Okay, having a bit of trouble with this question:
Eliminate the parameter t from the following parametric equations to find a cartesian equation of the curve they represent.
x = 5cos(6t)
y = 3sin(6t)
Any help greatly appreciated
Eliminate the parameter t from the following parametric equations to find a cartesian equation of the curve they represent.
x = 5cos(6t)
y = 3sin(6t)
Any help greatly appreciated
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x = 5cos(6t)
y = 3sin(6t)
x/5 = cos(6t)
y/3 = sin(6t)
(x/5)² + (y/3)² = cos²(6t) + sin²(6t)
(x/5)² + (y/3)² = 1
x² / 25 + y² / 9 = 1
There is your ellipse.
y = 3sin(6t)
x/5 = cos(6t)
y/3 = sin(6t)
(x/5)² + (y/3)² = cos²(6t) + sin²(6t)
(x/5)² + (y/3)² = 1
x² / 25 + y² / 9 = 1
There is your ellipse.
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Cartesian equations contain only x and y so naturally t has to be "eliminated"
Simply solve for t in both equations, the substitute:
x = 5cos(6t)
arccos(x/5)/6 = t
y = 3sin(6t)
arcsin(y/3)/6 = t
arcsin(y/3)/6 = arccos(x/5)/6
arcsin(y/3) = arccos(x/5)
Technically, this is Cartesian equation, but I'm going to try to make it neater:
Notice that arcsine is the inverse of sine,
so arcsin(y/3) can be represented as the angle in a right triangle facing a leg of length y and a hypotenuse of length 3.
(To help, draw a right triangle with hypotenuse 3 and one leg y)
Use pythagorean theorem to find the length of the other leg: sqrt(3^2 - y^2) = sqrt(9-y^2)
Now express the same angle as an expression with arccosine:
(remember arccosine undos cosine)
arcsin(y/3) = arccos( sqrt(9-y^2) / 3)
Substitute this back into the equation from before:
arccos( sqrt(9-y^2)/3)= arccos(x/5)
sqrt(9-y^2)/3 = x/5
9 - y^2 =(0.6x)^2
9 - y^2 = 0.36x^2
9 = 0.36x^2 + y^2 <-neater form
Simply solve for t in both equations, the substitute:
x = 5cos(6t)
arccos(x/5)/6 = t
y = 3sin(6t)
arcsin(y/3)/6 = t
arcsin(y/3)/6 = arccos(x/5)/6
arcsin(y/3) = arccos(x/5)
Technically, this is Cartesian equation, but I'm going to try to make it neater:
Notice that arcsine is the inverse of sine,
so arcsin(y/3) can be represented as the angle in a right triangle facing a leg of length y and a hypotenuse of length 3.
(To help, draw a right triangle with hypotenuse 3 and one leg y)
Use pythagorean theorem to find the length of the other leg: sqrt(3^2 - y^2) = sqrt(9-y^2)
Now express the same angle as an expression with arccosine:
(remember arccosine undos cosine)
arcsin(y/3) = arccos( sqrt(9-y^2) / 3)
Substitute this back into the equation from before:
arccos( sqrt(9-y^2)/3)= arccos(x/5)
sqrt(9-y^2)/3 = x/5
9 - y^2 =(0.6x)^2
9 - y^2 = 0.36x^2
9 = 0.36x^2 + y^2 <-neater form