The temperature of a liquid after being placed in a refrigerator is given by $T=5+95e^(-0.123t)$ degree Celsius where t is the time in minutes Show that $(dT)/dt=c(T-5)$ for some constant c. Find the value of c?

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Answer 1 · 2018-01-27T10:05:46

The value of $c=-0.123$

The equation is

$T=5+95*e^(-0.123t)$

$T-5=95*e^(-0.123t)$

Differentiating with respect to $t$

$(dT)/(dt)=-0.123*95e^(-0.123t)$

Plugging in the value of $95*e^(-0.123t)=T-5$

$(dT)/(dt)=-0.123(T-5)$

Comparing this to $(dT)/(dt)=c(T-5)$

The value of the constant $c=-0.123$

The temperature of a liquid after being placed in a refrigerator is given by T=5+95e^(-0.123t)T=5+95e^(-0.123t) degree Celsius where t is the time in minutes Show that (dT)/dt=c(T-5)(dT)/dt=c(T-5) for some constant c. Find the value of c?