Problem

Stiles-Crawford effect Light enters the eye throu…

06:13

Problem 50 Easy Difficulty

Foraging In Exercise 4.2 .2 we let $N(t)$ be the amount of nectar foraged from a flower by a bumblebee in $t$ seconds.
(a) What is the average rate of nectar consumption over a period of $t$ seconds?
(b) Find the average rate of nectar consumption for very short foraging visits by using I'Hospital's Rule to calculate the limit of the answer to part (a) as $t \rightarrow 0$ .

Answer

a. $\frac{N\left(t+t_{0}\right)-N\left(t_{0}\right)}{\left(t+t_{0}\right)-t_{0}}=\frac{N\left(t+t_{0}\right)-N\left(t_{0}\right)}{t}$
b. $\begin{aligned} \lim _{t \rightarrow 0} \frac{N\left(t+t_{0}\right)-N\left(t_{0}\right)}{t} &=\lim _{t \rightarrow 0} \frac{N^{\prime}\left(t+t_{0}\right)-0}{1} \\ &=\lim _{t \rightarrow 0} N^{\prime}\left(t+t_{0}\right) \\ &=N^{\prime}\left(t_{0}\right) \end{aligned}$

Top Calculus 1 / AB Educators

Watch More Solved Questions in Chapter 4

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Video Transcript

So in this problem we are given that researchers measured blood alcohol concentration of eight men Uh starting one hour after consumption of a couple of drinks. And we're giving a table rooty isn't ours. One 1.5, 2.2.5 and 3.0 and the concentration C of T in milligrams per millimeter 0.31 0.24, 0.18, 0.12 And 0.08. First of all, we're asked to find the average change of C with respect to T over each time interval. So going from 1.02, we have 0.18 -0.3, 1 Over 2 -1 Is a -0.18. This is milligrams per millimeter and time is in ours. Okay. And from 1.5 two, We have 0.18 -0.24 over 2 -1.5. So that a negative 0.06 over 0.5. So that's a negative 0.12 again, milligrams per mil later our Okay, The next one from 2.0 To 2.5. Well, from our table This will be 0.1, two -0.18. We're just using our average rate of change equation here between these two timeframes, keeping everything coordinated minus two. and so I have a negative 0.06 over 0.5, Which again is a negative 0.12 milligrams per million. Later our Okay. And the last one here, we're going from 2.0 23 oh. And so that's 0.08 minus 0.18 over 3.0 -2. And so that's a negative 0.10 over one. That's a negative 0.10 milligrams per million later. Power. All right. In part b says to estimate the instantaneous rate of change, estimate the instantaneous right of change. Ah T equals two. Okay, so then at T equals two. What we can do is come up here and think about what happens as we approach two from each direction left and right. Well from 1.5 to 2 we did right here A- -12. And from right here we did it from the right And got a negative 0.12. So we have the same rate coming in from both sides. So normally we would average these two. But since they're the same, that means we're going to have the same instantaneous rate as we have a constant rate coming in from both sides left and right.

Get More Help with this Textbook