Question

A warm object cools according to the function T(t)= 331e^-0.035 +82. Assuming a constant environmental temperature, what constant temperature, in F, will the object’s temperature eventually stabilize?

Answer 1

Explanation:

The function T(t) represents the temperature of the warm object at time t, and it is given by:

T(t) = 331e^(-0.035t) + 82

As time goes on, the exponential term e^(-0.035t) approaches zero, since the exponent becomes more and more negative. This means that the temperature of the object will eventually stabilize at a constant value, which we can find by taking the limit of the function as t approaches infinity.

lim T(t) as t → infinity = lim (331e^(-0.035t) + 82) as t → infinity

Since e^(-0.035t) approaches 0 as t approaches infinity, we can simplify the expression:

lim T(t) as t → infinity = lim 82 as t → infinity

Therefore, the temperature of the object will eventually stabilize at a constant value of 82°F.

Note that this assumes that the environmental temperature remains constant and does not affect the cooling of the object. If the environmental temperature were to change, it could affect the final stabilized temperature of the object.