Assignment #12



I would like you to be able to manipulate graphs in 2 and 3 dimensions. Today we will start with two dimensions. You should be able to translate and rotate a graph defined parametrically and implicitly. For polar equations, rotation about the origin by an angle $\phi$ is simply making the replacement $\theta \rightarrow \theta + \phi$, but translatation is probably easiest if you change from polar to parametric plot by representing the polar point $(\theta, r(\theta))$ as a point in rectangular coordinates $[r(\theta) cos(\theta), r(\theta) sin(\theta)]$ and then plotting the graph parametrically.

Question #1:
Write a function translate which accepts a pair of expressions in the parameter t, $x(t)$ and $y(t)$ (representing a parametric curve) and a point p = [px,py] and the result is an expression which translates the equation by px in the $x$-direction and py in the $y$-direction.

Test your function as follows:
> plot( [op(translate([cos(t),sin(t)],[1,-2])), t=0..2*Pi] );
and verify that this is a graph of a circle with origin at $(1,-2)$. Note that the op is there because plot expects you to give the parametic equation of the form plot( [xoft, yoft, t=a..b] ) but I would like your function translate to work so that translate([xoft, yoft], [a, b]) will return [newxoft, newyoft]. Your function should work on any parametric equation that I might give you.

Verify that
> A:=plot([t*cos(t),t^2,t=-1..1]);
> B:=plot([op(translate([t*cos(t),t^2],[-3,1])), t=-1..1], color="Blue");
> C:=plot([op(rotate_origin([t*cos(t),t^2],Pi/3)), t=-1..1], color="Green");
> DD:=plot([op(rotate_point([t*cos(t),t^2],[-3,1],Pi/3)), t=-1..1], color="Orange");
> display([A,B,C,DD]);
Yields the picture:
The following equation was rotated $\pi/6$ degrees clockwise around the point $(2,1)$ and then it was rotated $\pi/4$ degrees around $(-1,0)$. Find the equation after it was rotated back.

FYI, I had in this last sentence 'counter clockwise' but I meant 'clockwise'. This was a mistake that I have since corrected. Your final answer to this question should be very simple and in this case you will know you have the right answer.

[(1/2)*2^(1/2)*3^(1/2)*sin(t)*cos(t)-(1/2)*2^(1/2)*sin(t)*cos(t) -(3/4)*2^(1/2)*3^(1/2)+(9/4)*2^(1/2)+(1/4)*2^(1/2)*cos(t) +(1/4)*2^(1/2)*3^(1/2)*cos(t)-1, -(1/4)*2^(1/2)*(2*3^(1/2)*sin(t)*cos(t)-3^(1/2) +cos(t)+1+2*sin(t)*cos(t)-3^(1/2)*cos(t))]

Write a function rotate_origin which takes a parametric equation $x(t)$ and $y(t)$ (representing a parametric curve) and an angle $\theta$ and rotates a point about the origin by an angle $\theta$ (clockwise). Write a function rotate_point which rotates a parametric curve about a point by an angle $\theta$ (clockwise). Your function rotate_point should work by translating your function so that $p$ is at the origin and then rotating by $\theta$ degrees around the origin and then translating back so that $p$ is in its original place.

Question #2:
Do the same thing for an implicit equation, that is write a function translate_implicit, rotate_origin_implict and rotate_point_implicit so that you can translate and rotate an implicit equation. Test your functions on the following equations.



> A:=implicitplot(y=x^2*cos(y)^2,x=-1..1,y=0..0.8);
> B:=implicitplot(translate_implicit(y=x^2*cos(y)^2,[-3,1]),x=-4..-2,y=1..1.8,color="Blue");
> xmin:=min(rotate_origin([-1,0],Pi/3)[1],rotate_origin([1,0],Pi/3)[1],rotate_origin([-1,.8],Pi/3)[1],rotate_origin([1,.8],Pi/3)[1]);
> xmax:=max(rotate_origin([-1,0],Pi/3)[1],rotate_origin([1,0],Pi/3)[1],rotate_origin([-1,.8],Pi/3)[1],rotate_origin([1,.8],Pi/3)[1]);
> ymin:=min(rotate_origin([-1,0],Pi/3)[2],rotate_origin([1,0],Pi/3)[2],rotate_origin([-1,.8],Pi/3)[2],rotate_origin([1,.8],Pi/3)[2]);
> ymax:=max(rotate_origin([-1,0],Pi/3)[2],rotate_origin([1,0],Pi/3)[2],rotate_origin([-1,.8],Pi/3)[2],rotate_origin([1,.8],Pi/3)[2]);
> C:=implicitplot(rotate_origin_implicit(y=x^2*cos(y)^2,Pi/3),x=xmin..xmax,y=ymin..ymax, color="Green");
> xmin2:=min(rotate_point([-1,0],[-3,1],Pi/3)[1],rotate_point([1,0],[-3,1],Pi/3)[1],rotate_point([-1,.8],[-3,1],Pi/3)[1],rotate_point([1,.8],[-3,1],Pi/3)[1]);
> xmax2:=max(rotate_point([-1,0],[-3,1],Pi/3)[1],rotate_point([1,0],[-3,1],Pi/3)[1],rotate_point([-1,.8],[-3,1],Pi/3)[1],rotate_point([1,.8],[-3,1],Pi/3)[1]);
> ymin2:=min(rotate_point([-1,0],[-3,1],Pi/3)[2],rotate_point([1,0],[-3,1],Pi/3)[2],rotate_point([-1,.8],[-3,1],Pi/3)[2],rotate_point([1,.8],[-3,1],Pi/3)[2]);
> ymax2:=max(rotate_point([-1,0],[-3,1],Pi/3)[2],rotate_point([1,0],[-3,1],Pi/3)[2],rotate_point([-1,.8],[-3,1],Pi/3)[2],rotate_point([1,.8],[-3,1],Pi/3)[2]);
> DD:=implicitplot(rotate_point_implicit(y=x^2*cos(y)^2,[-3,1],Pi/3),x=xmin2..xmax2,y=ymin2..ymax2, color="Orange");
> display([A,B,C,DD]);
Yields the picture:


The following equation was rotated $\pi/6$ degrees clockwise around the point $(2,1)$ and then it was rotated $\pi/4$ degrees around $(-1,0)$. Find the equation after it was rotated back.

FYI, I had in this last sentence 'counter clockwise' but I meant 'clockwise'. This was a mistake that I have since corrected. Your final answer to this question should be very simple and in this case you will know you have the right answer.

(1/4)*x*2^(1/2)+(1/4)*2^(1/2)-(1/4)*y*2^(1/2)-1/2+(1/4)*3^(1/2)*2^(1/2)*x+(1/4)*2^(1/2)*3^(1/2)+(1/4)*3^(1/2)*y*2^(1/2)-(1/2)*3^(1/2) = (1/16)*(3^(1/2)*2^(1/2)*x+2^(1/2)*3^(1/2)-3^(1/2)*y*2^(1/2)-6*3^(1/2)-x*2^(1/2)-2^(1/2)-y*2^(1/2)+10)^2


You should open up a new worksheet and start from scratch. You will have to save your work in a file and upload that file on to the course moodle. Your solution should be a sequence of commands where it is easy to change the input string and after you execute the sequence of commands you should have the correct output string. Please add documentation to your worksheet to explain how it works. Just a few sentences is sufficient, but imagine that you were opening up the worksheet for the first time and wanted to know what it did. You will be marked down if what you write is not clear and coherent.

You should finish your assignment by Wednesday, November 29 by 11:59pm. Assignments submitted after this date will be assessed a penalty of 10% per day.