Please help me set up these problems.
1.) Find an equivalent equation in rectangular coordinates.
r(1 - 2 cos θ) = 1
2.) Find a rectangular equation for the plane curve defined by the parametric equations.
x = sec t, y = tan
3.) A projectile is fired with an initial velocity of 500 feet per second at an angle of 70° with the horizontal. To the nearest 10 feet, find the horizontal distance covered by the projectile.
Review the equation for a hyperbola in polar coordinates.
#2 Hint: sec^2 = 1 + tan^2
#3
y = 500 sin70° t - 16t^2
Solve for t when y=0 to find the time in flight
The x-velocity is constant at 700 cos70°
So using the time from step 1, distance = speed * time
Sure, I will guide you through the process of solving each problem step by step.
1.) Find an equivalent equation in rectangular coordinates.
To convert the given equation from polar coordinates to rectangular coordinates, you can make use of the following trigonometric identities:
r cos θ = x and r sin θ = y
Start by multiplying both sides of the equation by r:
r(1 - 2 cos θ) = 1
Simplifying the equation further, we have:
r - 2r cos θ = 1
Now replace r cos θ with x and r with √(x^2 + y^2):
√(x^2 + y^2) - 2x = 1
This is the equivalent equation in rectangular coordinates.
2.) Find a rectangular equation for the plane curve defined by the parametric equations.
Given the parametric equations:
x = sec t and y = tan t
We can eliminate the parameter t by expressing it in terms of x and y.
From the equation x = sec t, we know that sec t = x.
Since the reciprocal of secant is cosine, we can write cos t = 1/x.
Similarly, from the equation y = tan t, we know that tan t = y.
Since the reciprocal of tangent is cosine, we can write cos t = 1/√(1 + y^2).
Now, equating the two expressions for cos t, we get:
1/x = 1/√(1 + y^2)
To eliminate the fractions, we cross-multiply:
√(1 + y^2) = x
Squaring both sides of the equation, we have:
1 + y^2 = x^2
This is the rectangular equation for the plane curve defined by the parametric equations.
3.) To find the horizontal distance covered by the projectile, we can use the equation for horizontal projectile motion:
Range, R = (v^2 * sin 2θ) / g
Where:
- v is the initial velocity of the projectile (500 feet per second in this case),
- θ is the launch angle (70° in this case),
- g is the acceleration due to gravity (approximately 32.17 feet per second^2).
Plug in the given values into the equation and solve for R:
R = (500^2 * sin 2(70)) / 32.17
First, calculate the value of sin 2(70):
sin 2(70) ≈ sin 140 ≈ 0.94
Substitute the values:
R ≈ (500^2 * 0.94) / 32.17
R ≈ 7331.9
Therefore, the horizontal distance covered by the projectile is approximately 7330 feet (to the nearest 10 feet).