hitstudent

35 Reputation

5 Badges

5 years, 16 days

MaplePrimes Activity


These are questions asked by hitstudent

Hello.

When I m working on my laptop the plots wont show as it would on my stationnary machine. The surfaces is smooth without any grid and spacecurves dont display at all.

 

I have tried to uninstall and install maple without sucsess. This is maple 2016

 

Is it possible to get maple to show the system of equations as output?

 

Thanks

Hi.

In some cases when dealing with vectofields an such the are integral has to be expressed in terms of r(t).

the general form for r is r^2=(r*cos(t)-a)^2+(r*sin(t)-b)^2, When I solve this in maple it seems like I get the inverse of the desired result.

If I knew that was always the case I could just inverse my result to get the right expression for r, but im not sure if it only applies for this particular cas or all cases.

I would be happy if anyone took a quick look and suggested a way to obtain the desired solution for any center (a,b) for the circle.

 


 

Expression for radius, circle centred at (a,b)

RA := r^2 = (r*cos(t)-a)^2+(r*sin(t)-b)^2

r^2 = (r*cos(t)-a)^2+(r*sin(t)-b)^2

(1)

isolate(r^2 = (r*cos(t)-a)^2+(r*sin(t)-b)^2, r)

r = (1/2)*(a^2+b^2)/(cos(t)*a+b*sin(t))

(2)

eval(%, [a = -1, b = 0])

r = -(1/2)/cos(t)

(3)

plot3d([-2*x, x^2+y^2], y = -sqrt(-x^2-2*x) .. sqrt(-x^2-2*x), x = -2 .. 0, color = [green, red], orientation = [0, 0, 0])

 

(1/2)*Pi <= t and t <= 3*Pi*(1/2)

(1/2)*Pi <= t and t <= (3/2)*Pi

(4)

0 <= r and r <= -2*cos(t)

0 <= r and r <= -2*cos(t)

(5)

Area_off_center = int(r, [r = 0 .. -2*cos(t), t = (1/2)*Pi .. 3*Pi*(1/2)]); 1; Area_at_center = int(r, [r = 0 .. 1, t = 0 .. 2*Pi])

Area_off_center = Pi

 

Area_at_center = Pi

(6)

``

``


 

Download parametrization_of_r_not_centred_at_orgin.mw
 

I would also happily like to know how I can solve for the range r can take, obviously in the example i´m working with here r starts at 0, but that is not always the case i guess.

 

Thank you, your help is much apperciated

Hello.

In maple you can use the interactive menu and choose what axes you want to look at the plot, ie. I choose Z axes and see the plot from above, can i do this by using syntax as a plot option?

 


 

restart; with(plots); with(Student[Precalculus]); with(LinearAlgebra)

S1 := x^2+y^2 = 4;

x^2+y^2 = 4

 

z = 2-x

 

F := Vector(3, {(1) = cos(z), (2) = x^2, (3) = 2*y}, attributes = [vectorfield, coords = cartesian[x, y, z]])

(1)

F := VectorCalculus:-Curl(F)

F := Vector(3, {(1) = 2, (2) = -sin(z), (3) = 2*x}, attributes = [vectorfield, coords = cartesian[x, y, z]])

(2)

g := [diff(rhs(S2), x), diff(rhs(S2), y)]

[-1, 0]

(3)

intergrand := -F[1]*g[1]-F[2]*g[2]+F[3]

2+2*x

(4)

``

Integrasjons område er projeksjonen på z=0 planet, her sirkel med radius=2

Intersection := lhs(S1) = rhs(S2);

x^2+y^2 = 2-x

 

{x = x, y = (-x^2-x+2)^(1/2)}, {x = x, y = -(-x^2-x+2)^(1/2)}

(5)

X := solve(rhs(Y[1, 2]))

-2, 1

(6)

plot3d([lhs(S1), rhs(S2)], x = X[1] .. X[2], y = rhs(Y[2, 2]) .. rhs(Y[1, 2]), color = [red, green]);

 

plot3d([lhs(S1), rhs(S2)], x = X[1] .. X[2], y = rhs(Y[2, 2]) .. rhs(Y[1, 2]), color = [red, green])

 

``


 

Download stokes_start.mw

Given 3 surfaces:

x^2+y^2=1,z=0(the xy base plane) and z=1-x^2

To plot these I suggested to use cylindrical coordiates knowing x=r*cos(t) and y=r*sin(t)

Which leads to z=1-r^2*cos^2(t)

However i got problems knowing how to plot this object and dearly ask for help.

plothelp.mw

1 2 3 Page 2 of 3