Question

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

Answer 1

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

Answer 2

Cool, so make a tutorial!

Answer 3

example 2:
find explicit equation
\[x=t-1\\y=t^2+t+2\\ \rightarrow x=t-1 \rightarrow \\t=x+1\\ \rightarrow y=(x+1)^2+(x+1)+2\]

Answer 4

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

Answer 5

example 4:find explicit equation
\[x=\sqrt{t}\\y=2t-1\\\]

note that \[t \geq 0\]
\[x=\sqrt{t} \rightarrow t=x^2\\y=2t-1\\ \rightarrow y=2(x^2)-1\]

Answer 6

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

Answer 7

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

Answer 8

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
\[x^2=\cos^2t+\sin^2t+2sint cost \rightarrow x^2=1+2sint cost\\y^2=\cos^2t+\sin^2t-2sint cost \rightarrow y^2=x^2=1-2sint cost\\ \rightarrow \\x^2+y^2=1+2sint cost+1-2sint cost=2 \\ \rightarrow x^2+y^2=2\]
\[radius=\sqrt{2} ,center=(0,0)\]

Answer 9

example 8
: x=cost
y=cos (2t)
find explicit equation

note : you have to find a relation between cos t, cos 2t

\[x=\cos t\\y=\cos2t\\ \left\{ \cos2t=2\cos^2t-1 \right\}\rightarrow \\y=\cos2t=2\cos^2t-1 \\y=2x^2-1\]

Answer 10

example 9:
\[x=t+\frac{1}{t}\\y=t-\frac{1}{t}\]
find explicit equation

2 method will show
first
use \[(a+b)^2+(a-b)^2=2(a^2+b^2)\\(a+b)^2-(a-b)^2=4ab\]
\[x=t+\frac{1}{t} \rightarrow x^2=t^2+(\frac{1}{t})^2+2t \frac{1}{t}=t^2+(\frac{1}{t})^2+2\\y=t-\frac{1}{t} \rightarrow y^2=t^2+(\frac{1}{t})^2-2t \frac{1}{t}=t^2+(\frac{1}{t})^2-2\\find \\x^2-y^2\\x^2-y^2=(t^2+(\frac{1}{t})^2+2)-(t^2+(\frac{1}{t})^2-2)\\ \rightarrow x^2-y^2=4\]

Answer 11

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

Answer 12

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
\[x=2tant+3 \rightarrow \tan t=\frac{x-3}{2}\\y=3\cot t -5 \rightarrow \cot t =\frac{y+5}{3}\\ \rightarrow \tan t \times \cot t=1\\\frac{x-3}{2} \times \frac{y+5}{3}=1\\y+5=\frac{6}{x-3}\\y=\frac{6}{x-3}+5=\frac{6+5x-15}{x-3}=\frac{5x-9}{x-3}\]

Answer 13

example 11:
\[x=a_0+b_0t\\y=a_1+b_1t\]
find explicit equation

it is a line equation :

Answer 14

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

Answer 15

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