Question

The temperature of an object is given by T ( t ) = 280 + 1.5 t 2 e − 0.12 t where t is measured in minutes and T is measured in Kelvins (abbreviated K). (a) Find the rate of change of the rate of change of temperature at the instant t = 20 minutes.

Answer 1

Given

$$T(t)=280+1.5t^2e^{-0.12t}$$

$$T'(t)=(280+1.5t^2e^{-0.12t})'=$$

$$=1.5(2t)e^{-0.12t}+1.5t^2(-0.12e^{-0.12t})=$$

$$=3te^{-0.12t}-0.18t^2e^{-0.12t}$$

$$T''(t)=(3te^{-0.12t}-0.18t^2e^{-0.12t})'=$$

$$=3e^{-0.12t}+3t(-0.12e^{-0.12t})-$$

$$-0.18(2t)e^{-0.12t}-0.18t^2(-0.12e^{-0.12t})=$$

$$=3e^{-0.12t}-0.72te^{-0.12t}+0.0216t^2e^{-0.12t}$$

$$T''(20)=3e^{-0.12(20)}-0.72(20)e^{-0.12(20)}+$$

$$+0.0216(20)^2e^{-0.12(20)}\approx-0.25K/min^2$$

The rate of change of the rate of change of temperature at the instant t = 20 minutes is $-0.25\ K/min^2.$