The temperature T of food put in a freezer is t= 700/(t^2+4t+10) where t is the time in hours. Determine the rate of change of the temperature with respect to time when t=2

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The temperature of food put in a freezer is given by `T = 700/(t^2+4t+10)`

The rate of change of temperature is the derivative T'.

T = 700*(t^2+4t+10)^-1

`(dT)/(dt) = (-700*(2t + 4))/(t^2+4t+10)^2`

At t = 2

`(dT)/(dt) = (-700(4 + 4))/(4 + 8 + 10)^2`

=> `(-700*8)/484`

=> -11.57

The rate of change of temperature at t = 2 is -11.57

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