Asked by Sir KK
Using Parametric equations and vectors consider the following scenario:
Starting from an airport, an airplane flies 225 miles northwest, then 150 miles south-west.
Collaborate with your peers to answer the following:
Draw a graph or figure to represent this situation. (This graph must be digital on Desmos or equivelant not handdrawn.)
Describe how the concepts from this module can be applied in this case.
How far, in miles, from the airport is the plane?
Provide another example of a scenario that involves the same concept.
Starting from an airport, an airplane flies 225 miles northwest, then 150 miles south-west.
Collaborate with your peers to answer the following:
Draw a graph or figure to represent this situation. (This graph must be digital on Desmos or equivelant not handdrawn.)
Describe how the concepts from this module can be applied in this case.
How far, in miles, from the airport is the plane?
Provide another example of a scenario that involves the same concept.
Answers
Answered by
Anonymous
luckily NW is perpendicular to SW so d = sqrt (225^2 + 150^2)
Answered by
mathhelper
northwest in standard trig notation ---- 135°
southwest in standard trig notation ---- 225°
(x,y) = 225(cos135,sin135) + 150(cos225,sin225)
= 225(-√2/2, √2/2) + 150(-√2/2, -√2/2)
= (-225√2/2 , 225√2/2) + (-150√2/2 , -150√2/2)
= (-375√2/2 , 75√2/2)
distance from airport
= √( (-375√2/2)^2 + (75√2/2)^2) = appr 270.4 miles
southwest in standard trig notation ---- 225°
(x,y) = 225(cos135,sin135) + 150(cos225,sin225)
= 225(-√2/2, √2/2) + 150(-√2/2, -√2/2)
= (-225√2/2 , 225√2/2) + (-150√2/2 , -150√2/2)
= (-375√2/2 , 75√2/2)
distance from airport
= √( (-375√2/2)^2 + (75√2/2)^2) = appr 270.4 miles
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