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amoodarya

  • one year ago

I want to make short tutorial for parametric equation (not advanced )

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  1. amoodarya
    • one year ago
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    In mathematics, parametric equations of a curve express the coordinates of the points of the curve as functions of a variable, called a parameter for example : x=t+1 y=t-2 is a line equation if we want to find explicit equation first we find "t" from x or y and put it to other t=x-1 y=(t)-2=(x-1)-2 so y=x-3

  2. Kash_TheSmartGuy
    • one year ago
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    Cool, so make a tutorial!

  3. amoodarya
    • one year ago
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    example 2: find explicit equation \[x=t-1\\y=t^2+t+2\\ \rightarrow x=t-1 \rightarrow \\t=x+1\\ \rightarrow y=(x+1)^2+(x+1)+2\]

  4. amoodarya
    • one year ago
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    example 3:find explicit equation x=sint +1 y= cos t -2 we know \[\sin^2t+\cos^2t=1\] so \[x=sint +1\rightarrow sint =x-1\\y=cost-2\rightarrow \cot =y+2\\sin^2t+\cos^2t=1\\(x-1)^2+(y+2)^2=1\\\] it is a circle

  5. amoodarya
    • one year ago
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    example 4:find explicit equation \[x=\sqrt{t}\\y=2t-1\\\] note that \[t \geq 0\] \[x=\sqrt{t} \rightarrow t=x^2\\y=2t-1\\ \rightarrow y=2(x^2)-1\]

  6. amoodarya
    • one year ago
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    example 5:\[x=2cost \\y=3sint \\\]find explicit equation \[x=2cost \rightarrow cost =\frac{x}{2}\\y=3sint \rightarrow sint=\frac{y}{3}\\we-know \\sin^2t+\cos^2t=1\\so\\(\frac{x}{2})^2+(\frac{y}{3})^2=1\\\frac{x^2}{4}+\frac{y^2}{9}=1\] it is an ellipse

  7. amoodarya
    • one year ago
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    example 6: x=cost +sint y=3 sint find explicit equation \[x=cost +sint\\ y=3 sint \rightarrow sint =\frac{y}{3} \\ \rightarrow x=cost +sint =cost +\frac{y}{3} \\x-\frac{y}{3} =cost \\put -into \\sin^2t+\cos^2t=1\rightarrow \\(\frac{y}{3})^2+(x-\frac{y}{3})^2=1\]

  8. amoodarya
    • one year ago
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    example 7: x=cost +sint y=cost -sint find explicit equation \[x=cost +sint \\ y=cost -sint \\ \rightarrow \\x^2=\cos^2t+\sin^2t+2sint cost\\y^2=\cos^2t+\sin^2t-2sint cost\] if we find sum of them it will be a circle \[x^2=\cos^2t+\sin^2t+2sint cost \rightarrow x^2=1+2sint cost\\y^2=\cos^2t+\sin^2t-2sint cost \rightarrow y^2=x^2=1-2sint cost\\ \rightarrow \\x^2+y^2=1+2sint cost+1-2sint cost=2 \\ \rightarrow x^2+y^2=2\] \[radius=\sqrt{2} ,center=(0,0)\]

  9. amoodarya
    • one year ago
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    example 8 : x=cost y=cos (2t) find explicit equation note : you have to find a relation between cos t, cos 2t \[x=\cos t\\y=\cos2t\\ \left\{ \cos2t=2\cos^2t-1 \right\}\rightarrow \\y=\cos2t=2\cos^2t-1 \\y=2x^2-1\]

  10. amoodarya
    • one year ago
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    example 9: \[x=t+\frac{1}{t}\\y=t-\frac{1}{t}\] find explicit equation 2 method will show first use \[(a+b)^2+(a-b)^2=2(a^2+b^2)\\(a+b)^2-(a-b)^2=4ab\] \[x=t+\frac{1}{t} \rightarrow x^2=t^2+(\frac{1}{t})^2+2t \frac{1}{t}=t^2+(\frac{1}{t})^2+2\\y=t-\frac{1}{t} \rightarrow y^2=t^2+(\frac{1}{t})^2-2t \frac{1}{t}=t^2+(\frac{1}{t})^2-2\\find \\x^2-y^2\\x^2-y^2=(t^2+(\frac{1}{t})^2+2)-(t^2+(\frac{1}{t})^2-2)\\ \rightarrow x^2-y^2=4\]

  11. amoodarya
    • one year ago
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    second method : it easy to find "t" \[x=t+\frac{1}{t}\\y=t-\frac{1}{t}\\x+y=t+\frac{1}{t}+t-\frac{1}{t}=2t\\t=\frac{x+y}{2}\] then put t in 1st or 2nd equation

  12. amoodarya
    • one year ago
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    example 10: \[x=2tant+3 \\y=3\cot t -5\\\] find explicit equation we have tan , cot and we know tan t * cot t=1 so \[x=2tant+3 \rightarrow \tan t=\frac{x-3}{2}\\y=3\cot t -5 \rightarrow \cot t =\frac{y+5}{3}\\ \rightarrow \tan t \times \cot t=1\\\frac{x-3}{2} \times \frac{y+5}{3}=1\\y+5=\frac{6}{x-3}\\y=\frac{6}{x-3}+5=\frac{6+5x-15}{x-3}=\frac{5x-9}{x-3}\]

  13. amoodarya
    • one year ago
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    example 11: \[x=a_0+b_0t\\y=a_1+b_1t\] find explicit equation it is a line equation :

  14. amoodarya
    • one year ago
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    \[if \\b_0 \neq 0 \\t=\frac{x-a_0}{b_0} \\ \rightarrow y=a_1+b_1t\\y=a_1 +b_1(\frac{x-a_0}{b_0})\]

  15. amoodarya
    • one year ago
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    Now draw parametric curve if we do not eliminate the parameter we can put some value for parameter and find x, y as a point of (x,y) then with some point draw the curve for example \[x=\frac{t}{2}\\y=t+1\\t=0 \rightarrow x=0 ,y=1 \rightarrow (x,y)=(0,1)\\t=2 \rightarrow x=1 ,y=3 \rightarrow (x,y)=(2,3)\\...\\\] they are polynomial in degree 1 : so this is a line equation |dw:1436394739768:dw|

  16. amoodarya
    • one year ago
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    we need to draw \[x=\sqrt{t}\\y=t+1\] put some t into x,y equations \[x=\sqrt{t}\\y=t+1\\t \ge 0\\t=0 \rightarrow x=0 , y=1 \rightarrow point=(0,1)\\t=1 \rightarrow x=1 , y=2 \rightarrow point=(1,2)\\t=4 \rightarrow x=2 , y=5 \rightarrow point=(2,5)\\t=9 \rightarrow x=3 , y=10 \rightarrow point=(3,10)\] |dw:1436395050876:dw| note that \[t \geq 0 \rightarrow x \geq 0\]

  17. amoodarya
    • one year ago
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    Now : Parametric derivative Parametric derivative is a derivative in calculus that is taken when both the x and y variables (traditionally independent and dependent, respectively) depend on an independent third variable t, usually thought of as "time". for example : \[x=2t+cost\\y=t^2+sint+5\\\frac{dy}{dx}=?\] \[\frac{dy}{dx}=\frac{dy}{dt}*\frac{dt}{dx}=\\\frac{dy}{dt}*\frac{1}{\frac{dx}{dt}}=\\\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{y'_t}{x'_t}\\\] so ,in this case \[y'_x=\frac{dy}{dx}=\frac{y'_t}{x'_t}=\frac{(t^2+sint+5)'_t}{(2t+cost)'_t}=\frac{2t+cost}{2-sint}\]

  18. amoodarya
    • one year ago
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    @mukushla

  19. amoodarya
    • one year ago
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    @Donya2222 @dolloway97 @Donblue @Mehek14 @Mindblast3r @Marcorie @Moe95 @JFraser @superhelp101 @slade @sammixboo @Destinyyyy @Deeezzzz @Nicoleegilmoree @aaronq @AeroSmith @AG23 @Elsa213 @wil476003 @Rubyblades @razor99 @raggedy_roo @ritesh_asu @Rubyblades @Tinkerbell2001 @bruno102 @bunny256 @vera_ewing @Vocaloid @KAKES1967 @Hero @hendersonjen02 @iloveyou;* @IrishBoy123

  20. amoodarya
    • one year ago
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    we are in ,to help you

  21. Empty
    • one year ago
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    I was wondering if you've heard of the Leminscate curve: \[(x^2+y^2)^2=2a^2(x^2-y^2)\] Can we parametrize this?

  22. amoodarya
    • one year ago
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    a simple Leminscate curve: \[x=\frac{a \cos t}{1+\sin^2t}\\y=\frac{asint \cos t }{1+\sin^2 t}\]

  23. amoodarya
    • one year ago
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    Leminscate curve equation is nice in polar coordinate "Empty"

  24. amoodarya
    • one year ago
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    @YanaSidlinskiy @yomamabf @yingpeng @yacobtewolde @uybuyvf @uybuyvf @otaylor19 @pooja195 @peachpi @pattycake1 @pinkros @AbdullahM @xavierbo2 @Xaze @xamr @xitsaliciaa @calculusxy @valiant1 @brebre5564 @EmilyD22 @esam2 @Emilyf29 @radar @GeniousCreation @Ghostedly @gabgurl @GloGangg_Jayy @LegendarySadist @lizajune @lofi @lorrainetocuteherrera

  25. amoodarya
    • one year ago
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    I think ,maybe useful not for promoting I beg you ,if bother you

  26. YanaSidlinskiy
    • one year ago
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    *ahem* Mass tagging?;) Never knew anything about this. To me, this is like a whole new Italian language.

  27. amoodarya
    • one year ago
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    \[x=t\\y=2t-1\\0 \leq t \leq 2\\\] draw it note about domain of "t" \[x=t\\y=2t-1\\0 \leq t \leq 2\\ \rightarrow \left\{ (x,y):x=t,y=2t-1,0 \leq t \leq 2 \right\}\\0 \leq x \leq 2\\0 \leq t \leq 2 \rightarrow 0 \leq 2t \leq 4 \rightarrow 0-1 \leq 2t-1 \leq 4-1\\ -1\le y \le 3\] it is a line ,but restricted \[0 \leq t \leq 2\\t=0 \rightarrow x=t=0 ,y=2t-1=-1 \rightarrow (0,-1)\\t=2 \rightarrow x=t=2 ,y=2t-1=-1 \rightarrow (2,3)\] |dw:1436398201172:dw|

  28. amoodarya
    • one year ago
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    Trick to solve ,system of equation like this \[\frac{x}{2}=\frac{y-1}{3}=\frac{z}{5}\\x-y+z=11\] we can use the parameter to show all of variable by ,one variable \[\frac{ x }{ 2 }=\frac{ y-1 }{ 3 }=\frac{ z }{ 5 }=t\\x=2t\\y=3t+1\\z=5t\\\] now put them in x-y+z=11 \[x=2t\\y=3t+1\\z=5t\\x-y+z=11\rightarrow \\(2t)-(3t+1)+(5t)=11\\4t=12\\t=3\\\] now we can easily find x,y,z \[x=2t \rightarrow x=2(3)=6\\y=3t+1\rightarrow y=9+1=10\\z=5t \rightarrow z=15\\\]

  29. amoodarya
    • one year ago
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    some parametric curve Cycloid\[x=r(t-\sin t)\\y=r(1-cost)\] The cycloid represents the following situation. Consider a wheel of radius r. Let the point where the wheel touches the ground initially be called P. Then start rolling the wheel to the right. As the wheel rolls to the right trace out the path of the point P. The path that the point P traces out is called a cycloid and is given by the equations above. In these equations we can think of θ as the angle through which the point P has rotated.

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  30. amoodarya
    • one year ago
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    Here is a cycloid sketched out with the wheel shown at various places. The blue dot is the point P on the wheel that we’re using to trace out the curve.

  31. ganeshie8
    • one year ago
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    Nice!

  32. amoodarya
    • one year ago
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    Circle A more sophisticated example is the following. Consider the unit circle which is described by the ordinary (Cartesian) equation \[x^2+y^2=1\] This equation can be parameterized as follows \[ (\cos(t),\; \sin(t))\quad\mathrm{for}\ 0\leq t < 2\pi.\, \] With the Cartesian equation it is easier to check whether a point lies on the circle or not. With the parametric version it is easier to obtain points on a plot. In some contexts, parametric equations involving only rational functions (that is fractions of two polynomials) are preferred, if they exist. In the case of the circle, such a rational parameterization is \[x=\frac{1-t^2}{1+t^2}\\y=\frac{2t}{1+t^2}\] or \[x=\pm \sqrt{1-t^2}\\y=t\]

  33. amoodarya
    • one year ago
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    ellipse \[x=a sint \\y=b cost \\a \neq b \neq 0 \\(\frac{x}{a})^2+(\frac{y}{b})^2=\sin^2 t+\cos^t=1\] or \[ x=a(\frac{2t}{1+t^2})\\ y=b(\frac{1-t^2}{1+t^2})\]

  34. amoodarya
    • one year ago
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    Parabola The simplest equation for a parabola \[y=x^2\] can be parameterized by using a free parameter t, and setting \[x=t\\y=t^2\]

  35. amoodarya
    • one year ago
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    |dw:1436438108955:dw| note that \[t \geq 0\]

  36. amoodarya
    • one year ago
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    @mukushla

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