Draw a diagram. It is clear that the distance z between the kite and the spot directly over its anchor point obeys
100/z = tanθ
when 200 ft of string have been let out,
sinθ = 100/200 so θ = π/6
and z = 100√3
Now, we have
-100/z^2 dz/dt = sec^2θ dθ/dt
-100/300 = 4 dθ/dt
dθ/dt = -1/12 rad/s
A kite 100 ft above the ground moves horizontally at a speed of 3 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?
rad/s
2 answers
At a given time of t seconds, let the horizontal distance of the kite be x ft
and let the angle it makes with the ground be θ
let the length of string be d ft
given: dx/dt = 3 ft/s
find: dθ/dt when d = 200 ft
then
tanθ = 200/x
sec^ θ dθ/dt = -200/x^2 dx/dt
Your specific case:
when d = 200, sinθ = 100/200 = 1/2, and θ = 30°
tan30° = 200/x
x = 200/tan30 = 200/(√3/2) = 400/√3
also:
cosθ = √3/2
cos^2 θ = 3/4
sec^2 θ = 4/3 , so in
sec^ θ dθ/dt = -200/x^2 dx/dt
4/3 dθ/dt = -200/(400/√3)^2 * 3
your turn to do some arithmetic, solve for dθ/dt
(your answer will be in radians/sec, it did not matter that I used 30° or π/6 radians in the above calculations, but the end results is in radians because
the derivative rules for differentiating trig functions are valid only for radians)
and let the angle it makes with the ground be θ
let the length of string be d ft
given: dx/dt = 3 ft/s
find: dθ/dt when d = 200 ft
then
tanθ = 200/x
sec^ θ dθ/dt = -200/x^2 dx/dt
Your specific case:
when d = 200, sinθ = 100/200 = 1/2, and θ = 30°
tan30° = 200/x
x = 200/tan30 = 200/(√3/2) = 400/√3
also:
cosθ = √3/2
cos^2 θ = 3/4
sec^2 θ = 4/3 , so in
sec^ θ dθ/dt = -200/x^2 dx/dt
4/3 dθ/dt = -200/(400/√3)^2 * 3
your turn to do some arithmetic, solve for dθ/dt
(your answer will be in radians/sec, it did not matter that I used 30° or π/6 radians in the above calculations, but the end results is in radians because
the derivative rules for differentiating trig functions are valid only for radians)