Expert: Scotto - 2/24/2008

Question
I am looking for feedback for the most part- I worked hard on the problem, but I have doubts and can't find a better way.  Thank you for you help and explanations.

You light a candle.  As you walk away, the temperature (in degrees Fahrenheit) and illumination (in % of one candle power) decrease as your distance (in feet) from the candle increases.

Distance (feet)----Temp(degrees F)----Illumination(%)
0--------------------55-----------------100
1--------------------54.4----------------85
2--------------------53.5----------------75
3--------------------52------------------67
4--------------------50------------------60
5--------------------47------------------56
6--------------------43.5----------------53


Above, I have a given table.
Now, COLD is when it is less than 40 degrees.  DARK is when illumination is 50% or less.
A) don't need help.
B) What is the average rate at which the temperature is changing when the illumination drops from 75% to 56%?
My comments: I think I use the first derivative, but it's only an average rate, not an instantaneous rate.  So is it (56 - 75 / 6-2) = (-4 and 3/4)?

C) READING lighting is at 65% illumination and above.  Can you still read at 3.5 feet?
My Comments: I think: (67- 60/2) + 60 = 63.5 which is less than 65% so the answer is no.  This doesn't seem too hard unless there is a trick.

D)Suppose you know that at 6 feet the instantaneous rate of change of the temperature is -4.5 degrees per foot and the instantaneous rate of change of illumination is -3% candle power per foot.  Estimate the temperature and the illumination at 7 feet.
My Comments:  Once again this seems too simple to be correct!  Temp: (43.5 - 4.5) = 39 degrees F.  Illumination: (53 - 3) = 50

E) Are you dark before you are cold or vice versa?
My Comments: This is easy if the other answers are done. You are cold before you are dark.  If my answers were wrong, I may need to change this answer.

I am not confident in these answers.  Please help me!

Answer
B) Find the equation for the rate of change.  To find the average, integrate this equaition from 0.75 to 0.56 and divide by (0.56-0.75).

C) That's it as far as I know.

D) That's a very good estimate and is only linear.  Using numerical mathematics a better one could be found, but I don't think that's necessary.

E) That's right.  The rate of change of darkness is almost instantaneous whereas the rate of change of heat takes awhile.