Plotting contours of a function for different values of a parameter

I want to draw a picture of $(x-t)^2 + (y-t^2)^2 - t^2$ in $xy$ plane, with different values of $t$. I did it in a very silly way , namely,

ContourPlot[{
  (x - 1)^2 + (y - 1)^2 == 1, (x - 2)^2 + (y - 4)^2 == 4, 
  (x - 3)^2 + (y - 9)^2 == 9, (x - 4)^2 + (y - 16)^2 == 16, 
  (x - 5)^2 + (y - 25)^2 == 25, (x - 6)^2 + (y - 36)^2 == 36, 
  (x - 7)^2 + (y - 49)^2 == 49, (x - 8)^2 + (y - 64)^2 == 64, 
  (x - 9)^2 + (y - 81)^2 == 81, (x - 10)^2 + (y - 100)^2 == 100, 
  (x + 1)^2 + (y - 1)^2 == 1, (x + 2)^2 + (y - 4)^2 == 4, 
  (x + 3)^2 + (y - 9)^2 == 9, (x + 4)^2 + (y - 16)^2 == 16, 
  (x + 5)^2 + (y - 25)^2 == 25, (x + 6)^2 + (y - 36)^2 == 36, 

  (x + 7)^2 + (y - 49)^2 == 49, (x + 8)^2 + (y - 64)^2 == 64, 
  (x + 9)^2 + (y - 81)^2 == 81, (x + 10)^2 + (y - 100)^2 == 100
 }, 
 {x, -110, 110}, {y, -110, 110}
]

I am sure there must be a simpler way. Could you give me a hand?

Answer

Use Table to generate the values for different t:

ContourPlot[Evaluate@Table[(x - t)^2 + (y - t^2)^2 == t^2, 

    {t, -20, 20, 1}], {x, -110, 110}, {y, -110, 110}]

enter image description here

Comments

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...

Blog

Search This Blog