Acalf that weighs 60 pounds at birth gains weight at the rate dw/dt = k(1200 - w)
calculus question !

weight gain a calf that weighs 60 pounds at birth gains
weight at the rate
dwk1200w dt
where w is weight in pounds and t is time in years. solve the differential equation.
(a) use a computer algebra system to solve the differential equation for k 0.8, 0.9, and 1. graph the three solutions.
(b) if the animal is sold when its weight reaches 800 pounds, find the time of sale for each of the models in part (a).
(c) what is the maximum weight of the animal for each of the models?

Dw/dt = k(1200 - w)
dw/(1200 - w) = kdt
-ln(1200 - w) = kt + c

When t = 0, w = 60
-ln(1200 - 60) = c
c = -ln(1140)

a.) For k = 0.8: -ln(1200 - w) = 0.8t - ln(1140)
t =  [ln(1140) - ln(1200 - w)]/0.8

For k = 0.9:
t = [ln(1140) - ln(1200 - w)]/0.9

For k = 1:
t = [ln(1140) - ln(1200 - w)]
 
b.) For k = 0.8
t = [ln(1140) - ln(1200 - 800)]/0.8 = [ln(1140) - ln(400)]/0.8 = 1.3 years

For k = 0.9
t = [ln(1140) - ln(400)]/0.9 = 1.16 years

For k = 1
t = [ln(1140) - ln(400)] = 1.05 years

c.) For maximum weight,
dw/dt = 0
k(1200 - w) = 0
1200 - w = 0
w = 1200

Therefore, the maximum weight for each of the model is 1200 pounds.

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