Question

+0.049 where t is in hours after 6:00 AM last Sunday12The temperature in Middletown Park at 6:00 AM last Sunday was 434 degrees Fahrenheit. The temperature was changing at a rate given by r(t) = 3.27 cosROUND ALL ANSWERS TO 2 DECIMAL PLACESAt 10 00 AM last Sunday, the temperature in the park was increasing at a rate ofabout 1.68 degrees per hourFrom 6:00 AM to 1:00 PM last Sunday, the temperature in the park increasedby _degreesWhat was the temperature in the park at 1:00 PM last Sunday? _degreesWhat was the temperature in the park at 4:00 PM Last Friday (5 days later)? _degrees

Answer 1

1 ) According to the question, the temperature has been changing according to this function:

`r(t)\text{ =}3.27\text{ }\cos ((\pi t)/(12))+0.049`

Where t is the number of hours after 6:00 am last Sunday

b) From 6 am to 1 pm last Sunday, the temperature in the park increased by

6 am to 1 pm = 7 hours, let's plug into that:

`\begin{gathered} r(t)\text{ =}3.27\text{ }\cos ((\pi t)/(12))+0.049 \\ r(7)\text{ =}3.27\text{ }\cos ((7\pi)/(12))+0.049 \\ r(7)=3.31732 \\ r(7)\approx3.32 \end{gathered}`

The rate of change in 7 hours was approximately 3.32 degrees per hour

c)

The temperature in the park at 6 am was 43.4 ºF. To find the temperature we must find the value for y, 1 pm Last Sunday. Let's plug the value already found: 1.68 for r.

Considering that according to question a, the temperature increased by 1.68º per hour. 6 am to 10 pm: 4 hours

c)

Since the question wants the temperature from 6 am to 1 pm, and it has been increasing by 3.32 degrees per hour, in 7 hours

7 x 3.32 =23.24º

43.4º+23.24=110.04ºF

d) On Friday, 5 days later there was

5 x 24 at 6am + 10 hours =120+10=130 hours

`\begin{gathered} r(130)\text{=}3.27\text{ }\cos ((130\pi)/(12))+0.049 \\ r(130)\text{ =2.7}6 \end{gathered}`

Starting from 43.4º F +2.76 =46.16ºF