Examples for Cartesian and polar coordinates in two dimensions

Example 1

Find the polar coordinates of the point Q(-3, 5) and write down the vector -OQ--> in both Cartesian and polar forms.

Solution

Recall that if (x,y) are the Cartesian coordinates of Q and (r,

) are the polar coordinates, then the Cartesian and polar forms of ---> OQ

are xi + yj and r cos

i + r sin

j respectively, where

x = r cosh, y = rsinh,

and

V~ -------- x y r = x2 + y2, cos h = --, sin h = --. r r

The Cartesian form is therefore ---> OQ = -3i + 5j.

To find the polar form, first find r and . We have

V~ -----2----2 V~ --- r = (- 3) + 5 = 34

and

- 3 5 cosh = V~ --, sin h = V~ --. 34 34

Note that is in the second quadrant (x negative, y positive). Using the inverse cosine function on a calculator, we obtain (in radians)

- 1 --3-- h = cos ( V~ 34) ~~ 2.11.

Hence the polar form of - --> OQ is V~ --- 34 cos 2.11i + sin 2.11j

Example 2

Find the Cartesian form of the vector ---> OQ whose polar form is

-OQ--> = 3cos 200oi + 3 sin 200oj.

Solution

This time we find x and y from the polar coordinates. We have

o x = rcos h = 3cos 200 ~~ - 2.82

and

o y = rsinh = 3 sin 200 ~~ - 1.03.

Therfore the Cartesian form of -O-->Q is -2.82i - 1.03j. Find the polar form of the vector ---> OQ whose Cartesian form is

-2.82i - 1.03j.

Notice that this is just the reverse of the previous problem, included here to illustrate that care is needed to find the polar angle , especially when it’s in the third quadrant.

First,

V~ -------------------- r = (- 2.82)2 + (-1.03)2 ~~ 3.0022.

As cos = x- r and sin = y- r , we see that

cos h = - 0.939311, sin h = -0.343081.

Note also that

sinh y tan h = ----- = --= 0.365248. cos h x

The difficulty with using a calculator to find is that the inverse cosine function returns values between 0 and , while the inverse sine and inverse tan functions return values between -/2 and /2. So to obtain a positive angle in the third quadrant, we must make an adjustment to the calculator output. With the calculator in radian mode, any one of the three formulas

h = 2p - cos- 1(- 0.939311) -1 h = p - sin (- 0.343081) h = p + tan- 1(0.365248)

will give the right answer, which in radians is = 3.492 (or in degrees approximately 200°).

Example 3

Find the vector v of magnitude 2 in the direction of the vector r = 3i - j.

Solution

First calculate the length of r.

V~ -2--------2 V~ --- |r|= 3 + (- 1) = 10

Therefore the unit vector in the same direction as r is

^r = V~ 3-i- V~ 1-j 10 10

Multiplying by 2 gives the required vector,

v = V~ -6-i- V~ -2-j. 10 10