Question #a63a9

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Answer 1 · 2018-01-14T12:21:51

#r=1/5sec(theta)tan(theta)#

To convert to polar let:

#x= rcos(theta)#
#y= rsin(theta)#

and then preferably solve for #r#.

For #y=5x^2# we have:

#rsin(theta)=5(rcos(theta))^2#

#rsin(theta)=5r^2cos^2(theta)#

Since we know that #r# is not universally 0, we can divide through by #r#:

#sin(theta)=5rcos^2(theta)#

now we divide both sides by #5cos^2(theta)#:

#sin(theta)/(5cos^2(theta))=r#

Since #sin(theta)/cos(theta)=tan(theta)# and #1/cos(theta)=sec(theta)#, we can rewrite one more time:

#r=1/5sec(theta)tan(theta)#