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I seem to have lost my other account, so this one is new. Anyways, my questions is:

Is there a general way to produce an expression for a rotated curve that can then be integrated or differentiated like the original? I read through some of the other comments on other threads and I realize that, upon rotation, the resulting curve may not even be a function, but, if it is, how can one proceed?

For example, let $f$ be the function $f(x)=x^2$, $x\ge 0$, and let $g$ be the rotation of $f$ $135^\circ$ about the origin. If, say, $(1,1)$ is on the original curve, one can rotate to get:

$$ \begin{pmatrix} -\displaystyle \sqrt2/2 & -\sqrt2/2 \\ \sqrt2/2 & -\sqrt2/2 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$ Which gives a single coordinate as the rotated point. But, suppose I wanted to to find the derivative of $g$ at the rotated point? It isn't sufficient (is it?) to say:

$$ \begin{pmatrix} -\displaystyle \sqrt2/2 & -\sqrt2/2 \\ \sqrt2/2 & -\sqrt2/2 \end{pmatrix} \begin{pmatrix} x \\ x^2 \end{pmatrix}$$

Use the resulting $y$ coordinate, $\frac{\sqrt2}{2}x-\frac{\sqrt2}2x^2$, find the derivative of this curve, and plug in the $x$ value of the rotated point.

What is the appropriate way to do this?

P.S. No idea why the matrices are not appearing side by side. I used the same latex code I always do...

Nizar
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user299140
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2 Answers2

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If the graph of a differentiable function $f$ is rotated and the rotated curve is again the graph of a differentiable function $g$ (perhaps on a small domain) then the slope of the graph is increased by the angle of rotation $\theta.$ So the derivatives before and after rotation should satisfy

$$\arctan g'(x_1)=\theta + \arctan f'(x_0)$$

Also, the point immediately "below" your curve on the rotated $X$-axis satisfies $x_0-x_1=f(x_0)\tan\theta.$ Combine this to obtain

$$\arctan g'(x_0-f(x_0)\tan\theta)=\theta + \arctan f'(x_0)$$

Similarly, when integrating $g$ over an interval of the new independent variable $x_1,$ take into account that $dx_1=(1-f'(x_0)\tan\theta)dx_0:$

$$\int_{x_1=a}^bg(x_1)dx_1=\int_{x_0=a_0}^{b_0}f(x_0-f(x_0)\tan\theta)(1-f'(x_0)\tan\theta)dx_0$$

where $a_0$ and $b_0$ are solutions of $a_0-a=f(a_0)\tan\theta$ and $b_0-b=f(b_0)\tan\theta.$

Justpassingby
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  • This was my solution; to simply rotate the line. But, then, if I wanted to integrate the rotated function over a specific interval of points on the original say, $(0,0)$ to $(1,1)$, how could this be done? – user299140 Dec 16 '15 at 14:12
  • I do not understand what you mean by integrating over "points of the original". Is this a definite integral in the usual sense, i.e., an area bounded by vertical lines in the new coordinates? – Justpassingby Dec 16 '15 at 14:17
  • What I mean is $\int_0^1x^2=\frac13$. My questions was, is there a way to express this integral for the rotated curve, $g$? Suppose I wanted to rotate $g$ over a specific interval; is there a way to produce the answer without having an explicit form for $g$? – user299140 Dec 16 '15 at 14:28
  • See my latest edit. Note that this is not the same area, because an integral of $g$ measures an area with vertical sides in the rotated coordinate system. If you want the sides to be vertical in the original system then there is no point in rotating: just take the integral of $f$ because areas are invariant under rotations. – Justpassingby Dec 16 '15 at 14:30
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After rotation,

$$x'=Cx-Sy(x),\\y'=Sx+Cy(x)$$ and you would like to express $y'$ as a function of $x'$, right ?

Then you have to invert the equation

$$x'=Cx-Sy(x)=f(x),$$ i.e. solve it for $x$, and get

$$x=f^{-1}(x').$$

From there,

$$y'=Sf^{-1}(x')+Cy\left(f^{-1}(x')\right).$$

Quite often though, $f(x)$ isn't invertible, and when it is, quite often it is not invertible analytically.

The case of the straight line $y=Ax+B$ is easy.

$$x'=Cx-S(Ax+B)$$ gives

$$x=\frac{x'+SB}{C-SA},\\ y'=(S+CA)\frac{x'+SB}{C-SA}+B,$$ obviously another straight line.

The case of the parabola, $y(x)=Ax^2+Bx+D$ is tractable.

$$x'=Cx-S(Ax^2+Bx+D).$$

This is a quadratic equation which you solve for $x$ by the usual formula. As there can be two solutions, you will need to sacrifice one branch.

Actually this approach by means of explicit curve equations $y=f(x)$ is pretty impractical, and you should prefer the parametric approach,

$$x=f(t),\\y=g(t)$$

for which more flexible tools are available. But this is another story.