Find the shortest distance between r = 2 and r cos (Ø - 45º) = 3
Did you graph them?
r=2 is a circle.
And for:
r*cos(t-pi/4)=3 divide to get r=3*sec(t-pi/4)
Which is a line with slope -1.
See below.
Now to find the distance (which is minimum).
1)Draw a perpendicular from center of circle (the origin) to the line.
2)Find the equation of the perpendicular line (negative reciprocal of slope passing through the origin).
3)Find the intersection between the perpendicular line and the given line.
4)That point is the distance (which is minimum).
5)To find the distance find the distance to the origin and then subtract distance from circle (2 in this case).
Of course the problem is how to find the equation of perpendicular. To do that we need to know the equation of the straight line. Which is what?
We know that,
r=3*sec(t-pi/4)=3/cos(t-pi/4)=3/(cos t*cos pi/4 +sin t*sin pi/4)=3(2/sqrt(2))*(1/(cos t+sin t))=4.24*(1/cos t+sin t).
Divide by r.
1=4.24*(1/(r*cos t + r*sin t))=4.24*(1/(x+y))
Thus,
y=-x+4.24
This is equation of line represented by polar curve.
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The perpendicular has negative reciprocal and passes through the origin.
Thus, it has equation y=x.
The intersection between these two lines is as,
(x,y)=(2.12,2.12)
This distance until the origin is (by distance formula).
sqrt((2.12)^2+(2.12)^2)=3
Thus, the distance until the line is.
3 - 2 =1.