Question

I really need the answer! Find parametric equations for the curve y = x + 1, x ≥ 0, by using as parameter the distance of the point (x,y) from the origin.

Answer 1

I believe you'd start off by letting the new parameter be \(d\) and set it equal to
\[d=\sqrt{x^2+y^2}\]
Then, substitute \(y=x+1\):
\[d=\sqrt{x^2+(x+1)^2}\\
d=\sqrt{x^2+x^2+2x+1}\\
d=\sqrt{2x^2+2x+1}\]
Complete the square, you get
\[d=\sqrt{2\left(x^2+x+\frac{1}{2}\right)}\\
d=\sqrt{2\left(x^2+x+\frac{1}{4}+\frac{1}{4}\right)}\\
d=\sqrt{2\left(\left(x+\frac{1}{2}\right)^2+\frac{1}{4}\right)}\\
d=\sqrt{2\left(x+\frac{1}{2}\right)^2+\frac{1}{2}}\]
Now solve for x:
\[d^2=2\left(x+\frac{1}{2}\right)^2+\frac{1}{2}\\
d^2-\frac{1}{2}=2\left(x+\frac{1}{2}\right)^2\\
\frac{d^2}{2}-\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\\
\pm \sqrt{\frac{d^2}{2}-\frac{1}{4}}=x+\frac{1}{2}\\
\pm \sqrt{\frac{d^2}{2}-\frac{1}{4}}-\frac{1}{2}=x\]
I think you can drop the plus or minus and only use the positive root, due to its symmetry, but I'm not 100% sure. Also, since x is negative for \(d<1\), you'll have \(d\ge1\) as a restriction.

Here's a plot of the equation \(x=x(d)\): http://www.wolframalpha.com/input/?i=Plot%5B%7BSqrt%5Bd%5E2%2F2+-+1%2F4%5D-1%2F2%2C+-Sqrt%5Bd%5E2%2F2+-+1%2F4%5D-1%2F2%7D%2C%7Bd%2C-2%2C2%7D%5D

Answer 2

Actually, forget what I said about the negative root. You have to drop it because you have negative values of x. Only use the positive root.

Also, that thing I said about symmetry doesn't apply. I'm not sure what I was thinking.

Answer 3

make use of parametric plot

Answer 4

http://www.wolframalpha.com/input/?i=parametric+plot+x%28d%29+%3D+sqrt%28d^2%2F2+-+1%2F4%29-1%2F2%2C+y%28d%29+%3D+sqrt%28d^2%2F2+-+1%2F4%29%2B1%2F2