area of the triangle, (you said segment, not sector)
= (1/2)(10)(10)sinθ
d(area)/dθ = 50 cos θ dθ/dt
for the given data, 50° = 5π/18 radians
= 50 cos(5π/18)(1.2) cm^2/sec
= appr 38.57 cm^2/sec
A circular segment is a part of a 10cm radius circle whose subtended angle changes at 1.2 degrees per
seconds, what is the rate of change of its area when the subtended angle is 50 degrees?
2 answers
The segment is the circular hat on top of the triangle
a = 1/2 r^2 (θ - sinθ)
da/dt = 1/2 r^2 (1 - cosθ) dθ/dt
da/dt = 1/2 * 100 (1 - cos 5π/18) (1.2 * π/180) = 0.374 cm^2/s
a = 1/2 r^2 (θ - sinθ)
da/dt = 1/2 r^2 (1 - cosθ) dθ/dt
da/dt = 1/2 * 100 (1 - cos 5π/18) (1.2 * π/180) = 0.374 cm^2/s
1 answer