## 178 Reputation

17 years, 84 days

## question...

Are you trying to fill the region between the two curves? If you are, try replacing the last line display with this display([p1,p4,p5,p6,p7,p8,p2,p3]); I think that the problem is that your plots are given as a set to display. Maple may evaluate the plots in an arbitrary order, so some or all of the regions you assigned to be white may be getting covered and each time you reevaluate the worksheet the order may change. Using a list seems to work, but this may not be the best solution. Hopefully, one of the experts will comment.

## fix the parentheses...

There are two angles that work restart; 8=tan(m)*6-(9.8)/(2*20^2*cos(m)^2)*6^2: angle:=[solve(8=tan(m)*6-(9.8)/(2*20^2*cos(m)^2)*6^2,m)]; [1.487737664, 1.010353880, -1.653854989, -2.131238773] evalf(180/Pi*angle); evalf(180/Pi*angle); 85.24108914 57.88901312 You have the idea, here is how I did it with Maple (worksheet mode) restart: eq1:=y= tan(theta)*x - (9.8/(2*v^2*cos(theta)^2))*x^2; #Substitute values for y, x , and v in eq1. eq2:=subs(y=8,x=6,v=20,eq1); #Solve eq2 for theta. angle:=[solve(eq2,theta)]; #Convert to degrees to check. evalf(180/Pi*angle); evalf(180/Pi*angle); #I want to see the graph so I make one function for each angle. Note I used rhs(eq1) to get right hand side of eq1, that eliminates y and makes f1 and f2 functions of x. If I don't do that I will have two equations in two variables y and x and I will have to use implicitplot. f1:=subs(v=20,theta=angle,rhs(eq1)); f2:= subs(v=20,theta=angle,rhs(eq1)); plot([f1,f2] ,x=0..8); #To find the x coordinates of the points of intersection x = 0 and x = 6 do solve(f1=f2,x);

## Use the given information...

Sam, everything you need is given in the picture. Think of an xy-coordinate plane with the origin at A. The base of the triangle is the positive x-axis. If C has coordinates (x,y) you can use the equation you posted. You already have C's y coordinate from the given information C=(x, 8). Can you find x from the triangle? (hint: think geometry) After you find x you will have both x and y. Since v = 20 is given you can then substitute the values in the equation and solve for theta, which will be the only remaining variable. updated: I had inadvertently used the letter B in place of C. It is now fixed.

## solve...

Use solve, for example solve(5*(7*y + 3) = 39, y); 24/35 See the help page for solve for more details, you can enter "?solve" to get to this page directly from your worksheet or doc.

## piecewise...

I don't think Maple understands the piecewise construction in this context. p1,p2 := , : > r :=t-> piecewise( t<0, p1, p2); r := t -> piecewise(t < 0, p1, p2) > limit(r(t), t=1), r(1), r(-1); If you look at the output Maple literally uses the vector object as the value of the piecewise function. I tried to paste the output without success. I would love to know how you got that vector output to post.

## Missing Operator...

Insert the multiplication operator " * " in 2x as in 2*x. In worksheet mode Maple does not interpret 2x as 2 times x. In document mode it will usually add a space as in 2 x, which implies multiplication.

## Minimizing the maximum...

It seems this post was inadvertently overlooked so I am replying in part to bump it up and in part to offer a possible method. Since I am not a Maple expert my approach would be more of a "textbook" method (outlined below) which is probably not the best way to do this with Maple. My guess is that you already know this approach and are looking for a better method using Optimization. In that case, maybe the bump up will help it get noticed by one or the experts. It would also help if you posted the actual function and the ranges of the variables, since the complexity of the function is likely to factor into the choice of approach. Outline of a possible approach: I would try to locate possible local extrema of the function on the interior of the region by looking at the zeros of the partials and the Hessian to determine at which points if any the local maxima occur. Then I would check the boundaries - first the edges and then the four corners (from your post it appears the boundary is a rectangle). The successfulness of this approach will depend on the complexity of the function. If you would like more details let me know. Hopefully one of the experts will offer better advice first.

## 2-D plot with context menu...

I am not sure what the problem is, it works just fine for me. Have you assigned x anywhere else in the worksheet? Try using a new document and see if you get the same problem. If you do post your input exactly as you you entered it or upload the worksheet.

## IntTutor...

I see what you mean. Try to think of it as a tutor that is not going to do the problem for you, but will help you figure it out by doing most of it for you. Work in both directions: if you get stuck look at the hint, then click undo and see if you can fill in what is missing. Click Next step and see if you get to the same place as the hint had given. You will learn a lot more working this way since you will still have to figure some of it out on your own. If you are absolutely stuck post the problem here in the student forum and someone will help you. When you post it, try to explain some of the things you have already tried. People will be much more willing to help if they think you have made a sincere effort and are not just looking for someone to do your homework.

## implicitplot3d...

Interestingly, when I interchange x and z in the equation 16x^2=y^2+4z^2 to get 16z^2=y^2+4x^2 and then used implicitplot3d I expected to get a similar plot to the one I got above when I solved for x explicitly; just oriented along the z axis instead. But I did not. I am not sure why? Hopefully Acer, or one of the other experts will take a look and see what I am doing wrong.

## Try...

Does this give you an idea of how to proceed? For z=0, I let x=t in 3*x^2-5*y^2=0 to get 3*t^2-5*y^2=0, then solve for y and you have t->(t, sol to solve for y in terms of t, 0) as your curve. restart: with(plots): S:=[solve(3*t^2-5*y^2=0,y)]; p1:=plot3d(3*x^2-5*y^2, x=-2..2, y=-2..2): spc:=spacecurve({[t, S, 0],[t,S,0]}, t=-2..2, thickness=2, color=red): display([p1,spc]); Note: Since there were two solutions for y in terms of t, S and S, I entered a set of parameterized curves, {[t, S, 0],[t,S,0]} to get both lines.

## z=f(x,y)...

I think the problem is that implicit plot gives z=f(x,y) and you want x=g(y,z). You can try the following: restart: eq:=16*x^2=y^2+4*z^2; S:=[solve(eq,x)]; plot3d([S,S], y=-100..100,z=-100..100, view=-20..20); There are probably better ways that I am not aware of. You might try looking at plottools[transform] also.

## Pi for the constant...

Maple uses an uppercase P for the constant. You can type Pi/2 directly into the tutor.

## alternative...

You could also do the following:

> restart:

> solve({x^2+y^2+z^2=1, y=z}, [x,y,z]); > allvalues(%); >

If you like you can change the parameter to t, for example: z = t, y = t , x= sqrt(1-2*t^2)
for -1/sqrt(2) <= t <= 1/sqrt(2) gives one half of the circle.

>

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