I made a sketch showing the distance covered by the plane after it passed over the kangaroo as x miles
let the angle of elevation be Ø
I have :
tanØ = 2/x
xtanØ = 2
x sec^2 Ø dØ/dt + tanØ dx/dt = 0
when the distance between the kangaroo and the plance is 3
x^2 + 2^2 = 3^2
x = √5
when x = √5 , dx/dt = 600, tanØ = 2/√5, and sec^2 Ø = 49/5
√5(49/5) dØ/dt + (2/√5)(600) = 0
dØ/dt = (-1200/√5)/(49√5/5)
= -1200/49 radians/hr
= -20/49 radians/min
notice the angle is decreasing
check my arithmetic, I should write it out on paper first.
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
tan(Θ) = 2/x
dx/dt = -600
y = 2
Start with tan(Θ) = 2/x.Taking the derivative leaves you with:
sec^2(Θ)*dΘ/dt = -2/x^2*dx/dt
Solve for dΘ/dt.
dΘ/dt = -2/x^2 * dx/dt * 1/sec^2(Θ)
when the distance from the plane to the kangaroo is 3, x = sqrt(5); so Θ = arcsin(2/3).
Use the equation above and plug in dx/dt, x, and theta, then divide by 60 to get units to rads/min.
the answer should be about 2.22222 rad/min.
dx/dt = -600
y = 2
Start with tan(Θ) = 2/x.Taking the derivative leaves you with:
sec^2(Θ)*dΘ/dt = -2/x^2*dx/dt
Solve for dΘ/dt.
dΘ/dt = -2/x^2 * dx/dt * 1/sec^2(Θ)
when the distance from the plane to the kangaroo is 3, x = sqrt(5); so Θ = arcsin(2/3).
Use the equation above and plug in dx/dt, x, and theta, then divide by 60 to get units to rads/min.
the answer should be about 2.22222 rad/min.
6 answers