Find the intersection point of r cos(Ø - pi/4) = (2)^½ and r cosØ = 1

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- Mar 15th 2007, 11:24 PMcazimiintersection point
Find the intersection point of r cos(Ø - pi/4) = (2)^½ and r cosØ = 1

- Mar 16th 2007, 06:10 AMtopsquark
Must be a couple of people in the same class!

Call phi = t for convenience.

The first equation states:

rcos(t - pi/4) = rcos(t)cos(pi/4) + rsin(t)sin(pi/4) = sqrt(2)

or after some simplifying:

rcos(t) + rsin(t) = 2

Using the second equation gives:

1 + rsin(t) = 2

rsin(t) = 1

Now, consider the two equations:

rsin(t) = 1

rcos(t) = 1

Dividing the two gives:

tan(t) = 1

t = pi/4 + n*pi (where n = 0, 1, 2, 3, ...)

Now

r = 1/cos(t)

So

r = 1/cos(pi/4 + n*pi) = 1/[cos(pi/4)cos(n*pi) - sin(pi/4)*sin(n*pi)]

Now:

cos(pi/4) = sin(pi/4) = 1/sqrt(2)

cos(n*pi) = (-1)^n

sin(n*pi) = 0

Thus

r = (-1)^n*sqrt(2)

t = pi/4 + n*pi

-Dan