Yeast is cultivated in a nutrient solution for 15 hours

MATHEMATICS · Multi-part · AP · Calculus AB · Differential Equations · Differential Equations

Problem

Yeast is cultivated in a nutrient solution for 15 hours. During the first 6 hours ample sugar is available, and the population Y(t)Y(t) of yeast cells satisfies dYdt=k1Y\dfrac{dY}{dt}=k_1Y with constant k1>0k_1>0. At t=6t=6 hours the sugar supply diminishes, after which the growth continues proportionally but with a smaller constant k2>0k_2>0. The following observations are recorded: • Y(0)=500Y(0)=500 cells • Y(6)=3000Y(6)=3000 cells • Y(10)=3600Y(10)=3600 cells (a) Using the data for 0≤t≤60\le t\le6, determine the value of k1k_1 (per hour) correct to three decimal places. (b) Using the data for t≥6t\ge6, determine the value of k2k_2 (per hour) correct to three decimal places. (c) State a complete piecewise model for Y(t)Y(t) valid on 0≤t≤150\le t\le15, indicating each exponential expression and its applicable interval. (d) According to your model, calculate the increase in the yeast population between t=0t=0 and t=15t=15 hours. (e) Discuss one practical implication of the difference between k1k_1 and k2k_2 for managing nutrient levels in future experiments.

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