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course Mth 271
9/26 1030pIt was my understanding that this was not a major quiz and did not need to be proctored.
A man reading a newpaper walks at a constant speed away from the lamp which illuminates the paper he is reading. The illumination from the lamp at clock time t is given by I = 20 / (t+3)2, where I is illumination in watts/m2 and t is clock time in seconds.
At what average rate is the illumination from the lamp changing between clock times t = 15.5 and t = 15.6 seconds?
At what average rate is the illumination from the lamp changing between clock times t = 15.5 and t = 15.51 seconds?
At what average rate is the illumination from the lamp changing between clock times t = 15.5 and t = 15.501 seconds?
What do you estimate is the rate at which illumination is changing at clock time t = 15.5 seconds?
The rate at which the illumination a newspaper changes is given by Rate = 2830 / (t+3)3, where Rate is rate of change in watts/m2 per second and t is clock time in seconds. How much do you think illumination will change between t = 15.5 and t = 31 seconds?
I=100/(11.7+3)^2 = 100/216.09 = .4628
I = 100/(11.8+3)^2 = 100/219.04 = .4565
( .4628+.4565)/2= .4596
I = 100/(11.7+3)^2 = .4628
I = 100/(11.71+3)^2 = .4621
.9249/2 = .4624
.4628
I = 100/(11.701 +3)^2 = 100/216.119 = .4627
(.4628+.4627)/2 = .46275
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If I was the rate at which something is changing, the the average rate on an interval would be approximately equal to the average of the initial and final values, and your results would depict average rates.
However I is the illumination. Averaging two values of the illumination gives you the average value of the illumination, not the average rate at which the illumination is changing.
You'll need to find the average rates of change of the illumination for these three intervals.
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1850/(11.7+3)^3 = 1850/3176.523 = .5824 watts/m^2/sec
1850/(23.4+3)^3 = 1850/18399.744 = .1005 watts/m^2/sec"
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If the rate was 1850 / (t + 3)^3 your calculations would show the values of the rate at t = 11.7 sec and t = 23.4 sec.
If you averaged these two rates you would have an approximate average rate, which you could then use to find the change in the illumination.
However the given rate function is 2830 / (t+3)^3, and the given interval is from t = 15.5 sec to t = 31 sec.
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