Rotate $y=\sqrt {x}$

Question

I would like to rotate the function of $y=\sqrt {x}$ clockwise around the origin, such that it intercepts the x-axis at $x=4.63$.

How can I achieve this? Thank you.

Answer 1

You can rotate your coordinate system (counterclockwise) instead.

First, you need a new $x$ axis, such that it intercepts the curve at a distance $r=4.63$ from the origin. The equation of the straight line going through the origin is:

$$y=kx$$

We have the following conditions (for some $x_0$):

$$kx_0=\sqrt{x_0}$$

$$k^2x_0^2+x_0^2=r^2$$

From these 2 equations we can find both $k$ and $x_0$. Square the first equation:

$$k^2x_0^2=x_0$$

Subtract the result from the second:

$$x_0^2=r^2-x_0$$

$$x_0^2+x_0-r^2=0$$

Choose a positive root:

$$x_0=\frac{\sqrt{1+4r^2}-1}{2}$$

Now from the first equation:

$$k=\frac{1}{\sqrt{x_0}}=\sqrt{\frac{2}{\sqrt{1+4r^2}-1}}$$

$$k(4.63)=0.490472$$

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Now we know that for a line $k$ means also the tangent of the angle between the line and the x-axis:

$$y=kx=\tan (\alpha) x$$

$$\alpha= \arctan \left( \sqrt{\frac{2}{\sqrt{1+4r^2}-1}} \right)$$

$$\alpha(4.63)=0.455996=26.13 ^o$$

So this is the angle we need to rotate our coordinate system through.

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If our new coordinates are $x',y'$ then we can express the old ones through them as (see https://en.wikipedia.org/wiki/Rotation_matrix):

$$x=x' \cos \alpha - y' \sin \alpha \\ y=x' \sin \alpha + y' \cos \alpha$$

Now substitute these into the equation $y=\sqrt{x}$:

$$x' \sin \alpha + y' \cos \alpha=\sqrt{x' \cos \alpha - y' \sin \alpha}$$

This expression is implicit, but we can still plot it. Here's the plot made by WolframAlpha:

Is this what you wanted to get?

Answer 2

1. Rewrite your square root function with complex numbers as rotated to custom angle $a$ parabola (details there): $$e^{i a} \left(t+i t^2\right)$$

2. Extract real and imaginary components: $$\Re\left(e^{i a} \left(t+i t^2\right)\right)=t \cos (a)-t^2 \sin (a)$$ $$\Im\left(e^{i a} \left(t+i t^2\right)\right)=t^2 \cos (a)+t \sin (a)$$

3. Solve real component as $x$ for $t$: $$t \cos (a)-t^2 \sin (a)=x,$$ $$t\to \frac{1}{2} \left(\cot (a)\pm \csc (a) \sqrt{\cos ^2(a)-4 x \sin (a)}\right)$$

4. Replace $t$ with found value at imaginary component: $$y=\frac{1}{4} \csc ^2(a) \left(2 \cos (a)-2 x \sin (2 a) \pm \sqrt{2+2 \cos (2 a)-16 x \sin (a)}\right)$$ Now we have explicit formula for rotated parabola.

5. Solve for $a$ when $y=0$. Оne of all solutions would be $$a=-2 \tan ^{-1}\left(\frac{x \left(\sqrt{\frac{2 \left(1+x \sqrt{\frac{1}{x^2}+4}\right)}{x^2}}+\sqrt{\frac{1}{x^2}+4}\right)+1}{2 x}\right)$$

6. put $x=4.63$ and get $a=-2.026792495287185$