DJJerome1976

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These are answers submitted by DJJerome1976

Use plots:-implicitplot to plot equations such as x = 1. See my full solution attached.answer.mw
 

``

plot1 := plots:-implicitplot(x = 1, x = -5 .. 5, y = -1 .. 5)

plot2 := plot(x*exp(x^3), x = -1 .. 2, y = -1 .. 5)

plot3 := plots:-shadebetween(x*exp(x^3), 0, x = 0 .. 1)

plots:-display({plot1, plot2, plot3})

 

``


 

Download answer.mw

 

Try this:

plots:-implicitplot(theta=Pi/6,r=0..5,theta=0..2*Pi,coords=polar)

A range of values is specified using two periods: ..

So, for example, to specify a range of integration such as -sqrt(1-x^2) <= y <= sqrt(1-x^2), one would enter:

y=-sqrt(1-x^2)..sqrt(1-x^2)

Try implicitdiff(x*(x+y)^4 = y*(3*x-y)^2,y,x).

For example, to solve y'(x)=y(x)*x^2, do

dsolve(y'(x)=y(x)*x^2,y(x));

To get an implicitly defined solution, add the implicit option:

dsolve(y'(x)=y(x)*x^2,y(x),implicit);

I've tested the following commands in Maple 16-18, and it works fine.

f:=x->piecewise(0<x<1,x^2+1,1<x<2,x-x^2,2<x<3,x+1-x^2):

plot(f(x),x=0..4,discont=true)

I've been using Maple 17.02 on Mavericks since Mavericks was first released, and I've experienced no issues.

Using the numpoints and/or gridrefine options is one way to try to "smooth" out your implicltplot.

If you add the explicit option to your command, you'll see the two solutions.

solve([-sin(x) = cos(x), `and`(2*Pi > x, x > 0)], x, AllSolutions = true, explicit);

Here is a slight improvement to Carl's code. It turns out that Maple allows one to define uni- and bi-directional edges using [ ] and { }, respectively, in a single graph. Define the graph using:

G1:= Graph( {{G,C}, {T,G}, [A,T], [C,A], [T,C]});

Anything for efficiency, eh? :-)

Since the linalg package is deprecated, use the LinearAlgebra package, which then allows you to use to LinearSolve command:


with(LinearAlgebra):

A := Matrix([[2, 4], [1, -5]]):

b := Matrix([[2], [9]]):

LinearSolve(A, b)

Matrix(2, 1, {(1, 1) = 23/7, (2, 1) = -8/7})

(1)

``


Download linearsolve.mw

Here is one way to do it:

r:=[a,b,c]:

f:=x->lambda-x^2:

map(f,r)

Here is another way to achieve what you want:

A:=unapply(Matrix([[a(t),b(t),c(t)]]),t):

A(t) will yield the symbolic matrix, and A(1) will yield the matrix with each entry evaluated at t=1.

 

with(plots):

plota:=implicitplot(x^3-x*y+y^3=7,x=-5..5,y=-10..10):

plotb:=plot(-11*(x-2)-1,x=-5..5):

display({plota,plotb});

Using your definitions of x and y:

plot([x(t),y(t)],t=-5..5)

will give the desired plot.

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