Rate of Change Math Example 5

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Example 5

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The temperature in a room is T(t)=20+5cosโก(ฯ€t12)T(t) = 20 + 5\cos\left(\frac{\pi t}{12}\right) degrees Celsius at time tt hours. Find the instantaneous rate of change of temperature at t=6t = 6.

Solution

  1. 1
    Differentiate using the chain rule: Tโ€ฒ(t)=โˆ’5sinโก(ฯ€t12)โ‹…ฯ€12T'(t) = -5\sin\left(\frac{\pi t}{12}\right) \cdot \frac{\pi}{12}.
  2. 2
    At t=6t = 6: Tโ€ฒ(6)=โˆ’5ฯ€12sinโก(ฯ€โ‹…612)=โˆ’5ฯ€12sinโก(ฯ€2)=โˆ’5ฯ€12T'(6) = -\frac{5\pi}{12}\sin\left(\frac{\pi \cdot 6}{12}\right) = -\frac{5\pi}{12}\sin\left(\frac{\pi}{2}\right) = -\frac{5\pi}{12}.

Answer

Tโ€ฒ(6)=โˆ’5ฯ€12โ‰ˆโˆ’1.31ย ยฐC/hourT'(6) = -\frac{5\pi}{12} \approx -1.31 \text{ ยฐC/hour}
At t=6t=6 the temperature is decreasing (cooling) at a rate of 5ฯ€12\frac{5\pi}{12} degrees per hour. The chain rule introduces the factor ฯ€12\frac{\pi}{12} from the inner function.

About Rate of Change

A measure of how quickly one quantity changes with respect to another; the ratio of the change in output to the change in input.

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