Hello, I am attempting to eliminate the parameter from r'(θ) = r'(θ)=<-10sinθ,10cosθ> , but when I do, I get back to the same equation as I would for eliminating the parameter for r(θ)=<10cosθ,10sinθ>(adsbygoogle = window.adsbygoogle || []).push({});

(1) [itex]x^2 + y^2 = 10[/itex]

[itex]y = \sqrt{10-x^2}[/itex]

(2) [itex]y' = \frac{-x}{\sqrt{10-x^2}}[/itex]

then,

since [itex]x^2 + y^2 = 10[/itex], if we let

[itex]x=\sqrt{10}cosθ[/itex]

[itex]y=\sqrt{10}sinθ[/itex], then

(3) r(θ)=<10cosθ,10sinθ>

(4)r'(θ)=<-10sinθ,10cosθ>

then how come if I let,

[itex]x=-\sqrt{10}sinθ[/itex]

[itex]y=\sqrt{10}cosθ[/itex],

I get

[itex] \frac{-x}{\sqrt10}=sinθ[/itex]

[itex] \frac{y}{\sqrt10}=cosθ[/itex]

[itex] (\frac{-x}{\sqrt{10}})^2 + (\frac{y}{\sqrt{10}})^2= 1 [/itex]

[itex]x^2 + y^2 = 10[/itex]

But, my question is:

Why does eliminating the parameter in

r'(θ)=<-10sinθ,10cosθ> ⇒ [itex]x^2 + y^2 = 10[/itex]

instead of what seems more intuitive:

r'(θ)=<-10sinθ,10cosθ> ⇒[itex]y' = \frac{-x}{\sqrt{10-x^2}}[/itex]

It just feel like it would make more sense for a y(x) ⇒ r(θ) and y'(x) ⇒ r'(θ)

So I wonder why y(x) ⇒ r(θ) and y'(x) ⇒ r(θ)

In words, what I'm asking, is when I introduce the theta parameter for the semicircle (1), I get the paremetric/vector function (3). Then if I take the derivative of the semicircle I get (2), and when I introduce the theta parameter, I get (4). But if I eliminate the parameter in (3), I go back to (1), but when I eliminate the parameter of (4), I don't go back to (2), but instead go back to (1) as well.

Why does this occur?

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# Eliminating Parameter for <-rsinθ,rcosθ> ≠ assumed function

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