222 hxy by 2 gx 2 fy c 0 we get 55 a6 , h , b 6, g 7, f & c 4 22 Now, 5 67 2 5 5 25 5 35 25 6 6( 24 ) 10 7 42 2 2 4 2 2 4 5 74 2 25 5 35 25 6( 24 ) 10 7 42 4 2 2 4 25 5 35 25 6( 24 ) 10 7 42 4 2 2 4 150 175 175 ( 144 ) 25 294 4 4 4 576 150 100 175 175 1176 4 4 4 726 275 1001 4 4 4 1001 1001 0 44 Since 0 so the given equation represents a pair of straight lines.
18 Another process,
Given equation is, 6x22 5 xy 6 y 14 x 5 y 4 0 ( i ) Comparing this above equation with the standard equation ax222 hxy by 2 gx 2 fy c 0 we get 55 a6 , h , b 6, g 7, f & c 4 22 We know that, = + 2 ℎ − 2 − 2 − ℎ2 5 5 5 2 5 2 = − + 2 ( ) (− ) − ( ) − − 2 − (− ) 2 2 2 2 = 0 Since 0 so the given equation represents a pair of straight lines.
Problem 03:- Find the angle between the straight lines represented by the equation 6x22 5 xy 6 y 14 x 5 y 4 0 . Solution: Given equation is,
Comparing this above equation with the standard equation we get 55 a6 , h , b 6, g 7, f & c 4 22 Assume that be the angle between the straight lines then we have the followings 2 h2 ab tan ab 25 2 36 tan 4 66 tan tan1 2 Problem 04:- Find the equation of the straight lines represented by the equation x226 xy 9 y 4 x 12 y 5 0. Solution: Given equation is,
Arrange the above equation as a quadratic equation in x we get
x226 y 4 x 9 y 12 y 5 0 6y 4 6 y 42 4.1. 9 y2 12 y 5 x 2.1 6y 4 6 y 42 4 9 y2 12 y 5 x 2
19 6y 4 36 y22 48 y 16 36 y 48 y 20 x 2 6y 4 36 y22 48 y 16 36 y 48 y 20 x 2 6y 4 16 20 x 2 6y 4 36 x 2 6y 4 6 x 2 6y 4 6 Taking positive we get x 2 6y 4 6 x 2 62y x 2 2xy 6 2 xy 31 xy3 1 0 6y 4 6 Taking negative we get x 2 6y 4 6 x 2 6y 10 x 2 2xy 6 10 xy 35 xy3 5 0 Therefore, required equations of the straight lines and .
Problem 05:- Find the point of intersection of the straight lines represented by the equation 6x22 5 xy 6 y 14 x 5 y 4 0 . Solution: Given equation is,
Suppose f x, y 6 x22 5 xy 6 y 14 x 5 y 4 Now, Differentiating the function f x, ywith respect to x and y partially and equating with zero, we get f 12xy 5 14 x 12x 5 y 14 0 ( i )
20 And f 5xy 12 5 x 5xy 12 5 0 5x 12 y 5 0 ( ii ) Solving equation (i) and (ii) we get the point of intersection of lines represented by the given equation. Using cross multiplication method on equation (i) and (ii) xy1 25 168 70 60 144 25 xy1 143 130 169 143 11 130 10 xy & 169 13 169 13 11 10 Therefore, the coordinates of point of intersection is xy,, . 13 13
Problem-06: Test the nature of the equation 2 − − 2 + 0 − + = 0 and also find its center. Solution: 1st part: Given that, 2 − − 2 + 0 − + = 0 i Also the general equation of second degree is, 2 + 2ℎ + 2 + 2 + 2 + = 0 ii Comparing and we have,
= , ℎ = − , = − , = , = − , = 2 Now, = + 2 ℎ − 2 − 2 − ℎ2 13 13 = − + 2 (− ) − − − 2 − − 2 − 2 2 5 = − 2 + 2 0 − + − 2 4 33 = 4 33 Since, = 0 so the given equation represents a conic. 4 Again, ℎ2 − ab = + = 2 > 0 And, + = − = 0 Since, + = 0, ℎ2 − > 0, = 0 so the given equation represents a rectangular hyperbola.
2nd part: Let, , = 2 − − 2 + 0 − + = 0
= − + 0 = 0
And = + + = 0
The center of the conic is the intersection of two lines, − + 0 = 0 + + = 0 Solving and we have, 41 1 = − , = 25 5 41 1 Hence the center is at (− , ) 25 5
21 Problem-07: Test the nature of the equation 2 − 2 + 2 − − 0 + = 0.
Solution: Given that, 2 − 2 + 2 − − 0 + = 0 i Also the general equation of second degree is, 2 + 2ℎ + 2 + 2 + 2 + = 0 ii Comparing and we have, 0 = , ℎ = − 2, = , = − , = − , = 2 Now, = + 2 ℎ − 2 − 2 − ℎ2 1 1 1 1 2 = − 2 + 2 (− ) − − 2 − (− ) − − 2 − 2 2 − 2 2 1 = −20 2 − 0 0 − − 2 − 2 4 15 = − 4 15 Since, = − 0 so the given equation represents a conic. 4 Again, ℎ2 − ab = − 2 2 − = − = 0 Since, ℎ2 − = 0, 0 so the given equation represents a hyperbola.
Problem-08: Reduce the equation 8x22 4 xy 5 y 24 x 24 y 0 to the standard form.
Solution: Given general equation of second degree is 8x22 4 xy 5 y 24 x 24 y 0 ( i ) Comparing this above equation with the standard equation ax222 hxy by 2 gx 2 fy c 0 we get a8 , h 2, b 5, g 12, f 12 & c 0 Now, a h g 8 2 12 h b f 2 5 12 8(0 144) 2(0 144) 12( 24 60) g f c 12 12 0 8( 144) 2( 144) 12( 24 60) 1152 288 432 1296 0 And h22 ab 2 40 4 40 36 0 Since 0 and h2 ab 0. So the equation represents an ellipse. Suppose f x, y 8 x22 4 xy 5 y 24 x 24 y Now, Differentiating the function f x, ywith respect to x and y partially and equating with zero, we get f 16xy 4 24 x 16xy 4 24 0 4x y 6 0 ( ii ) And f 4xy 10 24 x 4xy 10 24 0 2x 5 y 12 0 ( iii )
22 Solving equation (i) and (ii) we get center of the conic represented by the given equation. Using cross multiplication method on equation (i) and (ii) xy1 12 30 12 48 20 2 xy1 Or, 18 36 18 18 36 xy 1 & 2 18 18 Therefore, the coordinates of center is , xy , 1,2 . Therefore, the equation of the conic referred to center as origin is 22 8x 4 xy 5 y c1 0 ( iv )
Where c1 g f c 12 24 0 36 So the equation ()iv becomes 8x22 4 xy 5 y 36 0 ( v ) When the xy term is removed by the rotation of axes then the reduced equation is 22 a11 x b y36 ( vi )
Then by invariants we have
a11 b a b 8 5 13 ( vii ) 22 a 1 b 1 h 1 ab h
ab 11 0 40 4 36
ab 11 36 We know, 22 a1 b 1 a 1 b 1 4 a 1 b 1
2 2 ab11 13 4 36 2 ab11 169 144 25 2 ab11 25
a11 b5 ( viii )
Solving equations ()vii and ()viii we have a1 9 and b1 4 The equation ()vi becomes 9xy22 4 36 xy22 1 49 Which is required equations.
Problem-09: Reduce the equation 32x22 52 xy 7 y 64 x 52 y 148 0to the standard form.
Solution: Given general equation of second degree is 32x22 52 xy 7 y 64 x 52 y 148 0 ( i ) Comparing this above equation with the standard equation ax222 hxy by 2 gx 2 fy c 0 we get a32 , h 26, b 7, g 32, f 26 & c 148
23 Now, a h g 32 26 32 h b f 26 7 26 162000 0 g f c 32 26 148
And h22 ab 26 224 900 0 Since 0 and h2 ab 0 . So the equation represents hyperbola. Suppose f x, y 32 x22 52 xy 7 y 64 x 52 y 148 Now, Differentiating the function f x, ywith respect to x and y partially and equating with zero, we get f 64xy 52 64 x 64xy 52 64 0 16x 13 y 16 0 ( ii ) And f 52xy 14 52 x 52xy 14 52 0 26x 7 y 26 0 ( iii ) Solving equation (ii) and (iii) we get center of the conic represented by the given equation. Using cross multiplication method on equation (ii) and (iii) xy1 112 338 416 416 112 338 xy1 450 0 450 450 0 xy 1 & 0 450 450 Therefore, the coordinates of center is , xy , 1,0. Therefore, the equation of the conic referred to center as origin is 22 32x 52 xy 7 y c1 0 ( iv )
Where c1 g f c 32 0 148 180 So the equation ()iv becomes 32x22 52 xy 7 y 180 0 ( v ) When the xy term is removed by the rotation of axes then the reduced equation is 22 a11 x b y180 ( vi ) Then by invariants we have
a11 b a b 32 7 25 ( vii ) 22 a 1 b 1 h 1 ab h
ab 11 0 224 676 900
ab11900 We know, 22 a1 b 1 a 1 b 1 4 a 1 b 1
2 2 ab11 25 4 900
24 2 ab11 4225
a11 b65 ( viii )
Solving equations ()vii and ()viii we have a1 45 and b1 20 The equation ()vi becomes 45xy22 20 180 xy22 1 49 Which is required equations.
Problem 10:- What are represented by the following equations?
2 2 (1) 12x 7xy 10y 13x 45y 35 0
22 (2) x2 xy y 2 x 1 0 (A) If it will be a straight line then find, (i) The angle in between the lines; (ii) find the equation of each line. (B) If it will be a curve then find, (i) Find a rotation angle by which the xy-term will be eliminated; (ii) find the standard form.
Solution:- (1) Given equation is,
Comparing this above equation with the standard equation ax222 hxy by 2 gx 2 fy c 0 we get
= 2, = − 0, = − , = , = , ℎ = 2 2 2 Now we have, = + 2 ℎ − 2 − 2 − ℎ2
45 13 45 2 13 = 2 − 0 − + 2 ( ) ( ) − 2 ( ) − − 0 2 − 2 2 2 2 2 2 − ( ) 2 = 0 Since 0 so the given equation represents a pair of straight lines.
The angle between two lines:- If be the angle between the straight lines then we know that 2 h2 ab tan ab
2√ −12 −1 tan = 12 −1 tan = = tan−1 = ′
25 Separation of lines:-
Given equation is, 2 2 12x 7xy 10y 13x 45y 35 0 12x2 (7y 13)x (10y 2 45y 35) 0 (7y 13) (7y 13)2 4 (10y 2 45y 35) 12 x 212 24x (7y 13) (23y 43)2 24x 7y 13 (23y 43)
Taking positive, Taking negative, 24x 7y 13 (23y 43)
3x 2y 7 0
Hence, and 4x 5y 5 0 are the required equations of two straight lines represented by .
Solution:- (2)
Given general equation of second degree is x222 xy y 2 x 1 0 ( i ) Comparing this above equation with the standard equation ax222 hxy by 2 gx 2 fy c 0 we get a1, h 1, b 1, g 1, f 0& c 1 Now, a h g 1 1 1 h b f 1 1 0 1 0 g f c 1 0 1 And h2 ab 1 1 0 Since 0 and h2 ab 0 . So the equation represents parabola.
To remove ‘ ’ term we need a rotation of ‘ ’ where − cot 2 = 2ℎ 1−1 Or, cot 2 = = 0 2 1 Or, 2 = 0 = This is the required angle. Now we know that, = cos − sin and = sin + cos = ′cos − ′sin = sin + cos 1 1 1 1 = ′ − ′ = ′ + ′ √2 √2 √2 √2 1 1 = ′ − ′ = ′ + ′ √2 √2
26
Putting this in (i) we get,
22 x2 xy y 2 x 1 0 2 Or, (x y) 2x 1 0 2 1 1 1 Or, (x y) (x y) 2 (x y) 1 0 2 2 2 2 1 1 Or, (x y x y) 2 (x y) 1 0 2 2 Or, 2(x)2 2x 2y 1 0 Or, 2(x)2 2x 2y 1 This can be written as 2(x)2 2x 2y 1 Standard form, 2(x)2 2x 2y 1 1 1 1 Or, (x)2 x y (Both side multiply by 2) 2 2 2 1 1 1 1 1 Or, (x)2 2 x ( )2 y 2 2 2 2 8 2 2 1 1 5 Or, (x )2 y 2 2 2 8 1 1 5 Or, (x )2 (y 2 ) 2 2 2 8 1 1 5 Or, (x )2 (y ) 2 2 2 4 2 1 1 5 Or, (x )2 4 (y ) 2 2 4 2 4 2 1 1 5 X 2 4AY where X (x ), A &Y (y ) 2 2 4 2 4 2 That is the standard form of Parabola.
H.W
What are represented by the following equations? (1) x2 + xy + y2 + x + y = 0 . (2) 6 x2 – 5 xy – 6 y2 = 0 ; (3) 6 x2 – 5 xy – 6 y2 = 0
(4) 4 x2 –15 xy – 4 y2 – 46 x + 14 y + 60 = 0 ; (5) x2 – 5 xy + y2 + 8 x – 20 y + 15 = 0 ;
(A) If it will be a straight line then find, (i) The angle in between the lines ; (ii) find the equation of each line. (B) If it will be a curve then find,
(i) Find a rotation angle by which the xy-term will be eliminated; (ii) find the standard form.
27 ◙ Equation in 3 variables:
● 1st degree general equation : ax + by + cz + d = 0 It always represents a plane. ● The 2nd degree general equation : It is given by: It represents a quadratic surface (in brief, quadrics). Some Surfaces in standard forms:
Equations Represents (a second order polynomial containing quadratic terms in x, y, and z) = constant x 2 y 2 z 2 0 an elliptic cone a 2 b2 c 2 x 2 y 2 z 2 1 an ellipsoid. a 2 b2 c 2 x 2 y 2 z 2 1 a hyperboloid of one sheet a 2 b2 c 2 x 2 y 2 z 2 1 a hyperboloid of two sheets. a 2 b2 c 2
(Equations containing two quadratic terms and one linear term) x 2 y 2 z an elliptic paraboloid. a 2 b 2 c x 2 y 2 z a hyperbolic paraboloid. a 2 b2 c x 2 y 2 1 an elliptic cylinder a 2 b 2 x 2 y 2 1 a hyperbolic cylinder a 2 b 2 x2 2kz 0 a parabolic cylinder
28 Coordinates Transformation
■ TWO DIMENSIONAL (2D): ● Rectangular/Cartesian ( x,y ) system ● Polar ( r, θ ) system. ■ THREE DIMENSIONAL (3D): ● Rectangular/Cartesian ( x,y,z ) system ● Cylindrical ( r,θ,h ) system.
● Spherical ( ρ, φ, θ ) system. Θ is called the azimuthal angle, φ the zenith angle.
■ TWO DIMENSIONAL SYSTEM (2D): Cartesian /Rectangular coordinate System:- In the Cartesian coordinate system in 2D, the point in a space or in three dimensional systems be represented by the symbol (x, y)where x is the distance on X axis and y is the distance on Y axis of the point(x, y). Figure: Y
P(x, y) 2 2 2 r x y y X O(0.0)
■ Mutual relation (2D Cases)
x = r cosθ , y = r sin θ ; r2 = x2 + y2 , There are 3 formulas to find θ for some given point P(x,y). These are:
P P
P P
θ = tan-1( y/x ) θ = tan-1 ( y/x ) + 1800 , θ = tan-1 ( y/x )+ 3600
29 ■ THREE DIMENSIONAL SYSTEM (3D): Cartesian /Rectangular coordinate System (RS): In the Cartesian coordinate system in 3D, the point in a space or in three dimensional systems be represented by the symbol x,, y z where x is the distance on x axis, y is the distance on y axis and z is the distance on z axis of the point . Figure:
Cylindrical coordinate System (CS): In the Cylindrical coordinate system in 3D, the point in a space or in three dimensional systems be represented by the symbol , , ℎ where r is the distance of the point from origin or length of radial line, is the angle between radial line and x axis and ℎ is the distance of the point , , ℎ from the xy plane.
푃 푟, 휃, ℎ h
Spherical coordinate system (SS): In the Spherical coordinate system in 3D, the point in a space or in three dimensional systems be represented by the symbol ( ρ, φ, θ ) where ρ is the distance of the point from origin or length of radial line, is the angle between radial line and z axis and the angle between radial line (joining with the foot point of the perpendicular from the given point on the xy plane) and the x axis.
( ρ, φ, θ ) ρ
30 .. Relation between Cartesian/Rectangular and Cylindrical System:
CS RS RS CS x r cos r x2 y2
1 y y r sin tan ( ) x z h h z
.. Relation between Cartesian/Rectangular and Spherical System:
SS RS RS SS x2 y2 z2 x sin cos z z cos1( ) cos1( ) y sin sin 2 2 2 x y z z cos y tan1( ) x
.. Relation between Cylindrical and Spherical System: SS CS CS SS r 2 h2 r sin 1 r tan ( ) h h cos
Restriction:
31 Mathematical Problems
Problem no: 01 Convert 3, , 4 from Cylindrical to Cartesian Coordinates. 3 Solution: Given that,
Cylindrical coordinates of a point is (r,,h)
We know that,
x r cos, y r sin, z h 13 Now, xrcos 3cos 3 3 2 2 3 3 3 y 3sin 3 3 2 2 z h 4 3 3 3 Therefore the Cartesian coordinates of the given point is . x, y , z , , 4 23 H.W:
Convert the followings cylindrical coordinates to the Cartesian Coordinates system:
0 0 1. 4 3 , , 4 2. 4 3,0, 5 3. ( 5,55 ,3) 4. (3,70 ,2) 4
Problem no: 02 Convert 2,2,3 from Cartesian to Cylindrical Coordinates.
Solution: Given that, Cartesian coordinates of a point is x, y , z 2,2,3 We know that, r x2 y2 y tan1( ) x h z Now, 2 2 2 2 r x y ( 2) 2 4 4 8 2 2
1 yy2 tan tan tan 1 xx2
32 Here tan 1 tan tan 4 tan tan( ) 4 3 tan tan( ) 4 3 4 And h z 3 3 Therefore the Cylindrical coordinates of the given point isrz, , 2 2 , ,3 . 4
H.W:
Convert the followings Cartesian coordinates to the Cylindrical Coordinates system:
1. 4 3,4, 4 2. 3, 4,4 3. 3,4,2 4. 4 2 , 1, 4 5. 4 3,0, 5 6. 0,4,9
Problem no: 03 Convert 8, , from Spherical to Cartesian Coordinates. 46 Solution: Given that,
Spherical coordinates of a point is (,,) =
We know that, x sin cos, y sin sin, z cos Now, 1 3 4 3 x sin cos 8sin cos 8 2 6 4 622 2 1 1 4 y sin sin 8sin sin 8 2 2 4 622 2 And 1 z cos 8cos 8 4 2 4 2 Therefore the Cartesian coordinates of the given point isx, y , z 2 6 ,2 2 ,4 2
H.W: Convert the followings Spherical coordinates to the Cartesian Coordinates system:
oo oo 1. 4 3, , 2. 3,124 ,75 3. 3,140 ,140 44
33 Problem no: 04 Convert 2 3,6, 4 from Cartesian to Spherical Coordinates. Solution:
Given that, Cartesian coordinates of a point is x, y , z 2 3,6, 4 We know that, x2 y 2 z 2
1 z 1 z cos ( ) cos ( ) x 2 y 2 2 2 2 Or, tan 1 ( ) x y z z y tan1 ( ) x Now , x2 y 2 z 2(23) 2 6 2 (4) 2 123616 64 8
2 2 2 2 2 2 1 x y x y (2 3) 6 휑 tan ⇒ tantan 휑 tan zz4 12 36 48 4 3 tan 3 4 4 4 Here tan 3 tan tan 3 tan tan 3 2 tan tan 3 2 3
1 yy63 And 휃 tan tan푡푎푛휃 tan 3 xx 2 3 3 Here 푡푎푛휃tan 3 tan푡푎푛휃 tan 3 휃 3 2 Therefore the Spherical coordinates of the given point is 휌,, 휑, 휃 8, , . 33
34 H.W:
Convert the followings Cartesian coordinates to the Spherical Coordinates system: 1. 4 3,4, 4 2. 3, 4,4 3. 3,4,2 4. 4 2 , 1, 4 5. 3,0,0 6. 0,4,9 7. 4 3,0, 5
Problem no: 05 Convert 1 , , 1 from Cylindrical to Spherical Coordinates. 2 Solution:
Given that,
Cylindrical coordinates of a point is (r,,h)
We know that, r 2 h2 r tan1( ) h Now,
2 2 2 2 zr 1 1 1 1 2
1 rr1 tan⇒ tan tan휑 1 zz1 Here tantan휑 1 tantan휑 tan 4 휑 4 And = 2 Therefore the Spherical coordinates of the given point is (,,) ( 2, , ) . 4 2
H.W:
Convert the followings Cylindrical coordinates to the Spherical Coordinates system:
1. 4 3,42o , 4 2. 4 3,0o ,5 3. 3,134o , 4
35 Problem no: 06 Convert 4 3 , , from Spherical to Cylindrical Coordinates. 44 Solution:
Given that,
Spherical coordinates of a point is (,,)
We know that, r sin
h cos Now, 1 r sin 4 3sin 4 3 2 6 4 2 4 1 And z cos 4 3 cos 4 3 2 6 4 2 Therefore the Cylindrical coordinates of the given point isrz 푟,,휃, ℎ 2 6 , ,2 6 . 4
H.W:
Convert the followings Spherical coordinates to the Cylindrical Coordinates system:
1. ( 5, , ) 2. ( 5,1400 ,750 ) 3. (3,400 ,150 ) 4 3
Example:- Convert each of (i) , − , − (ii) , 20 , 0 to the other two system.
Solution:- (i)
Hear the given coordinate system is , − , − which is RS, so we need to convert it to CS and SS.
In CS: r x 2 y 2 82 (3) 2 8.54 y 3 3 3 tan 1 ( ) tan 1 ( ) tan 1 ( ) 3600 tan 1 ( ) 3390 2638 x 8 8 8 h z 7
Hence the point is, (r,,h) (8.54,33902638,7)
36 In SS: x 2 y 2 z 2 82 (3)2 (7)2 11.045 z z 7 cos 1 ( ) cos 1 ( ) cos 1 ( ) 12901827 x2 y 2 z 2 11.045 y 3 3 3 tan1 ( ) tan1 ( ) tan1 ( ) 3600 tan1 ( ) 33902638 x 8 8 8
Hence the point is, , , =(11.045 ,12901827, 33902638).
(ii) Hear the given coordinate system is , 20 , 0 which is SS, so we need to convert it to CS and RS.
In RS: x sin cos 5sin1200 cos 3300 3.75 y sin sin 5sin1200 sin 3300 2.17
z cos 5cos1200 2.5
Hence the point is, , , =( 3.75, 2.17 , 2.5).
In CS: r sin 5sin1200 4.43 3300
h z 2.5
Hence the point is, , , ℎ = ( 4.43, 3300 , ).
H.W:
Convert each of from below to other 2 systems. (i) (4√3, 4, -4) ; (ii) (-√3, 4, 2) ; (iii) (-√3, -4, 4); (iv) (5, 1200 , 3300 ) ; (v) (4√2, -1, -4); (vi)(0, 4, 9) ; (vii) (√3, 0, 0) ; (viii) (4√3, 420 , -4) ; (ix) (4√3, 00, 5) ;
(x) (-√3, 1340, -4); (xi) (4√3, 450,450) ; (xii) (-√3, 1240,750) ;
(xiii) (√3, 1400,1400) ;
37 Transformation of Equations Mathematical problems
Problem 01:- Express Cartesian Equation xy2225 in Cylindrical Equation.
Solution:
Given Cartesian Equation is We have x rcos , y r sin and z푧 = ℎ z
Replacing x and y from the given equation we get desired cylindrical equation as follows, rrcos22 sin 25 rr2cos 2 2 sin 2 25 r2cos 2 sin 2 25 r2 cos 2 25 r2 25 Sec (2 )
This is the required cylindrical equation.
Problem 02:- Express Cartesian Equation x2 y 2 z 2 0 in Cylindrical Equation.
Solution: Given Cartesian Equation is We have, 푧 = ℎ Replacing x, y and z from the given equation we get desired Cylindrical equation as follows, rcos22 r sin z2 0 r2cos 2 r 2 sin 2 z 2 0 rz2cos 2 sin 2 2 0 rz220 This is the required cylindrical equation.
H.W: Transform the following Cartesian equations into the Cylindrical Equations:
1. x2 y 2 23 z 2 x 2. x2 y 2 z 2 2 z 3. z2 y 2 x 2 4. x y z 1
38 Problem 03:- Transform Cartesian Equation x2 y 2 z 2 1 to Spherical Equation.
Solution: Given Cartesian Equation is We have, x sin cos y sin sin z cos Replacing x, y and z from the given equation we get desired Cylindrical equation as follows,
( sin cos)2 ( sin sin)2 ( cos)2 1
2 sin 2 cos 2 2 sin 2 sin 2 2 cos 2 1 2 sin 2 (cos 2 sin 2 ) 2 cos 2 1 2 sin 2 2 cos 2 1
2 (sin 2 cos 2 ) 1
2 (cos 2 sin 2 ) 1
2 cos(2) 1
2 sec(2)
This is the required Spherical Equation.
H.W: Transform the following Cartesian equations into the Spherical Equations:
1. x2 y 2 23 z 2 x 2. x2 y 2 z 2 2 z 3. z2 y 2 x 2 4. x y z 1
Problem 04:- Transform Spherical Equation 2cos to Cylindrical Equation.
Solution: Given Spherical Equation is 2cos We have,
2 2 1 r r h , tan ( ), h Replacing and from the given equation we get desired Cylindrical equation as follows,
r 2 h2 = 2cos
h = 2 [h cos ] h r 2 h2 = 2 r 2 h 2 r 2 h2 = 2h This is the required Cylindrical equation.
39 H.W:
Transform the following Spherical Equations into the Cylindrical Equations: 1. 2. 2sec 3. cosec 4
Problem 05:- Transform Cylindrical Equation r 2 cos 2 h to Cartesian/Rectangular Equation.
Solution:
Given Cylindrical Equation is We have, x rcos , y r sin and z푧 = ℎ z Given equation is
rz2(cos 2 sin 2 ) r2cos 2 r 2 sin 2 z rcos22 r sin z x22 y z [Putting values] This is the required Cartesian/Rectangular Equation.
H.W: Transform the following Cylindrical Equations into the Cartesian/Rectangular Equations: 1. r 2sin 2. z 5sin
Problem 05:- Transform Spherical Equation sin 1 to Cartesian/Rectangular Equation. Solution: Given Spherical Equation is sin 1 We have, x sin cos, y sin sin, z cos and x2 y2 z2 Now, sin 1 2 sin 2 1 2 (1 cos 2 ) 1 2 ( cos)2 1 x2 y 2 z 2 z 2 1 xy221
H.W: Transform the following Spherical Equations into the Cartesian/Rectangular Equations: 1. sin 1 2. 2sec 3. csc 4. sin 2cos
40