N=400/2t+1 +100t+1000 find the rate of change when t=2 the answer is 68 mosquiotes per hour

Answer 1

To find the rate of change when t=2, we need to differentiate the given equation with respect to t.

Differentiating N = (400/2t+1) + 100t + 1000 with respect to t:

dN/dt = d/dt (400/2t+1) + d/dt (100t) + d/dt (1000)

Applying the power rule for differentiation:

dN/dt = (-400)(2t+1)^(-2)(2) + 100 + 0

Simplifying:

dN/dt = (-800)/(2t+1)^2 + 100

Substituting t = 2 into the equation:

dN/dt = (-800)/(2(2)+1)^2 + 100
= (-800)/(4+1)^2 + 100
= (-800)/(5)^2 + 100
= (-800)/25 + 100
= -32 + 100
= 68

Therefore, the rate of change when t=2 is 68 mosquitoes per hour.