Problem

An object starts above room temperature and follows \(T(t)~=~20~+~A\cdot~b^{t}~^{\circ}\text{C}\), where \(A~>~0\) and \(0~<~b~<~1\). Interpret the input and output, horizontal asymptote, direction of approach, and the difference \(T(t)~-~20\).

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input = t time units; output = T(t)°C; asymptote = x=20 time units; approach = T(t) reaches 20°C immediately; difference = T(t)-20 is always 0

input = t time units; output = T(t)°C; asymptote = y=A°C; approach = T(t) stays a fixed distance above 20°C; difference = T(t)-20=A

input = t time units; output = T(t)°C; asymptote = y=20°C; approach = T(t) increases toward 20°C from below; difference = T(t)-20 is negative and grows away from 0

input = t time units; output = T(t)°C; asymptote = y=20°C, the room temperature; approach = T(t) decreases toward 20°C from above; difference = T(t)-20=A·bᵗ is positive excess temperature shrinking toward 0

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An object starts above room temperature and follows \(T(t)~=~20~+~A\cdot~b^{t}~^{\circ}\text{C}\), where \(A~>~0\) and \(0~<~b~<~1\). Interpret the input and output, horizontal asymptote, direction of approach, and the difference \(T(t)~-~20\).

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