Question

The rate at which the temperature is dropping is 6t 5t2 degrees Celsius per hour t hours after sundown. How much had the temperature decreased between sundown and 3 hours after sundown

Answer 1

Using a differential equation, it is found that the temperature dropped 72 degrees Celsius between sundown and 3 hours after sundown.

What is the differential equation that describes the temperature in t hours after sundown?

The rate at which the temperature is dropping is $6t + 5t^2$ degrees Celsius per hour t hours after sundown, hence the differential equation is:

$\frac{dT}{dt} = 6t + 5t^2$

Applying separation of variables, we find the solution as follows:

$\frac{dT}{dt} = 6t + 5t^2$

$dT = (6t + 5t^2) dt$

$\int dT = \int (6t + 5t^2) dt$

$T(t) = \frac{5t^3}{3} + 3t^2 + T(0)$

In which T(0) is the temperature at sundown.

In 3 hours, the change will be of:

$C = T(3) - T(0) = \frac{5t^3}{3} + 3t^2|_{t = 3}$

Hence:

$\frac{5(3)^3}{3} + 3(3)^2 = 45 + 27 = 72$

The temperature dropped 72 degrees Celsius between sundown and 3 hours after sundown.

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