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July 31, 2014

July 31, 2014

Posted by **Paige** on Monday, April 18, 2011 at 1:24pm.

- Calc 2 -
**MathMate**, Monday, April 18, 2011 at 1:29pmWould you like to post your answer if you just wanted to check?

- Calc 2 -
**Mgraph**, Monday, April 18, 2011 at 1:38pmy=... or r=...?

- Calc 2 -
**MathMate**, Monday, April 18, 2011 at 1:52pmIt would be in the form:

x=f(θ),

y=g(θ)

where f and g are functions of θ.

- Calc 2 -
**Mgraph**, Monday, April 18, 2011 at 2:13pmRead the terms of your problem!

- Calc 2 -
**Mgraph**, Monday, April 18, 2011 at 2:26pmIf r=tan(theta)sec(theta)

then r(cos(theta)^2)=sin(theta)

(r*cos(theta))^2=r*sin(theta)

x^2 = y

- Calc 2 -
**Mgraph**, Monday, April 18, 2011 at 2:26pmIf r=tan(theta)sec(theta)

then r(cos(theta)^2)=sin(theta)

(r*cos(theta))^2=r*sin(theta)

x^2 = y

- Calc 2 -
**Paige**, Monday, April 18, 2011 at 2:57pmI am sorry, it is r = 5tan(theta)sec(theta)

- Calc 2 -
**Paige**, Monday, April 18, 2011 at 3:03pmMggraph: but (r*cos(theta))^2 is not x^2, is it? I thought rcos(theta) = x^2

This is what I got:

1 = 5y/[(x^2)(square root of(x^2 + y^2))]

- Calc 2 -
**Mgraph**, Monday, April 18, 2011 at 3:08pmThen all right sides multiply by 5:

x^2=5y or

y=0.2x^2

- Calc 2 -
**Paige**, Monday, April 18, 2011 at 3:48pmAha I got it! Thanks so much! :)

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