Calc 2: Curves Defined by parametric equations

MillerGenuine
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these are fairly simple questions that only require you to plot points and then find a Cartesian equation of the curve. My question is when trying to solve for the Cartesian Equation, whether to solve for X first or Y? I will give an example.

X=3t-5 , y=2t + 1
Graphing this is simple, solving for Cartesian equation of the curve by eliminating the parameter is fairly simple,

x=3t-5 therefore t=(x+5)/3

Subsitute t into y=2t+1 and you get y=2/3(x+5) + 1

easy enough.

Now in the following problem

x=(t^2)-2 , y=5-2t

again, graphing is simple, but this time to find the Cartesian equation they first solve for t like so..

y=5-2t therefore t=(y-5)/2

then x= 1/4 (y-5)^2 - 2

So does it matter whether I have my Cartesian equation in the Form of X=... or Y=...
or do i just take the easiest path to solve for "t" and substitute?
 
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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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