Stitz-Zeager_College_Algebra_e-book

# 7 circles and semi circles have been used many times

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Unformatted text preview: ls, N , in thousands, after t days. Solution. We begin with N (t) = N0 ekt . Since N is to give the number of cells in thousands, we have N0 = 12, so N (t) = 12ekt . In order to complete the formula, we need to determine the growth rate k . We know that after one week, the number of cells has grown to ﬁve million. Since t 9 The average rate of change of a function over an interval was ﬁrst introduced in Section 2.1. Instantaneous rates of change are the business of Calculus, as is mentioned on Page 121. 382 Exponential and Logarithmic Functions measures days and the units of N are in thousands, this translates mathematically to N (7) = 5000. 1250 t We get the equation 12e7k = 5000 which gives k = 1 ln 1250 . Hence, N (t) = 12e 7 ln( 3 ) . Of 7 3 course, in practice, we would approximate k to some desired accuracy, say k ≈ 0.8618, which we can interpret as an 86.18% daily growth rate for the cells. Whereas Equations 6.3 and 6.4 model the growth of quantities, we can use equations like them to describe the decline of quantities. One example we’ve seen already is Example 6.1.1 in Section 6.1. There, the value of a car decli...
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