the height of the stuntman at time t seconds after falling is
y = 100-16t^2
dy/dt = -32t
When the stuntman is at height y,
tanθ = y/200
sec^2 θ dθ/dt = 1/200 dy/dt
(1+tan^2θ) dθ/dt = 1/200 dy/dt
(1+(y/200)^2) dθ/dt = 1/200 dy/dt
at t=2, y=36, dy/dt = -64, so
(1+(36/200)^2) dθ/dt = 1/200 (-64)
dθ/dt = -.3 rad/s or -17.8°/s
Question 1.
A stuntman falls from a hot air balloon that is hovering at a constant altitude,100 feet above a lake. A television camera on shore, 200 feet from a point directly below the balloon,follows his descent. At what rate is the angle of elevation, theta, of the camera changing 2 seconds after the stuntman falls? ( neglect the height of the camera)
It seems like I'm missing a rate of change to solve this problem. I think I need to do something with the 2 seconds,but I don't know what. Any help would really be appreciated.
Question 2
An isosceles triangle has equal sides of length 12m. If the angle theta between these sides is increased from 30 to 33 degrees, use differentials to approximate the change in the area of the triangle.
I know how to find the area, missing sides, and the height. My intial thought was to do a derivative of the area and multiply by 3 degrees. However, i don't have any given rates so I am lost on what to do. Any help would really be appreaciated.
2 answers
The length of half the base is 12 sin(θ/2), so the area of the triangle is
a = 1/2 bh = 1/2 (24 sin(θ/2)) (12 cos(θ/2))
= 144 sinθ
da = 144 cosθ dθ
θ increases by 3θ, so dθ = π/60
da ~= 144 cos π/6 * π/60 = 144 * √3/2 * π/60 = 6π/5 √3
a = 1/2 bh = 1/2 (24 sin(θ/2)) (12 cos(θ/2))
= 144 sinθ
da = 144 cosθ dθ
θ increases by 3θ, so dθ = π/60
da ~= 144 cos π/6 * π/60 = 144 * √3/2 * π/60 = 6π/5 √3