#1

A rowboat is pulled toward a dock from the bow through a ring on the dock 12 feet above the bow. If the rope is hauled at 10/13 ft/sec, fow fast is the boat approaching the dock when 13 feet of rope are out?

#2

A rowboat is pulled toward a dock from the bow through a ring on the dock 12 feet above the bow. If the rope is hauled at10/13ft/sec, fow fast is the boat approaching the dock when 13 feet of rope are out?

this spot confuses me. anyways you'll probably have to use pythagores theorem to find the relation of the values but you probably already knew that...

#3

(x^2) + (12^2) = (13^2)

she then said to get the derivative or something but didnt knowhow to set it up.

she then said to get the derivative or something but didnt knowhow to set it up.

#4

how fast is the rope being hauled? (10/13)ft/s?

#5

Who has a 12 foot high dock

#6

the answer is 1/0 oh shi-

#7

x = 12

y = 13

dx

-- = 10/13 ft/sec

dt

dy

-- = ?

dt

Equation

(x)^2 + (5)^2 = (y)^2

2x dx/dt = 2y dy/dt

2(12) (10/13) = 2(13)(dy/dt)

240/13 = 26(dy/dt)

240 = (338)(dy/dt)

120/169 = dy/dt

(120/169)ft/sec

i think this is right, i cant remember some of this stuff though

y = 13

dx

-- = 10/13 ft/sec

dt

dy

-- = ?

dt

Equation

(x)^2 + (5)^2 = (y)^2

2x dx/dt = 2y dy/dt

2(12) (10/13) = 2(13)(dy/dt)

240/13 = 26(dy/dt)

240 = (338)(dy/dt)

120/169 = dy/dt

(120/169)ft/sec

i think this is right, i cant remember some of this stuff though

*Last edited by kristjantomasso at May 27, 2008,*

#8

x = 12

y = 13

dx

-- = 10/13 ft/sec

dt

dy

-- = ?

dt

Equation

(x)^2 + (5)^2 = (y)^2

2x dx 2y dy

-- = --

dt dt

2(12) (10/13) = 2(13)(dy/dt)

240/13 = 26(dy/dt)

240 = (338)(dy/dt)

120/169 = dy/dt

(120/169)ft/sec

i think this is right, i cant remember some of this stuff though

This looks right, although it has been a while since I last did related rates.

By the way, isn't it a little late in the year to be doing related rates? I remember doing them early in my first semester of calculus.

#9

This looks right, although it has been a while since I last did related rates.

By the way, isn't it a little late in the year to be doing related rates? I remember doing them early in my first semester of calculus.

review...final next week

#10

I think that kristjantomasso is using the wrong variables. With the triangle, the 12 is constant through time, so it doesn't apply to the related rates.

If I'm right, the triangle should be - x=13 (the ropes length), z=12 (the dock's height), with "y" on the bottom (the distance of the boat from the dock).

Set the problem up as this:

Have: z=12

dx/dt = 10/13

Want: dy/dt @ x=13

Using the pythagorean theorem, @ x=13, y=5.

Now, the equation is, leaving x and y as variables, and entering "z" as 12, since it is constant - y^2 + z^2 = x^2 ; this becomes y^2 + 144 = x^2. Now you take the derivative using implicit differentiation, giving you

2y(dy/dt) = 2x (dx/dt).

Now place in what you know - 2(5)(dy/dt) = 2(13)(10/13). Now just do the math -

10(dy/dt) = 20

Which gives you dy/dt = 2fps.

I think that is right, and explains it somewhat.

If I'm right, the triangle should be - x=13 (the ropes length), z=12 (the dock's height), with "y" on the bottom (the distance of the boat from the dock).

Set the problem up as this:

Have: z=12

dx/dt = 10/13

Want: dy/dt @ x=13

Using the pythagorean theorem, @ x=13, y=5.

Now, the equation is, leaving x and y as variables, and entering "z" as 12, since it is constant - y^2 + z^2 = x^2 ; this becomes y^2 + 144 = x^2. Now you take the derivative using implicit differentiation, giving you

2y(dy/dt) = 2x (dx/dt).

Now place in what you know - 2(5)(dy/dt) = 2(13)(10/13). Now just do the math -

10(dy/dt) = 20

Which gives you dy/dt = 2fps.

I think that is right, and explains it somewhat.

#11

I think that kristjantomasso is using the wrong variables. With the triangle, the 12 is constant through time, so it doesn't apply to the related rates.

If I'm right, the triangle should be - x=13 (the ropes length), z=12 (the dock's height), with "y" on the bottom (the distance of the boat from the dock).

Set the problem up as this:

Have: z=12

dx/dt = 10/13

Want: dy/dt @ x=13

Using the pythagorean theorem, @ x=13, y=5.

Now, the equation is, leaving x and y as variables, and entering "z" as 12, since it is constant - y^2 + z^2 = x^2 ; this becomes y^2 + 144 = x^2. Now you take the derivative using implicit differentiation, giving you

2y(dy/dt) = 2x (dx/dt).

Now place in what you know - 2(5)(dy/dt) = 2(13)(10/13). Now just do the math -

10(dy/dt) = 20

Which gives you dy/dt = 2fps.

I think that is right, and explains it somewhat.

this is the right way, i screwed my answer up

#12

Obviously, it is moving at 10/13 ft/s, whatever the hell that is. Duh.

-SD

-SD

#13

I don't see how this could be relevant to real life.

Danny: Eh! Tony how fast do you think this here boat is moving right now that 13 of stupid rope has already been uhh pulled.

Tony: Gee Danny, I really cant tell let me use my Calculas degree from Jersey Tech: Home of the MEATBALLS. w00t GO GREASERS.

Danny: GREASERS!

Besides if two dock workers were really bored.

Danny: Eh! Tony how fast do you think this here boat is moving right now that 13 of stupid rope has already been uhh pulled.

Tony: Gee Danny, I really cant tell let me use my Calculas degree from Jersey Tech: Home of the MEATBALLS. w00t GO GREASERS.

Danny: GREASERS!

Besides if two dock workers were really bored.

#14

I don't see how this could be relevant to real life.

Danny: Eh! Tony how fast do you think this here boat is moving right now that 13 of stupid rope has already been uhh pulled.

Tony: Gee Danny, I really cant tell let me use my Calculas degree from Jersey Tech: Home of the MEATBALLS. w00t GO GREASERS.

Danny: GREASERS!

Besides if two dock workers were really bored.

haha i asked my calculus teacher the same thing, he tried to think of something but came up with nothing

#15

thats what im saying but its just one problem out of 30 that equal to like 2 test grades so i dont really mind boosting up my grade...

BTW thank you very much. i was never good at related rates.

BTW thank you very much. i was never good at related rates.