Total area of rectangle rotation

Question

Question:

The rectangle $ABCD$ is rotated into a rectangle $A′B′C′D′$.

Initial state $A$ is at the origin, $B$ is on the y-axis $D$ is on the x-axis.

During the rotation, $A$ travels vertically along the $y$ -axis until it meets $A′$.

While $D$ simultaneously travels horizontally along the $x$ -axis until it reaches the origin.

(The x-axis and y-axis can be understood as impenetrable walls).

The length of the short side is $h$ and the length of the long side is $w$.

Find the total area swept during this process.

My Attemptions

I drew the following diagram, $ABOC'$ area is $h^2$.

But I don't quite understand what the arc $\overset{\frown}{B'C}$ looks like, I don't know how to get an expression for this curve, either in parametric equation form or implicit equation form.

Answer 1

After our exchange, I have understood you mean the type of "rotation" given in the figure below. Caution : it isn't a rotation in the usual sense of the word because this transformation hasn't a center (which should be an invariant point) : see the second interpretation in the "Comment" below.

Fig. 1 : The sides of the moving rectangle "envelope" curves. The North-East curve generated by line segment $[B_tC_t]$ is a piece of astroid : see the classical "sliding ladder image" for example here. The North curve is divided into two parts : an arc of the black curve with equation given below in (1) and an arc of the red curve which is an ellipse with parametric equations given in (2).

Using the technique of envelopes it is possible to get the equation of the envelope of line $A_tB_t$ (black curve on fig. 1). Here is the way I have proceeded :

I start from the equation of line $A_tB_t$ which is :

$$x \cos t - y \sin t = -w (\sin t)^2 \tag{a}$$

I differentiate (1) wrt parameter $t$, considering variables $x$ and $y$ as constant, giving, after a sign change :

$$x \sin t + y \cos t = 2 w \sin t \cos t \tag{a'}$$

Solving system (a) + (a') gives the parametric equations of the envelope (black line) of moving line $A_tB_t$ :

$$\begin{cases}x&=&\tfrac12 w \sin(t)\sin(2t)\\y&=&\tfrac12 w \sin(t)(3+\cos(2t))\end{cases}\tag{1}$$

But in fact, we deal with a moving segment which is not the same as a moving line ; as a consequence, there is a kind of "competition" (I haven't found a better word) with the (small) elliptic arc issued from the locus of $B_t$, itself a quarter of an ellipse (red curve) with equation :

$$\begin{cases}x&=&h \sin(t)\\y&=& w \sin(t)+ h \cos(2t)\end{cases}\tag{2}$$

Please note that in (2), both parameters $w$ and $h$ are present whereas in (1), only $w$ is present.

In fact, we have to do the same kind of calculations for $B_tC_t$. Can you take it from here ?

Fig. 1 has been generated by the following Matlab program :

 clear all;close all;hold on;
 axis([0,2.4,0,2.4]);
 axis equal tight
 w=2;h=1; % rectangle's height and width
 for t=0:pi/50:pi/2
     b=w*cos(t);c=i*w*sin(t);d=h*i*exp(-i*t);
     R=[b,b+d,c+d,c,b]; % rectangle : complex coord.
     c='g';
     if t==0;c='b';end;
     plot(real(R),imag(R),'color','c','linewidth',2);
 end
     t=0:pi/50:pi/2;
     plot(w*(sin(t).^2).*cos(t),w*sin(t).*(1+cos(t).^2),'k','linewidth',8);
     plot(h*sin(t),w*sin(t)+h*cos(t),'r','linewidth',8)
     text([0.,0.,w,w,h+0.05],[0.1,1.1,1.1,0.1,w],{'A_0','B_0','C_0','D_0','B_1'})       

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Comment : here is a different interpretation of your question with a true rotation around center $(w/2,w/2)$ :

Fig. 2 : Axes are in black. The center of rotation is materialized by a little circle.

The swept area can be computing rather easily because it can be decomposed into circular sectors.

(in fact, I don't think it is the kind of transformation you had in mind).

Matlab program having generated figure 2 :

 clear all;close all;hold on;axis equal;
 w=3;h=1; % length and width of rectangle
 C=[w/2;w/2]; % center
 plot(C(1),C(2),'ok')
 R=[0,w,w,0,0
    0,0,h,h,0]; % rectangle's vertices
 for t=0:-pi/50:-pi/2
    M=[cos(t),-sin(t);sin(t),cos(t)]; % rot. matrix
    X=M*R+((eye(2)-M)*C)*ones(1,5); using transl. by C
    plot(X(1,:),X(2,:),'linewidth',4);
 end
 plot([0,w],[0,0],'linewidth',8,'color','k');
 plot([0,0],[0,w],'linewidth',8,'color','k'); % axes' plots

Answer 2

The figure below is a .GIF image showing the sliding/rotation of the a rectangle that has $w = 10$ and $h = 5$.

The upper black curve is the trajectory of vertex $B$ and the right black curve is the trajectory of vertex $C$. The green curve is the envelope of $AB$ while the pink curve is the envelope of $CD$. The blue curve is the envelope of $BC$ while the red curve is the envelope of $AD$.

If $t$ is the angle (in radians) that side $AD$ makes with the $x$ axis, then the equation of the trajectories of $B$ and $ C $ are

$B(t) = (h \sin t , h \cos t + w \sin t ) $

$ C(t) = (w \cos t + h \sin t , h cos t ) $

The green curve is given by

$ P_{AB}(t) = (w \cos t - w \cos^3 t , 2 w \sin t - w \sin^3 t ) $

Then pink curve is given by

$ P_{CD} = (2 w \cos t - w \cos^3 t , w \sin t - w \sin^3 t ) $

The blue curve is given by

$ P_{BC} = (h \sin t + w \cos^3 t , h \cos t + w \sin^3 t ) $

To find the area swept by the sliding/rotating rectangle, we need to find numerically two points:

1. The intersection point between $P_{AB}(t)$ and $B(s)$. I got the intersection point numerically to be $(3.828361, 10.87288)$ at $t = 0.997414, s = 0.872086$

2. The intersection point between $P_{BC}(t)$ and $B(s)$. I got the intersection point numerically to be $(4.981225, 10.39535)$, at $t = 1.358781, s = 1.484109$.

The area enclosed is determined by the integral

$ A = \dfrac{1}{2} \displaystyle \int x(t) y'(t) - y(t) x'(t) \ dt $

where the integral is segmented into three parts corresponding to $P_{BC}(t), B(t)$, and $P_{AB}(t)$.

With the found values of intersection between the curves, the area integral is

$\text{Area} = \displaystyle \int_{\pi/4}^{1.358781} (-h^2 + 3 w^2/8 ) + hw \sin(2 t) \\ \displaystyle - 3/8 w^2 \cos(4 t) \ d t \\ \displaystyle + \int_{1.484109}^{0.872086} - h^2 \ dt \\ \displaystyle + \int_{0.997414}^{\pi/2} w^2 (\frac{1}{8} - \frac{1}{2} \cos(2t) + \frac{3}{8} \cos(4t) ) \ dt $

This integral evaluates to

$\text{Area} = 89.26953 $