10.5. Applications of Parametric Equations http://www.ck12.org
Next, it is important to note the starting point, center point and direction. You should already have the graphs of
sine and cosine memorized so that when you see a pattern in words or as a graph, you can identify what you see as
+sin,−sin,+cos,−cos.In this example, the vertical component starts at a low point of 6, travels to a middle point
of 26 and then a height of 46 and back down. This is a−cos pattern. The amplitude of the−cos is 20 and the
vertical shift is 26. Lastly, the period is 60. You can use the period to help you findb.
60 =^2 bπ
b= 30 πThus the vertical parameterization is:
y=−20 cos( 30 πt)+ 26
Try to find the horizontal parameterization on your own. The solution will be discussed in the “Concept Problem
Revisited” section.
Projectile motion has a vertical component that is quadratic and a horizontal component that is linear. This is
because there are 3 parameters that influence the position of an object in flight: starting height, initial velocity, and
the force of gravity. The horizontal component is independent of the vertical component. This means that the
starting horizontal velocity will remain the horizontal velocity for the entire flight of the object.
Note that gravity,g,has a force of about− 32 f t/s^2 or− 9. 81 m/s^2 .The examples and practice questions in this
concept will use feet.
If an object is launched from the origin at a velocity ofvthen it has horizontal and vertical components that can be
found using basic trigonometry.
sinθ=vvV→v·sinθ=vV
cosθ=vvH→v·cosθ=vHThe horizontal component is basically finished. The only adjustments that would have to be made are if the starting
location is not at the origin, wind is added or if the projectile travels to the left instead of the right. See Example A.
x=t·v·cosθ