## amoodarya one year ago I want to make short tutorial for parametric equation (not advanced )

1. amoodarya

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

Cool, so make a tutorial!

3. amoodarya

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

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

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

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

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

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

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

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

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

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

example 11: $x=a_0+b_0t\\y=a_1+b_1t$ find explicit equation it is a line equation :

14. amoodarya

$if \\b_0 \neq 0 \\t=\frac{x-a_0}{b_0} \\ \rightarrow y=a_1+b_1t\\y=a_1 +b_1(\frac{x-a_0}{b_0})$

15. amoodarya

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

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

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

@mukushla

19. amoodarya

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20. amoodarya

21. Empty

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

a simple Leminscate curve: $x=\frac{a \cos t}{1+\sin^2t}\\y=\frac{asint \cos t }{1+\sin^2 t}$

23. amoodarya

Leminscate curve equation is nice in polar coordinate "Empty"

24. amoodarya

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25. amoodarya

I think ,maybe useful not for promoting I beg you ,if bother you

26. YanaSidlinskiy

*ahem* Mass tagging?;) Never knew anything about this. To me, this is like a whole new Italian language.

27. amoodarya

$x=t\\y=2t-1\\0 \leq t \leq 2\\$ 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

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

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.

30. amoodarya

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

Nice!

32. amoodarya

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

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

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

|dw:1436438108955:dw| note that $t \geq 0$

36. amoodarya

@mukushla