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Brums [2.3K]
2 years ago
9

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

ture decreased between sundown and 3 hours after sundown
Mathematics
1 answer:
Whitepunk [10]2 years ago
5 0

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

<h3>What is the differential equation that describes the temperature in t hours after sundown?</h3>

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

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

Applying <em>separation of variables</em>, 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.

You can learn more about differential equations at brainly.com/question/24348029

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