5. What is the polar equation that is equivalent to the following?
x = 2 cos^2 θ/sin θ
y = 2 cos θ
6. An ant is climbing up a branch that is 60°to the ground. It moves at a constant velocity of 0.5 cm/sec. Find the set of parametric equations that describes the path of the ant’s travel. Assume the ant starts at the origin.
Simplify each expression.
3. 4 (1 point)
-16
16
16 0-1 16
16
5. To convert the Cartesian coordinates to polar coordinates:
r^2 = x^2 + y^2 = (2 cos^2 θ/sin θ)^2 + (2 cos θ)^2
r^2 = 4 cos^4 θ/sin^2 θ + 4 cos^2 θ
r^2 = 4 cos^2 θ (cos^2 θ/sin^2 θ + 1)
r^2 = 4 cos^2 θ (cot^2 θ + 1)
r^2 = 4 cos^2 θ csc^2 θ
r = 4 cos θ/sin θ
To convert the y-coordinate:
y = 2 cos θ = r sin θ
So the polar equation equivalent to the given equation is:
r = 4 cos θ/sin θ
y = 2 cos θ
6. Let's assume that the length of the branch is 1 (this won't affect the equations, since we can scale them later). The ant's position can be described by its distance from the origin (r) and the angle it makes with the branch (θ).
At time t, the ant has climbed up a vertical distance of 0.5t (since its velocity is 0.5 cm/sec). Since the branch makes a 60° angle with the ground, the ant has also climbed up a horizontal distance of 0.5t*sin(60°) = 0.5t*sqrt(3)/2 = 0.25t*sqrt(3). So the ant's position is:
x = 0.25t*sqrt(3)
y = 0.5t
To convert to polar coordinates:
r^2 = x^2 + y^2
r^2 = (0.25t*sqrt(3))^2 + (0.5t)^2
r^2 = 0.25t^2 + 0.25t^2*3/4
r^2 = 0.25t^2*5/4
r = 0.5t*sqrt(5)
θ is the angle the ant makes with the branch, so we can use tangent:
tan θ = y/x
θ = arctan(y/x) = arctan(2/(0.5sqrt(3)t)) = arctan(4/sqrt(3)t)
So the parametric equations are:
x = 0.25t*sqrt(3)
y = 0.5t
r = 0.5t*sqrt(5)
θ = arctan(4/sqrt(3)t)
Note that we could simplify the equations by setting the coefficient of t to 1 (by dividing by 0.5), but we'll leave them as they are for clarity.
Answer c is supposed to be 1/16 is your answer still accurate
If answer choice c is 1/16, then the correct answer would be:
3. 4 (1 point)
-16
16
1/16
Thank you for bringing this to my attention.
So what is the answer
The answer is 1/16.
Simplify 5^0
Any number (except 0) raised to the power of 0 is equal to 1. Therefore, 5^0 = 1.
To find the polar equation that is equivalent to the given rectangular equation, we can make use of the conversion formulas between polar and rectangular coordinates.
Given:
x = 2 cos^2 θ/sin θ
y = 2 cos θ
Step 1: Express x and y in terms of r and θ.
Using the formula x = r cos θ and y = r sin θ, we can solve for r in terms of θ:
x = r cos θ
2 cos^2 θ/sin θ = r cos θ
2 cos θ = r sin θ
r = 2 cos θ / sin θ
Step 2: Rewrite the equation in polar form.
Replace r with the expression obtained in Step 1:
r = 2 cos θ / sin θ
Therefore, the polar equation that is equivalent to the given rectangular equations is r = 2 cos θ / sin θ.
For the second question, to find the parametric equations that describe the path of the ant's travel, we need to use the angle of inclination and the constant velocity.
Given:
Angle of inclination = 60°
Constant velocity = 0.5 cm/sec
Step 1: Convert the angle of inclination to radians.
60° = (60π)/180 = π/3 radians
Step 2: Set up the parametric equations.
The parametric equations for the path of the ant can be written as:
x(t) = velocity * time * cos(θ)
y(t) = velocity * time * sin(θ)
Since the ant moves at a constant velocity of 0.5 cm/sec, we can replace velocity with 0.5 in the equations above. Also, the starting point of the ant is at the origin, so the initial time (t) is 0.
Therefore, the parametric equations for the path of the ant are:
x(t) = 0.5 * t * cos(π/3)
y(t) = 0.5 * t * sin(π/3)