Solve the problem.
-Newton's law of cooling indicates that the temperature of a warm object will decrease exponentially with time and will approach the temperature of the surrounding air. The temperature $T ( t )$ is modeled by $T ( t ) = T _ { a } + \left( T _ { 0 } - T _ { a } \right) e ^ { - k t }$. In this model, $T _ { a }$ represents the temperature of the surrounding air, $T _ { 0 }$ represents the initial temperature of the object and $t$ is the time after the object starts cooling. The value of $k$ is the cooling rate and is a constant related to the physical properties of the object.
A cake comes out of the oven at $335 ^ { \circ } \mathrm { F }$ and is placed on a cooling rack in a $70 ^ { \circ } \mathrm { F }$ kitchen. After checking the temperature several minutes later, it is determined that the cooling rate $k$ is $0.050$. Write a function that models the temperature $T ( t )$ (in ${ } ^ { \circ } \mathrm { F }$ ) of the cake $t$ minutes after being removed from the oven.
A) $T ( t ) = 335 + 70 e ^ { 0.050 t }$
B) $T ( t ) = 70 + 265 e ^ { 0.050 t }$
C) $T ( t ) = 70 + 265 e ^ { - 0.050 t }$
D) $T ( t ) = 70 + 335 e ^ { 0.050 t }$

Question

Quizplus · Accepted Answer

The correct model is $T(t) = T_a + (T_0 - T_a)e^{-kt}$, where $T_a = 70^\circ F$, $T_0 = 335^\circ F$, and $k = 0.050$. Substituting these values gives $T(t) = 70 + (335 - 70)e^{-0.050t}$, which simplifies to $T(t) = 70 + 265e^{-0.050t}$.

Solve the Problem T(t)T ( t )T(t) Is Modeled By T(t)=Ta+(T0−Ta)e−ktT ( t ) = T _ { a } + \left( T _ { 0 } - T _ { a } \right) e ^ { - k t }T(t)=Ta​+(T0​−Ta​)e−kt

Solve the Problem $T ( t )$ Is Modeled By $T ( t ) = T _ { a } + \left( T _ { 0 } - T _ { a } \right) e ^ { - k t }$