I need to convert y=x^2 to polar form preferably with r in terms of theta. So far I have:

y=x^2

x^2-y=0

(r^2cos^2θ)-(rsinθ)=0

after that I get lost, help me please

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- April 3rd 2007, 08:22 PMshadowsirenrectangular to polar conversions
I need to convert y=x^2 to polar form preferably with r in terms of theta. So far I have:

y=x^2

x^2-y=0

(r^2cos^2θ)-(rsinθ)=0

after that I get lost, help me please - April 3rd 2007, 08:40 PMJhevon
- April 3rd 2007, 08:44 PMJhevon
Going from what you did:

... (r^2cos^2θ)-(rsinθ)=0

=> rcos^2θ - sinθ = 0 .............divided both sides by r

=> rcos^2θ = sinθ ..................add sinθ to both sides

=> r = sinθ/cos^2θ .................divide both sides by cos^2θ

=> r = sinθsec^2θ ..................change 1/cos^2θ to sec^2θ by trig identity - April 7th 2007, 04:56 PMshadowsiren
Thanks 4 the help.

Could someone explain to me how to convert (x^2)-(y^2)=25 to polar? - April 7th 2007, 05:41 PMtopsquark
- April 7th 2007, 06:10 PMSoroban
Hello, shadowsiren!

Quote:

Convert y = x² to polar form.

We have: . . . . . . x² .= .y

Substitute: .r²·cos²θ .= .r·sinθ

Divide by r: . r·cos²θ .= .sinθ

Divide by cos²θ: . . r .= .sinθ/cos²θ

Therefore: . . . . . . r .= .secθ·tanθ

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We have: . . . . . . . . . . .x² - y² .= .25

Substitute: .(r·cosθ)² - (r·sinθ)² .= .25

Then: . . . . . r²·cos²θ - r²·sin²θ .= .25

Factor: . . . . . r²(cos²θ - sin²θ) .= .25

and we have: . . . . . . r²·cos2θ .= .25

Therefore: . . . . . . . . . . . . .r² .= .25·sec2θ