Question

A thin rod of length L cm and a small particle lie on a line separated by a distance of A cm. If K is a positive constant and F is measured in Newtons the gravitational force btwn them is:
F=K/((a(a+L))

Answer 1

If a is increasing at the rate 2cm/min when a =15 and L=5, How fast is F decreasing?

Answer 2

Thanks

Answer 3

sorry i don't know a single thing about physics

Answer 4

it isn't physics this is calc

Answer 5

it is related rates

Answer 6

oooooooooooh ok let me try to read carefully

Answer 7

is this
\[F=\frac{K}{a(a+L)}\]?

Answer 8

yup

Answer 9

ok so we want F' and K and L are fixed. so we take the derivative wrt time right?

Answer 10

well we fisrt take the derivative assuming that l is fixed and a isn't

Answer 11

since we see that a is changing

Answer 12

no a is not fixed. K is a constant and L is a constant. a is changing and so is F

Answer 13

ya

Answer 14

I got an answer but a real ugly one

Answer 15

so i get
\[F'=\frac{-K(2aa'+a'L)}{(a^2+aL)^2}\]

Answer 16

oh ok

Answer 17

i see hwere i went wrong already

Answer 18

now you know all the numbers so it should just be a substitution.
i did not really use the "quotient rule' i used the fact that
\[(\frac{1}{f})'=\frac{-f'}{f^2}\] and the chain rule

Answer 19

oh ok

Answer 20

also i see that your answer will have a K in it right? because we don't know K
we know L = 5, a i= 15, a' = 2

Answer 21

ya

Answer 22

I still got the same ugly number

Answer 23

i would use a calculator

Answer 24

what did u get?

Answer 25

let me try it and see

Answer 26

i think i got
\[-\frac{7K}{9000}\]

Answer 27

I got -7k/900

Answer 28

oh so maybe this is it

Answer 29

I am always doubting myself

Answer 30

i mean i am sure i got it. maybe i put in the numbers wrong, but this is what i computed
\[-\frac{K(2\times 15\times 2+2\times 5)}{(15^2+15\times 5)^2}\]

Answer 31

ya ok

Answer 32

check to see that i put the right numbers in the right spot. i could have messed up

Answer 33

so that must be the answer

Answer 34

looks good to me

Answer 35

k there is another part to the question

Answer 36

I just need you to tell me if something i scorrect

Answer 37

a simple one

Answer 38

the next part says that L is decreasing by 2 cm so I guess a is constant and now L is varying?

Answer 39

Related Rates suck. She's posting like all of the hardest questions. That's the last question in the section I bet

Answer 40

Nope it isn't

Answer 41

It is one of the first

Answer 42

Sattelite is gone with the wind

Answer 43

If that's the first question, I'd hate to see the last one

Answer 44

It was a stupid question and he just ran off on me

Answer 45

Ya I am stuck on one

Answer 46

I have 4 more left to go

Answer 47

You should go to bed. It's getting late

Answer 48

ok now what? a is fixed and L is increasing??

Answer 49

Oh ur back?????

Answer 50

What happened to u?

Answer 51

Well neways I used my own brain but I have another question and i will post it in another post

Answer 52

how can L change? it is a length?

Answer 53

if l is decreasing at the rate 2 cm /min when a =15 and and L=5 how fast is a increasing

Answer 54

so I guess L is decreasing

Answer 55

ok so now we have to start over and take the derivative wrt L right?

Answer 56

I already solved this one. ya that is what i did

Answer 57

u don't have to do it but i will post another question but can u pls come and help?

Answer 58

Cherry please ;-)

Answer 59

oh ok. i was hoping that a and L were not both moving, but i guess we could deal with that too if we had to. sure post and i will be there

Answer 60

hey wait a sec

Answer 61

ok i will wait

Answer 62

A isn't moving in the second part?

Answer 63

I don't think a is varying here. Do u think so?

Answer 64

I solved it as a being a constant

Answer 65

i don't know i can't see the question. we took the derivative the first time as if "a" was a variable and K and L were constants.
if L is a function then we have to take the derivative differently

Answer 66

A thin rod of length L cm and a small particle lie on a line separated by a distance of A cm. If K is a positive constant and F is measured in Newtons the gravitational force btwn them is:
F=K/((a(a+L))

Answer 67

but the way the first question was phrased it looked like L was constant

Answer 68

ok then what?

Answer 69

A) If a is increasing at the rate 2cm/min when a =15 and L=5, How fast is F decreasing?

Answer 70

ok we did that right i believe

Answer 71

B) if l is decreasing at the rate 2 cm /min when a =15 and and L=5 how fast is a increasing

Answer 72

yes we did part a and this is part b

Answer 73

i am not sure how a thin rod can be decreasing in length. does this make any sense?

Answer 74

B) if L is decreasing at the rate 2 cm /min when a =15 and and L=5 how fast is a increasing

Answer 75

it must be

Answer 76

in this case you would have
\[F'=\frac{-akL'}{(a^2+aL)^2}\] i believe

Answer 77

ya ok so that is what i got. I made a a constant

Answer 78

now
\[L'=-2,a=15,L=5\] and replace

Answer 79

whtvr it is fine. If i get it wrong then I will learn what to do for next time

Answer 80

looks good to me. i think the idea is to introduce you to partial derivatives, where you hold one thing constant and take the derivative wrt the other variable. i think this looks right.
post the next one and i will look.

Answer 81

ok thanks u r the best!!!!