Question

the function S(d)=93logd+65 relates the speed of the wind, S, in mile per hour, near the centre of a tornado to the distance that the tornado travels, d, in miles.

a)Calculate the avg. rate of change for the speed of the wind at the centre of a tornado from mile 10 to mile 100.
b) Calculate the approx. rate of change of the wond at the centre of the tornado at the moment it has travelled 10th mile and its 100th mile.

Answer 1

your last few posts all deal with basically the same method, only the function equation differs.

a) find S(10) = 93log10 + 65 = 158
then find S(100)

avg rate = (S(100) - S(10))/(100-10)

b) from a) S(10) = 158
consider S(10.01) = 93log(10.01) + 65
= 158.04

so appr. rate of change at 10th mile
= (158.04-158)/(10.01-10)
= appr 4 mph

repeat the same steps for the 100th mile
(I got 0.4)

Answer 2

Almost all correct!

After you get 158 from S(10) and 251 from S(100)
*Do the y2-y1/x2-x1

(y2=251, y1=158, x1=100, and x2=10)

SO...
251-158 = 93
100-10 = 90

Then the answer is 93/90 = 1.03 miles/hour/mile

Answer 3

Sorry Reiny your correct I didn't see the end of part a so I thought the end answers were 4 and 0.4