Which is the polar form of the parametric equations x=5cos(theta) and y=5sin(theta) ?
a. r= 5(theta)
b. r= 5
c. r= 25 cos (theta) sin (theta)*****
d. r= 25cos^2 (theta) + 25sin^2 (theta)
Which is the polar form of the parametric equations x=4t and y=t^2?
a. r= 16 tan (theta) sec (theta)*****
b. r= 16 tan^2 (theta)
c. r= 16 sec^2 (theta)
d. r= 16 sec (theta)
A bowling ball is rolled down the alley with a constant velocity of 1.5m/sec. at an angle of 87 degrees to the starting line. The position of the person throwing the bowling ball can be represented by the point (0,0). Where is the ball after 7 seconds?
a. (0.5,10.5)
b. (0.1,1.5)
c. (10.5,10.5)*****
d. (29,200)
Reiny is being rude as hell
To find the polar form of the parametric equations x = 5cos(theta) and y = 5sin(theta):
1. Recall that in polar coordinates, the Cartesian coordinates (x, y) can be expressed as (r, theta), where r is the distance from the origin to the point and theta is the angle formed with the positive x-axis.
2. Use the Pythagorean identity sin^2(theta) + cos^2(theta) = 1 to eliminate theta from the equations.
x = 5cos(theta)
y = 5sin(theta)
=> x^2/25 + y^2/25 = cos^2(theta) + sin^2(theta) = 1
=> (x^2 + y^2)/25 = 1
=> r^2/25 = 1
=> r^2 = 25
=> r = 5
3. The answer is option c. r = 5(theta).
Therefore, the polar form of the parametric equations x = 5cos(theta) and y = 5sin(theta) is r = 5.
To find the polar form of the parametric equations x = 4t and y = t^2:
1. Substitute x and y in terms of r and theta:
x = 4t
=> x = 4r(cos(theta))
y = t^2
=> y = r^2(sin^2(theta))
2. Combine the equations:
r(cos(theta)) = 4r
=> cos(theta) = 4
r^2(sin^2(theta)) = r^2
=> sin^2(theta) = 1
3. Solve for theta:
Since cos(theta) cannot be greater than 1, there is no solution for cos(theta) = 4.
However, sin^2(theta) = 1 has two solutions: sin(theta) = 1 and sin(theta) = -1.
4. The answer is option a. r = 16tan(theta)sec(theta), since r is determined by the tangent and secant functions.
To find the position of the ball after 7 seconds:
1. Given that the ball rolls with a constant velocity of 1.5 m/sec at an angle of 87 degrees to the starting line, we can determine the x and y coordinates using trigonometry.
x = velocity * time * cos(angle)
= 1.5 * 7 * cos(87)
≈ 0.1 (rounded to one decimal place)
y = velocity * time * sin(angle)
= 1.5 * 7 * sin(87)
≈ 1.5 (rounded to one decimal place)
2. The answer is option b. (0.1, 1.5), representing the x and y coordinates of the ball after 7 seconds.
To find the polar form of parametric equations, we need to use the conversion formulas:
r = sqrt(x^2 + y^2) and theta = arctan(y/x)
Let's apply these formulas to each set of parametric equations:
1. For x = 5cos(theta) and y = 5sin(theta):
Plug these values into the conversion formulas:
r = sqrt((5cos(theta))^2 + (5sin(theta))^2)
r = sqrt(25cos^2(theta) + 25sin^2(theta))
r = 5sqrt(cos^2(theta) + sin^2(theta))
r = 5
Theta can be found using arctan(y/x):
theta = arctan((5sin(theta))/(5cos(theta)))
theta = arctan(sin(theta)/cos(theta))
theta = arctan(tan(theta))
Therefore, the polar form for this set of parametric equations is r = 5.
2. For x = 4t and y = t^2:
Plug these values into the conversion formulas:
r = sqrt((4t)^2 + (t^2)^2)
r = sqrt(16t^2 + t^4)
r = sqrt(t^2(16 + t^2))
r = t*sqrt(16 + t^2)
Theta can be found using arctan(y/x):
theta = arctan((t^2)/(4t))
theta = arctan(t/4)
Therefore, the polar form for this set of parametric equations is r = t*sqrt(16 + t^2) with theta = arctan(t/4).
3. Position of the ball after 7 seconds with constant velocity of 1.5m/sec and an angle of 87 degrees:
Using the velocity vector representation, we can find the displacements in x and y directions:
Delta x = velocity * cos(angle) * time = 1.5m/s * cos(87 degrees) * 7s
Delta x ≈ 0.1995m
Delta y = velocity * sin(angle) * time = 1.5m/s * sin(87 degrees) * 7s
Delta y ≈ 10.4859m
The position of the ball relative to the starting point is given by the coordinates (Delta x, Delta y), which in this case is approximately (0.1995, 10.4859).
Therefore, the ball is at approximately (0.1995, 10.4859) after 7 seconds.
The correct answer is (0.1995, 10.4859), which is closest to option b. (0.1, 1.5).
There are 4 main conversion relations that you need
r^2 = x^2 + y^2
x = rcosØ
y = rsinØ
tanØ = y/x
Your given parametric equations represent a circle with radius 5
http://www.wolframalpha.com/input/?i=plot+x%3D5cos(theta),+y%3D5sin(theta)
r = 5 would do the same
#2. convert to Cartesian first , then to polar