Question

Atmospheric Pressure decays exponentially as altitude increases. With pressure, P, in inches of mercury and altitude h in feet above sea level we have:
P=30e^(-3.23*10^(-5)h)

Answer 1

At whta altitude is the atmospheric pressure 25 inches of mercury?

Answer 2

I got h as a negative number

Answer 3

I don't think that makes sense

Answer 4

Hey I got a better answer this time

Answer 5

is it 5644

Answer 6

Ya that is what I got the second time

Answer 7

ok have to do nothing just plug in the value of p and u will get the answer

Answer 8

where does u get into the picture?

Answer 9

means??????

Answer 10

I just plugged in P as 25 and then solved it using natural logaritthm

Answer 11

It means where did u get the "u" from

Answer 12

http://www.wolframalpha.com/input/?i=25%3D30e^%28-3.23*10^%28-5%29*x%29+

Answer 13

Oh i see i gotta solve it by hand

Answer 14

yeah i dint hv calculator so i solved it using that website and also i dont have log table otherwise i might have..............lol

Answer 15

Thanks for ur help ;-)

Answer 16

thnk that web portal not me

Answer 17

So then thanks for your time

Answer 18

m just a messenger of that almighty try to help others as much as i can..so thank that almighty......................lol

Answer 19

U R nuts

Answer 20

Hey it wasn't meant to be an offensive comment

Answer 21

m not nuts.....here every1 on this planet is having a defined duty and i feel this one is meant for me.................

Answer 22

Yup at least ur helping others and performing acts of charity

Answer 23

no it is not act of charity........every human being is a part of that almighty so how their can be any difference between two living beings............so how i can perform an act of charity in front of almighty.............and as everything belongs to him only so i m just caretaker of his property so how can i do anything with his belongings...........

Answer 24

I see that u r pretty religious

Answer 25

yeah family values

Answer 26

There is another part of the question that I need help with can u help me?

Answer 27

yeah sure will try to

Answer 28

Do u know related rates?

Answer 29

pls clarify lil bit

Answer 30

ok I will type the question and then you will see

Answer 31

i feel it must be related to two or more variables relation right???????

Answer 32

A glider measures the pressure to be 25 inches of mercury and experiences a pressure increase of .1 inches of mercury per minute. At wwhat rate is the changing altitude

Answer 33

Yes that is what it is

Answer 34

first task is to find out what things are changing and with relative to which things i.e., functions

Answer 35

here the pressure is increasing with the time
it means that pressure is a function of time

Answer 36

I got an answer but I don't know If it is correct

Answer 37

right but when the presuures is increasing the altitude must be decreasing?

Answer 38

but u have not mentioned in this question

Answer 39

well it is part of the first question look at the post

Answer 40

A glider measures the pressure to be 25 inches of mercury and experiences a pressure increase of 0.1 inches of mercury per minute. At wwhat rate is the changing altitude

Answer 41

No the original post

Answer 42

ok
it means that pressure increases with the decrease in altitude
thus pressure is a function of altitude

Answer 43

Do we first find the derivative of the equation with respect to time?

Answer 44

so it would be dp/dt= 30e^(-3,23*10^-5h)*(-3.23*10^-5)?

Answer 45

yeah u can but according to first post pressure is a function altitude
and according to second post it is a function of time

Answer 46

oh whoops i made a mistake. It should really be:
dp/dt= 30e^(-3,23*10^-5h)*(-3.23*10^-5)*(dh/dt)

Answer 47

dp/dt= 30e^(-3,23*10^-5h)*(-3.23*10^-5)
this is impossible here we dont have t in the RHS so if we'll get 0

Answer 48

yeah now ok

Answer 49

so according to second post pressure is function of time

Answer 50

ya

Answer 51

dp/dt = 0.1 inches

Answer 52

right

Answer 53

now solve for h

Answer 54

ok wait a sec

Answer 55

0.1 in/min = 30e^(-3,23*10^-5h)*(-3.23*10^-5)*(dh/dt)

Answer 56

I got an ugly number

Answer 57

wt u got

Answer 58

wait what do i plug in for h?

Answer 59

4655?

Answer 60

0.1 in/min = 30e^(-3,23*10^-5h)*(-3.23*10^-5)*(dh/dt)
can u solve this one

Answer 61

Well I am put I have to plug a value into h and I assumed it would be 4655

Answer 62

since in the equatiion it mentions abt 25 inches of mercury

Answer 63

and if this is correct then the answer i got is -.012

Answer 64

see after diff u get
-0.000969*e^(-3.23*10^(-5)*h)(dh/dt) = dp/dt = 0.1 inches /min

Answer 65

huh?

Answer 66

did u got the answer

Answer 67

but what did you do with h?

Answer 68

LIke how can u solve when you have two unknown varaibles?

Answer 69

http://www.wolframalpha.com/input/?i=e^(-3.23*10^(-5)*x)(dx%2Fdt)+%3D+0.1%2F(-0.000969)+

Answer 70

go to this place u will get the answer

Answer 71

I so didn't get that. I went there but it didn't give me any answer

Answer 72

on this post?

Answer 73

ya

Answer 74

just skim thru the whole post and try to follow what is going on

Answer 75

there are 2 questions that are really part of one

Answer 76

the original post give sthe equation and then there was question # 1 that asked abt the altitude which I got the answer

Answer 77

and then there is a second part to th eequation with the question abt the glider

Answer 78

That is where I need help

Answer 79

Atmospheric Pressure decays exponentially as altitude increases. With pressure, P, in inches of mercury and altitude h in feet above sea level we have:
P=30e^(-3.23*10^(-5)h)

did you get the first answer to be
P(5644)=25
?

Answer 80

yup

Answer 81

A glider measures the pressure to be 25 inches of mercury and experiences a pressure increase of 0.1 inches of mercury per minute. At wwhat rate is the changing altitude

This is where we are, right?

Answer 82

yup

Answer 83

ya i got stuck here

Answer 84

so our function is\[P=30e^{-3.23\times10^{-5}h}\]what do you get when you differentiate this w/respect to time?

Answer 85

(dp/dt)=30e^(-3.23*10^(-5)h)*(-.323*10^-5)*(dh/dt)

Answer 86

right
\[{dp \over dt}=-9.69\times10^{-4}e^{-3.23\times10^{-5}h}{dh \over dt}\]we know all these variables except dh/dt, so we can solve this
dp/dt=?
h=?
let me see what I get and we can check our answers.

Answer 87

is h=5644?

Answer 88

dp/dt-.1

Answer 89

yes, because we know that we are at P=25 which only happens at h=5644
dp/dt is correct

Answer 90

whoops dp/dt=.1

Answer 91

right, positive, my bad

Answer 92

So where is the problem? I get dh/dt=-124
Do you know the answer?

Answer 93

i didn't get the answer i got an ugly number

Answer 94

well let's see what happened.
We agree the derivative is

(dp/dt)=30e^(-3.23*10^(-5)h)*(-3.23*10^-5)*(dh/dt)

notice that with h=5644 this can be written as

dp/dt=25(-3.23*10^-5)(dh/dt)=0.1

because

30e^(-3.23*10^(-5)5644)=25

according to our earlier work.
now what do you get?

Answer 95

Hey I got the same answer

Answer 96

I guess i did some wrong calculation on the way

Answer 97

Probably just a typo when you had too many numbers floating around. I didn't see the simplification at first either, but it's important to take advantage of that kind of thing to avoid errors.

I saw you original post had a typo, perhaps you used it on accident?

You wrote
(dp/dt)=30e^(-3.23*10^(-5)h)*(-.323*10^-5)*(dh/dt)
^decimal should be after the three
Should be
(dp/dt)=30e^(-3.23*10^(-5)h)*(-3.23*10^-5)*(dh/dt)

maybe that was your mistake.

Answer 98

K thanks for ur help