6) To convert a polar equation into parametric equations, we use the following formulas:
x = r cos θ
y = r sin θ
We have:
r(θ) = 2 cos θ cot θ
Using the formula for cotangent:
cot θ = cos θ / sin θ
We can rewrite the equation as:
r(θ) = 2 cos θ (cos θ / sin θ)
Simplifying:
r(θ) = 2 cos^2 θ / sin θ
Now we can use the formulas to get the parametric equations:
x = r cos θ = (2 cos^2 θ / sin θ) cos θ = 2 cos^3 θ / sin θ
y = r sin θ = (2 cos^2 θ / sin θ) sin θ = 2 cos^2 θ
So the parametric equations are:
x = 2 cos^3 θ / sin θ
y = 2 cos^2 θ
7) To eliminate the parameter, we need to express one variable (x or y) in terms of the other. We have:
x(t) = 2t
y(t) = -4t^2 + 2
We can rearrange the first equation to get t in terms of x:
t = x/2
Substituting into the second equation:
y(x) = -4(x/2)^2 + 2
y(x) = -x^2 + 2
So the equation without parameter is:
y = -x^2 + 2
6) Convert the following polar equation into prametric equation/ Show all your work. Simplify the equations.
r(θ)=2 cos θ cot θ
7) Eliminate the parameter for the following parametric equations. Show all your work. Simplify as far as you can.
x(t) = 2t
y(t) = -4t^2+2
3 answers
Which direction would the arrows be in a graph to represent the orientation of the following parametric equations?
x(t) = 3 cos 1/2 t
y(t) = -5sin 1/2 t
x(t) = 3 cos 1/2 t
y(t) = -5sin 1/2 t
The orientation of the arrows on a graph depends on the direction of increasing values of the parameter t.
For the given parametric equations:
x(t) = 3 cos (1/2)t
y(t) = -5 sin (1/2)t
Note that the coefficient of t in both equations is positive (1/2), indicating that the parameter increases as the angle increases in a counterclockwise direction. Therefore, on the graph, the arrows would be oriented counterclockwise.
For the given parametric equations:
x(t) = 3 cos (1/2)t
y(t) = -5 sin (1/2)t
Note that the coefficient of t in both equations is positive (1/2), indicating that the parameter increases as the angle increases in a counterclockwise direction. Therefore, on the graph, the arrows would be oriented counterclockwise.