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with(plots):

a:=polarplot(3-3*cos(theta),theta=0..2*Pi):

c:=plot((3*sqrt(2)+3)/2 + ((-3*sqrt(2))/(-3*sqrt(2)-6))*(x+((3*sqrt(2)+3)/2)),x=-10..10):

display(a,c,view=[-10..10,-10..10]);

 

a:= is the polar plot of the cardiod (3-3cos(theta)


In order to plot the tangent line to the cardiod in theta= 3Pi/4, I find the point (x,y) in rectangular coord x=(3-3cos(theta)cos(theta) and y=(3-3cos(theta)sin(theta); then I find the derivative of dx/dy=

[(3-3*cos(theta)*cos(theta)+sin(theta)(3sin(theta)]/[-(3-3cos(theta)sin(theta)+cos(theta)(3sin(theta)]and from here I get the slope.So I can plot c:= tangent line to the cardiod in 3Pi/4.

How can I avoid having to convert everyting to rectangular coords, and plot the tangent line in polars?

 

 

Hi All,

I have a problem with regard to partial differential equations. I am using Lagrangian dynamics for a problem. First i have a function First i defined a function with two speeds of angles (first derivatives):

ODE := 5*(diff(theta1(t), t))+diff(theta2(t), t). This gives:

Now this gives an output. Lagrange (just a simple example now) demands that i now derive the obtained function with regard to the first derivative of theta1. In this case, the answer i want is 5. Now, if i give the command: 

diff(ODE, diff(theta1(t),t)), maple says go home. Does anybody know how to solve this? I have been searching for a solution all afternoon.

 

Thnx in advance!

Hi everyone,

I have a question regarding the derivation of tensors/matrices.
Let's assume for simplicity, that I have a vector (6x1) s and a matrix A (6x6)defining Transpose(s)*Inverse(A)*s. From this function I want to calculate the derivative w.r.t. s. My approach would be

restartwith(Physics):
with(LinearAlgebra):
Define(s,A)

Diff(
Transpose(s)*Inverse(A)*s, s)

As a result I get

though I'd rather expect something like Inverse(A)*s + Transpose(s)*Inverse(A)

Now as I'm pretty new to Maple, I can imagine that my approach is wrong, but I don't know any better and can't seem to get any information out of the help documents.

Thanks in advance for any of your suggestions!

Hi

I need you to help me in writting procedure with input "r" ( order of derivative) and some coefficients seq(alpha[i],i=1..N).  My code work very well, need only put all the element in procedure with output The Taylor series obtained in last line of my code and the order of error.  I want the procedure return the coefficients beta used in the series and the order of Error  and the coefficients beta[i]

May thinks.

Second_Question.mw

 

I will not modify my previous question: only I add this remark. I tried to write these lines of procedure.
Add_display.mw

I need only to add( if the procedure is true) these lines in my procedure.  These lines gives the degree of error computed in the code, but when I put these lines in the codes, there is an error. Thanks for your help.

Degree := degree(Error,stepsize);
    if (showorder) then:
       print(cat(`This stencil is of order `,Degree));
    fi:
    if (showerror) then:
       print(cat(`This leading order term in the error is `,Error));
    fi:
    convert(D[r$n](f)(vars) = stencil,diff);

 

 

 

Dear All, I need your help to plot the numerical solution. many thanks.

The variable t in [0,T], x in [0,1], b in [0,2].

Difference finie for waves equation is :

pde:=diff(u(x, y,t), t$2) = c^2*(diff(u(x, y,t),x$2)+diff(u(x,y,t),y$2));

i: according to x, j according to y, and k according to t.

u[i,j,k+1]=2*u[i,j,k]-u[i,j,k-1]+(c*dt/dx)^2*(u[i-1,j,k]-2*u[i,j,k]+u[i+1,j,k])+ (c*dt/dy)^2*(u[i,j-1,k]-2*u[i,j,k]+u[i,j+1,k])

 

Boundary condition: u(t=0)=1, diff(u(x,y,t),t=0)=0, and the normal derivative on the boundary of Omega =0.

How can solve this problem and plot the numerical solution.

 

 

 

I'm simplifying this Reynolds Equation starting from here:

Reynolds:=Diff(p(x)*h(x)^3/(12*mu)*Diff(p(x),x),x)-u(x)/2*Diff(p(x)*h(x),x)+Diff(p(x)*h(x)^3/(12*mu)*Diff(p(x),z),z)=Diff(p(x)*h(x),t):

 

Then I apply the dchange command:

dchange({p(x)=P(X)*Pa,x=Lx*X,h(x)=H(X)*h2},Reynolds,{P,h,X,u,H});

 

The problem is that it expands the new derivatives and I need the equation in its compact form.

How do I tel Maple only to make the substitution but not to expand the result?

I know that it expands by default, can I modify that?

 

 

I'm writing a simple Maple program to test the Generalized Finite Element Method: main_screened_Poisso.mw

When trying to define the Neumann boundary conditions, I have to define a directional derivative dudn=dudx*n. However, I can't seem to define a unit vector normal to Gamma, which is defined by a LineSegments objects.

Other than that, the row reduction is very slow, even though I'm using floating point arithmatic and not exact arithmatic, I believe.

How can I solve these problems? Thanks in advance!

 

I want  to verify the following function expression

is indeed an antiderivative of the function expression

where  A>0 with B^2-4*A*C<0.

I have tried the command

>diff((2),x);simplify(%);

where (2) is the label of the antiderivative expression.  But the result is very awkarward and lengthy. How could I verify the antiderivative expression is indeed an antiderivative of the other function by using Maple 17? 

 

I have following expression

f:=t->((1/8)*s^2*sinh(4*t)+t+(1/2)*s^2*t+s*sinh(2*t))/(1+s*cosh(2*t))

which is 1 solution of the ODE

ode2 := -(diff(y(t), t, t))+(4-12/(1+s*cosh(2*t))+(8*(-s^2+1))/(1+s*cosh(2*t))^2)*y(t) = 0

Now I wanted to construct 2 linear independent solutions via:

f1:=f(t_b-t)

f2:=f(t-t_a)

and calculate the Wronskian:

with(LinearAlgebra); with(VectorCalculus)

Determinant(Wronskian([f(t_b-t), f(t-t_a)], t))

Since I know these functions are solutions of the second order ODE which does not contain any first order derivative the Wronskian should be a constant. Unfortunately Maple has a hard time to simplify it since the epxression is a little big. Is it my fault or has anyone an idea what to do?

I assign a function like so: f:=x->0.2*x^2*(x-3)^3

Then using d/dx in the expression palette I differentiate: d/dx f(x)

Get the following: 0.6x^2 (x-3)^2 D(x^2)(x-3)

What is the "D" in the last expression? What hasn't the whole differentiation been completed?

Hello, Suppose I have two sinusoids with the same amplitude and frequency. By changing the phase of one of them it is possible to 'align' them. When the phases are exactly the same the difference between the two sinusoids becomes zero. Now lets define a variable (U[int]) that gives a measure of the difference between the two sinusoids and change the phase phi to generate an animation and 3D plot and see if it is possible to mathematically determine that the difference becomes zero when the phases of the waveforms are equal. Chosing a time t>0 and solving for the derivative does the job.

My question is, can (and how can) Maple determine for which phase phi we have the minimum function without chosing a time t.
 

restart

with(plots):

u[G] := sin(omega*t)

u[L] := sin(omega*t+phi)

U[int] := int((u[G]-u[L])^2, t)+C

C := solve(subs(t = 0, U[int]) = 0, C)

omega := 9:

animate(plot, [[u[G], u[L], (1/10)*U[int]], t = 0 .. 2, legend = ["Public grid voltage", "Local grid voltage (control goal)", "Cum. Actuator voltage (scaled)"]], phi = -Pi .. Pi, gridlines = true, labels = ["Time [s]","Voltage [V]"], labeldirections = ["horizontal", "vertical"], labelfont = ["ARIAL", "bold", 12])

plot3d(U[int], t = 0 .. 2, phi = -Pi .. Pi, shading = zhue, orientation = [-150, 70, 15])

t := 1:

phi = fsolve(diff(U[int], phi))


Download 20131114_Finding_min.mw

 

Thanks a lot!

Consider the curve defined by f(x, y) = 3+2x+y+2x^2+2xy+3y^2 = 0.Locally on the curve we can view y as a function of x, i.e. y = y(x).Compute formulas for the first and second derivative of y with resoect to x.

I have an equation x2y-3y3x=0. I want to find the slope of the graph at the point (3,1) , that is the derivative at that point. I was told it was possible to do this in one line of command, so I was just wondering if anyone knoew how to do this. Any ideas would be helpful. Thanks in advance.

I would like to find the derivative for f(x)=min(x^2 +1, 2x+3) and plot f(x) and its derivative on the same graph. I know the "diff" command works on functions, but I'm not sure how to use it on this one. Suggestions would be appreciated. Thanks!

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