Newton's Law of cooling states that the temperature changes with a speed proportional to the difference of the temperatures. The corresponding formula is `T ( t ) = T_R + ( T_0 - T_R ) e^( - k t ) , ` where `T_R = 20^@ C ` is the...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

Newton's Law of cooling states that the temperature changes with a speed proportional to the difference of the temperatures. The corresponding formula is `T ( t ) = T_R + ( T_0 - T_R ) e^( - k t ) , ` where `T_R = 20^@ C ` is the room temperature, `T_0 = 80^@ C ` is the initial cup temperature, `k = 0.09 min^( -1 ) ` is the cooling constant, and `t ` is the time in minutes.

The formula itself gives the answer `T ( 6 ) = 20 + 60 e^( - 0.54 ) ` `approx 55^@ C .`

But the question is about linear approximating the formula, which is the same as to use the tangent line at `t = 0 . ` This tangent line has the equation `T_( lin ) ( t ) = T ( 0 ) + t * T ' ( 0 ) . `

Naturally, `T ( 0 ) = T_0 = 80^@ C . ` Also, `T ' ( t ) = -k ( T_0 - T_R ) e^( - k t ) = - 5.4 e^( - 0.09 t ) , ` thus `T ' ( 0 ) = - 5.4 ( (^@ C ) / min ) .`

This way, the linear approximation gives us `T ( 6 ) approx 80 - 5.4 * 6 = 47.6 (^@ C ) .`

It's not a surprise that the approximated temperature is lower: the linear approximation does not take into account the slowdown of cooling as the difference of temperatures decreases.