make a sketch to get a right-angled triangle
Let Ø be the angle of elevation, the height is 2,
the horizontal distance is x and the hypotenuse is h
tanØ = 2/x
xtanØ = 2
xsec^2 Ø dØ/dt + tanØ dx/dt = 0
when h = 3
x^2 + 4 = 9
x = √5 , and dx/dt = -600 , (x is decreasing)
also when h = 3, tanØ = 2/√5 ,
cosØ = √5/3
secØ = 3/√5
sec^2 Ø = 9/5
plug into the derivative equation:
√5(9/5) dØ/dt + (2/√5)(-600) = 0
dØ/dt = 1200/√5) (5/(9√5) )
= 1200/9 radians/hr
= 20/9 radians/min
An airplane flys at a constant altitude of 2 miles and a constant speed of 600 miles per hour on a straight course that will take it directly over a kangaroo on the ground. How fast is the angle of elevation of the kangaroo's line of sight increasing when the distance from the kangaroo to the plane is 3 miles? Give your answer in radians per minute.
2 answers
Thank you
2 answers