Question

微积分的问题~~~

shows the temperature of a cup of coffee during a ten minute period. (Each of your answers in (a) – (c)
should have the units "degrees per minute.")
(a) What was the average rate of cooling
from minute 0 to minute 10?
(b) What was the average rate of cooling
from minute 7 to minute 8? from
minute 8 to minute 9?
(c) What was the rate of cooling at
minute 8? at minute 2?
(d) When was the cold milk added to
the coffee?

谁能给我解下这题目啊？？要有过程的~~

Answer 1

a) It takes 10 minutes for the temperature to fall from 150 to 10. So the average cooling rate is (150-70)/10 = 8 degrees per minute;
b) Same method. From minute 7 to 8, it's 80 to 75, so (80-75)/(8-7) = 5;
From minute 8 to 9, it's 75 to 70, so (75-70)/(9-8) = 5;
c) At minute 8, to be exact you have to draw a tangent at the point on the curve, I'm sure you can do this by yourself, then just extent it long enough to measure its tangent, that's the rate.
Another approximate method is just to take the average of the two numbers you got in question b) which is 5.
At minute 2, same methods apply. The approximate way gives you 10.
d) Apparently, it's at minute 6, cas the way how the temperature lowers exhibits a turning point here.

追问

i understand now~~ thanks!

追答

Yep.
For 5, you got 5s in question b), right ? So (5+5)/2 = 5, that's at minute 8;
For 10, I'm sorry that I got it wrong previously, the answer is 7.5. Here's how:
first, calculate the average cooling rate from minutes 1 to 2, which is:
(130-120)/(2-1) = 10;
then calculate the same thing for minute 2 to 3, which is:
(120-115)/(3-2) = 5;
Finally, take the average of above two numbers, (10+5)/2 = 7.5. This is the cooling rate at minute 2. The basic trick is to calculate the rates around it, then take average.

Again sorry for the confusion, I was not careful enough.