Question

a 10-foot ladder is leaning against a wal. If a person pulls the base of the ladder away from the wall at the rate of 1 foot per second, how fast is the top going down when the base of the ladder is 6 feet from the wall?

Answer 1

@ganeshie8 @hartnn The roc in movement is the derivative, right?

Answer 2

@mathmale Is this a good way to look at it?

Answer 3

I've drawn a triangle to represent it

Answer 4

and x should equal 6, right?

Answer 5

we are given the change of x with respect to time

which we can write as dx/dt = 6

Answer 6

nope, they are giving you a rate

Answer 7

Ah yes, of course, because it is changing with respect to time

Answer 8

so think hmmm change of distance over change of time.... if distance is x and time is t then we write dx/dt

Answer 9

6 is the final pos it is approaching

Answer 10

yes that is usually the tricky part... putting their words into derivative form

Answer 11

oh gosh I am sorry I must be too tired

Answer 12

you are correct, dx/dt is 1ft/s

Answer 13

hmm, ok so we have some intervals, but I'm not sure how to use this info

Answer 14

okay well we have a right angle triangle... what do we know that relates all the sides

Answer 15

now when the ladder moves away at 1ft/s, the top should move at 1ft/s too then?

Answer 16

x^2+y^2=10^2

Answer 17

hmm wait yes I was about to say use pyth.

Answer 18

and just to make sure you know about this... the reason I am subbing in the 10feet now is because it is the length of the ladder, which is a constant (the ladder isn't getting longer or shorter)

Answer 19

yes, it makes sense, the others vary over time

Answer 20

Okay so step
1. find all variables
2. find an equation that relates all the variables (phythagorus works here!)
3. find the derivative of that equation with respect to time

Answer 21

because we already know dx/dt... so if we take the derivative of respect to time we can just sub that in!

Answer 22

do I need to calculate the distance from the top of the ladder to the ground for each interval?

Answer 23

hmm I am not quite sure

Answer 24

nope they just want to know when the base of the ladder is 6 feet from the wall (so when x=6)

Answer 25

ok

Answer 26

do lets take the derivative of x^2+y^2=0 with respect to time

Answer 27

that should be a 10^2 at the end there

Answer 28

wait why 0?

Answer 29

ok lol

Answer 30

heh actually it wouldn't matter because the derivative is 0 anyways

Answer 31

yes, is it 2x/dt +2y/dt =0 ?

Answer 32

Just a bit more info about that 10 thing (the length of the ladder). If you forgot to sub in the 10 and just made it a variable (z) you would find the derivative to be (2z)(dz/dt)

Answer 33

hmm almost... is should be (2x)(dx/dt) + (2y)(dy/dt) = 0

Answer 34

ok

Answer 35

but yeah, what I was saying about the (2z)(dz/dt). If you end up doing this and start to panic because you don't know either, just think that the ladder (z) is not changing, so (dz/dt)=0 and so (2z)(dz/dt)=0

Answer 36

ok, Ill remember that thanks

Answer 37

I hope this isn't confusing. This really messed me up when I started doing these questions... I was like when do I plug in the values??? now or later???

Answer 38

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Answer 39

so did you see where you went wrong? when you take the derivative with respect to time of x^2 you must use the chain rule

(2x)(dx/dt)

Answer 40

hmm because it is x*x

Answer 41

2 chain rules and a sum rule

Answer 42

I guess you could look at it like that

Answer 43

I think of it as start with the outside (the 2 exponent) and move it in front of the x. Then you must take the derivative of the x with respect to t

thus 2x(dx/dt)

Answer 44

this is confusing at the start because up till you learn implicit differentiation you were taking the derivative with respect to x. so (d/dx) of (x^2) is 2x(dx/dx) and the dx cancels out leaving 2x

Answer 45

ah, the way I did it was: g=x g'=1 f=u^2 f'=2u

Answer 46

yes we need to keep in mind we are taking the derivative with respect to time

Answer 47

so do you understand how to get to 2x(dx/dt) + 2y(dy/dt) = 0 now?

Answer 48

then used the chain rule: 2(x)(1) and then the dx/dt confused me

Answer 49

let me look over what you said

Answer 50

why did you put that 1 there

Answer 51

with the 2(x)(1)

Answer 52

derivative of x is 1

Answer 53

g=x g'=1 f=u^2 f'=2u

Answer 54

the derivative of x with RESPECT TO x might be 1.... but we want the derivative of x with RESPECT TO t

Answer 55

ah ok

Answer 56

this factor still confuses me

Answer 57

up until now you have probably been asked to find dy/dx=blah blah blah right

Answer 58

more or less

Answer 59

I used to hate that notation and always used y' or whatever

Answer 60

but the dy/dx shows you what you are taking the derivative of with respect to

Answer 61

using the chain rule:
(d/dt) (x^2)
first move the 2 over to the front of the x

THEN

you must multiply that by the DERIVATIVE OF THE INSIDE FUNCTION WITH RESPECT TO TIME!

Answer 62

I dislike those words " with respect to" >_<

Answer 63

hehe it just means what is x doing when t is whatever

Answer 64

when you take dy/dx, you are finding out what y is doing when x is doing something else...

there is probably a better way to explain this

Answer 65

so the inside function is x

Answer 66

so x times d/dt is dx/dt

Answer 67

yes... and up till now you have been finding the derivative of x with respect to x (which means what is x doing when x is doing something... well they are the same variable! as x increase so does x so it would just be in a 1/1 ratio simplifying to 1)

Answer 68

ok

Answer 69

which explains why it's not always written

Answer 70

don't just think of it as times though... think of it as what is this variable (x) doing when t is doing something else

Answer 71

ok

Answer 72

yes because at the beginning they just want you to get familiar with all the rules

Answer 73

I guess I mean what you are doing arithmetically to get this result using the chain rule

Answer 74

What sometimes helps me is to try using these rules on simple problems that I already know I can solve other ways... it proves to me that these other weirder ways work

Answer 75

yep, just the chain rule

Answer 76

and when you get to a variable that you don't know the answer to, just put it with respect to the other variable and move on!

Answer 77

but this dx/dt is our g' in the chain rule then, right?

Answer 78

so back to x^2+y^2=10^2

Answer 79

what is the formula for chain rule again

Answer 80

just to clarify

Answer 81

ah yes d/dx(fog)(x)=d/dx f(g(x))=f'(g(x)) g'(x)

Answer 82

I would say that g'(x) = dx/dt in our situation

Answer 83

ok :) cool, that's how I visualized it, ok, I understand this part, thanks

Answer 84

good so we have

2x(dx/dt) + 2y(dy/dt) = 0

Answer 85

yes

Answer 86

we can sub in dx/dt, right?

Answer 87

lets just remember what each thing is

x is the length the ladder is away from the wall
dx/dt is the rate that the ladder is moving away from the wall

y is the height of the ladder from the ground
dy/dt is the rate that the ladder is moving down the wall