Solving Cartesian Equation: r= 9 cosθ

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1. Find a Cartesian equation to represent the curve r = 9 cosθ



2. I know that rcosθ= x and cos θ= x/r



3. I got (x-9/2)^2 +y^2 = (9/2)^2 but its coming up wrong when I put it into our online homework. Can anyone help me?
 
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hi knv! :smile:

(try using the X2 button just above the Reply box :wink:)
knv said:
1. Find a Cartesian equation to represent the curve r = 9 cosθ
I got (x-9/2)2 +y2 = (9/2)^2

how?? :confused:

anyway, just multiply by r …

r2 = 9rcosθ :smile:
 
knv said:
1. Find a Cartesian equation to represent the curve r = 9 cosθ



2. I know that rcosθ= x and cos θ= x/r



3. I got (x-9/2)^2 +y^2 = (9/2)^2 but its coming up wrong when I put it into our online homework. Can anyone help me?

Your result looks correct to me. It probably has something to do with how you "simplify" it before you type in the answer. Maybe something simple like writing the right side as 81/4?
 
the answer was more simple than I thought.

x2+y2=9x


Thanks!
 
knv said:
the answer was more simple than I thought.

x2+y2=9x


Thanks!

So your original answer was correct, just not in the form the software wanted.
 
yes. I always get things wrong just because its not in the form the software wanted. Miss when we could turn homework in on paper haha
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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