Answer:
According to this model, the rate of increase of bees spotted with respect to time when 100 bees are spotted = 80 bees per day.
Step-by-step explanation:
√B(x) = 50 k + 2x
where x is the daily high temperature, in degrees Farenheit.
k is a positive constant.
When the number of bees spotted = 100,
The daily high temperature is increasing at a rate of 2°F per day; (dx/dt) = 2 °F/day
How quickly is the number of bees changing with respect to time when 100 bees are spotted?
√B(x) = 50k + 2x
Taking a time derivative of both sides,
(d/dt) [√B] = (d/dt) [50 k + 2x]
(1/2) (B⁻⁰•⁵) (dB/dt) = (d/dt)(50k) + (d/dt)(2x)
½(B⁻⁰•⁵) (dB/dt) = 50 (dk/dt) + 2 (dx/dt)
B = 100 bees
(dB/dt) = ?
(dk/dt) = 0 (since k is a positive constant)
(dx/dt) = 2°F/day
½(100⁻⁰•⁵) (dB/dt) = 50 (0) + 2 (2)
½(10⁻¹) (dB/dt) = 4
(1/2)(1/10) (dB/dt) = 4
(dB/dt) = 4×2×10 = 80
(dB/dt) = 80 bees per day
Hope this Helps!!!
