This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chapter 2
(a) We are given that M = 100. To ﬁnd C we must solve the equation T (0) = 40 = C + 100. This gives C = −60. Thus the equation becomes T (t) = −60e−0.02180t + 100. We want to solve for t when T (t) = 90. This gives us 90 = −60e−0.02180t + 100 ⇒ −0.0218t = ln 1 6 ⇒ ⇒ 1 = e−0.02180t 6 0.0218t = ln 6 . Therefore t = ln 6/0.0218 ≈ 82.2 min. (b) Here M = 140, so we solve T (0) = 40 = C + 140 As above, solving for t in the equation T (t) = −100e−0.02180t + 140 = 90 (c) With M = 80, we solve 40 = C + 80, yielding C = −40. Setting T (t) = −40e−0.02180t + 80 = 90 ⇒ 1 − = e−0.02180t . 4 ⇒ t ≈ 31.8 . ⇒ C = −100. This last equation is impossible because an exponential function is never negative. Hence it never attains desired temperature. The physical nature of this problem would lead us to expect this result. A further discussion of Newton’s law of cooling is given in Section 3.3. 37. The diﬀerential equation dP r = P dt 100 38 ...
View
Full
Document
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

Click to edit the document details