1. Problem 8.2.6 from textbook Below is a table of predicted temperatures. Suppose the function f(t) gives the temperature at t hours after noon in degrees Fahrenheit. a) What are the units of f’(t)? b) What does the statement f’(7)=-3 mean? c) What is the significance of f’(t) being positive? versus being negative? t012345678910 f(t)7579807875716764626059

2. Problem 8.2.6 from textbook Below is a table of predicted temperatures. Suppose the function f(t) gives the temperature at t hours after noon in degrees Fahrenheit. a) What are the units of f’(t)? b) What does the statement f’(7)=-3 mean? c) What is the significance of f’(t) being positive? versus being negative? t012345678910 f(t)7579807875716764626059 Units for derivatives are always function/variable. In this case we have °F/hr.

3. Problem 8.2.6 from textbook Below is a table of predicted temperatures. Suppose the function f(t) gives the temperature at t hours after noon in degrees Fahrenheit. a) What are the units of f’(t)? b) What does the statement f’(7)=-3 mean? c) What is the significance of f’(t) being positive? versus being negative? t012345678910 f(t)7579807875716764626059 Units for derivatives are always function/variable. In this case we have °F/hr. f’(7)=-3 is a statement about the derivative of f(t) at t=7. In this context the derivative tells us about how the temperature is changing. At 7pm the temperature is dropping at 3 °F/hr.

4. Problem 8.2.6 from textbook Below is a table of predicted temperatures. Suppose the function f(t) gives the temperature at t hours after noon in degrees Fahrenheit. a) What are the units of f’(t)? b) What does the statement f’(7)=-3 mean? c) What is the significance of f’(t) being positive? versus being negative? t012345678910 f(t)7579807875716764626059 Units for derivatives are always function/variable. In this case we have °F/hr. f’(7)=-3 is a statement about the derivative of f(t) at t=7. In this context the derivative tells us about how the temperature is changing. At 7pm the temperature is dropping at 3 °F/hr. f’(t) being negative tells us that the temperature is decreasing, rather than increasing.

5. t012345678910 f(t)7579807875716764626059 Let’s try one more thing with this problem. d) Find the average rate of change of f(t) between 1pm and 4pm.

6. t012345678910 f(t)7579807875716764626059 Let’s try one more thing with this problem. d) Find the average rate of change of f(t) between 1pm and 4pm. An average rate of change is just a slope of a secant line. This is the type of slope you have been doing for years. Just find rise/run. slope

7. t012345678910 f(t)7579807875716764626059 Let’s try a few more things with this problem. d) Find the average rate of change of f(t) between 1pm and 4pm. An average rate of change is just a slope of a secant line. This is the type of slope you have been doing for years. Just find rise/run. slope e) Would the slope you found in part d) be a good estimate for f’(1)?

8. t012345678910 f(t)7579807875716764626059 e) Would the slope you found in part d) be a good estimate for f’(1)? Take a look at the graph of our data. The average slope we found joins the 1pm and 4pm points. f’(1) is the tangent line slope at 1pm. The tangent at 1pm slopes upwards, but we got a negative value for the slope we calculated from 1pm to 4pm. The average is clearly not close to the slope at 1pm (because the points we used were so far apart).

9. t012345678910 f(t)7579807875716764626059 f) Use a tangent line approximation at t=7 to approximate the temperature at 7:20pm.

10. t012345678910 f(t)7579807875716764626059 f) Use a tangent line approximation at t=7 to approximate the temperature at 7:20pm. We know the temp. at 7pm is 64°F and we are given f’(7)=-3°F/hr. Since 7:20pm is 1/3 hr after 7pm we can write the following formula: