Calulus Help!!!

[AACG]

Hello everyone,

I was wondering if you could help me out by helping me get started on this problem I have in calculus. The question is for related rates. Here is the question.

A 13 foot ladder is leaning against the side of a house. The foot of the ladder is pulled away from the house at a rate of 2 feet per second. How fast is the top of the ladder descending down the side of the house when the foot of the ladder is 5 feet from the house?

Any help would be greatly appreciated.

Thanks AACG

 

[Kyteland]

42

Oh, and do your own homework!

 

[Juno]

Oh boy, I hate all math courses even I passed both Calculus with a grade of C-.

 

[royaldank]

Did you have folks give you the answers for English and Spelling classes as well?

 

[AACG]

The only reason I asked the question is because I was not in class for the lesson we went over due to my grandfather passing away. I would just like to know how to start the problem that is all. I do not even want the answer.

 

[Kyteland]

OK, I'm a sucker for sob stories, so take a look at this link. It sets up the problem.

mathhelp.jpg

Now that you have an equation for ft, you need to do your integration magic to get an equation for ft/s

 

[Legendary]

c = 15
a^2+b^2=c^2
da/dt = 2 ft/sec
dc/dt = 0 (ladder doesn't change length)
db/dt = what you're looking for given a = 5

Now think REALLY REALLY REALLY hard.

 

[AACG]

Thank you Kyteland and Legendary
for you help that is all I wanted.

 

[Amused]

Calulus???

 

[Electric Amish]

Do you know how many years it's beed since I studied or used and calulus???

 

[Amused]

Originally posted by: Electric Amish
Do you know how many years it's beed since I studied or used and calulus???

Nope. How many?

 

[JulesMaximus]

Originally posted by: Electric Amish
Do you know how many years it's beed since I studied or used and calulus???

Same here. I'm an accountant and I wouldn't know how to begin trying to solve that question. Nor do I care to. :roll:

 

[Fingolfin269]

Originally posted by: JulesMaximus

Originally posted by: Electric Amish
Do you know how many years it's beed since I studied or used and calulus???

Same here. I'm an accountant and I wouldn't know how to begin trying to solve that question. Nor do I care to. :roll:

Thank God. I was starting to wonder if I was the only one who had forgotten most, if not all, of my calculus knowledge. I'd say it's still up there somewhere but I'd really rather not bring it back.

 

[Electric Amish]

Originally posted by: Amused

Originally posted by: Electric Amish
Do you know how many years it's beed since I studied or used and calulus???

Nope. How many?

Over 10 years.

 

[TuxDave]

Sorry, never studied calulus

 

[Orsorum]

Damn, I have a minor in mathematics and I can't remember how to do that. Haven't done any calc in two years.

 

[TTM77]

Originally posted by: Legendary
c = 15
a^2+b^2=c^2
da/dt = 2 ft/sec
dc/dt = 0 (ladder doesn't change length)
db/dt = what you're looking for given a = 5

Now think REALLY REALLY REALLY hard.

Holy-calculus.

Another way I see this is the sin wave. The top slowly falling while the bottom is pulling away.

I think Legendary is right. It's derivative.

 

[TuxDave]

Originally posted by: TTM77

Originally posted by: Legendary
c = 15
a^2+b^2=c^2
da/dt = 2 ft/sec
dc/dt = 0 (ladder doesn't change length)
db/dt = what you're looking for given a = 5

Now think REALLY REALLY REALLY hard.

Holy-calculus.

Another way I see this is the sin wave. The top slowly falling while the bottom is pulling away.

I think Legendary is right. It's derivative.

Related rate problems are usually done using derivatives. I don't see any simple sine wave solution.

 

[civad]

In case your calculus problem is solved, you can start working on your spelling problem...

 

[ColdFusion718]

Originally posted by: Legendary
c = 15
a^2+b^2=c^2
da/dt = 2 ft/sec
dc/dt = 0 (ladder doesn't change length)
db/dt = what you're looking for given a = 5

Now think REALLY REALLY REALLY hard.

Where did you get 15 from? The length of the ladder is 13 feet, genius. And why are you using a, b, and c when using X and Y would be more intuitive since they basically tell you what the coordinate system is?

Since it the length does not change, you do not need to treat it as a variable and add confusion with its rate of change (dc/dt). Don't post partial solutions that aren't even correct and try to sound

REALLY REALLY REALLY hard

smart.

Solution:

Let X = distance ladder is away from the wall
Let Y = veritical distance ladder is above the ground

X^2 + Y^2 = 13^2

Now, differentiate both sides:

2X* dX/dt + 2Y*dY/dt = 0

Go back to your triangle to when the ladder is 5 feet above the ground, figure out how far away it is from the wall.

You have:

X^2 + 5^2 = 13^2
X = 12 feet from the wall
Y = 5 feet from the ground when dX/dt = 2 ft/s

Now you have all the necessary information

Plug and chug to get:

2X*dX/dt + 2Y*dY= 0 ---> 2(12)2 + 2(5)dY/dt = 0

10*dY/dt = -48
dY/dt = -4.8 ft/s or 4.8 ft/s downwards

Btw, did you have to think

REALLY REALLY REALLY hard

to solve that problem?

Just in case if everyone is wondering why I'm tripping over this: when you give someone academic help, especially in Calculus, make sure you know what it is you are talking about--if you fvck up, you fvck up the person you are trying to help. Maybe you don't know how hard it is to un-fvck someone's brain after they learn something the wrong way in a subject as difficult as Caculus, but I do.

 

[TuxDave]

Wow... someone pissed in your pants... he posted a 'partial solution' because he believed that the OP should learn how to figure out the rest. I'm pretty sure that Legendary made a simple typo for the c=15... but he posted enough information to get started on the problem.

 

[ColdFusion718]

Originally posted by: TuxDave
Wow... someone pissed in your pants... he posted a 'partial solution' because he believed that the OP should learn how to figure out the rest. I'm pretty sure that Legendary made a simple typo for the c=15... but he posted enough information to get started on the problem.

First off, if you are going to flame me, do it right. It's "Wow... someone pissed in their pants."

 

[AACG]

Originally posted by: civad
In case your calculus problem is solved, you can start working on your spelling problem...

I have a spelling problem where? Also thank you ColdFusion718

 

[DrPizza]

coldfusion - I agree completely with Legendary. I don't know what your problem with his solution is... Furthermore, the OP had already thanked him for the help.
Note the word "HELP." The OP didn't ask someone to do the problem for him, just to help him with it. The relationship necessary to solve the problem is, of course, the pythagorean theorem, which 99% of the time is written in terms of a,b, and c. The only real difference between your solution and his is the substitution of x and y for a and b. Forcing all problems to be in terms of x & y is laziness and reduces a students ability to deal with relationships with other variables. I do not, though, disagree with kyteland's picture that he provided - a picture is worth 1000 words, and lacking a picture, a^2 + b^2 = c^2, take the derivative with respect to t and substitute known values is probably the briefest way to explain the problem. There's absolutely nothing wrong with his inclusion that dc/dt = 0, in fact, IMHO (the opinion of a calculus teacher), dealing with three "variables" now with this problem is probably a better stepping stone to other typical related rates problems, especially those with more than 2 variables, ex. sand falling into a conical pile where V = 1/3*pi*r^2*h. Would you advocate switching the r to x and h to y?? I certainly hope not.

 

[OulOat]

Originally posted by: ColdFusion718

Originally posted by: Legendary
c = 15
a^2+b^2=c^2
da/dt = 2 ft/sec
dc/dt = 0 (ladder doesn't change length)
db/dt = what you're looking for given a = 5

Now think REALLY REALLY REALLY hard.

Where did you get 15 from? The length of the ladder is 13 feet, genius. And why are you using a, b, and c when using X and Y would be more intuitive since they basically tell you what the coordinate system is?

Since it the length does not change, you do not need to treat it as a variable and add confusion with its rate of change (dc/dt). Don't post partial solutions that aren't even correct and try to sound

REALLY REALLY REALLY hard

smart.

Solution:

Let X = distance ladder is away from the wall
Let Y = veritical distance ladder is above the ground

X^2 + Y^2 = 13^2

Now, differentiate both sides:

2X* dX/dt + 2Y*dY/dt = 0

Go back to your triangle to when the ladder is 5 feet above the ground, figure out how far away it is from the wall.

You have:

X^2 + 5^2 = 13^2
X = 12 feet from the wall
Y = 5 feet from the ground when dX/dt = 2 ft/s

Now you have all the necessary information

Plug and chug to get:

2X*dX/dt + 2Y*dY= 0 ---> 2(12)2 + 2(5)dY/dt = 0

10*dY/dt = -48
dY/dt = -4.8 ft/s or 4.8 ft/s downwards

Btw, did you have to think

REALLY REALLY REALLY hard

to solve that problem?

Just in case if everyone is wondering why I'm tripping over this: when you give someone academic help, especially in Calculus, make sure you know what it is you are talking about--if you fvck up, you fvck up the person you are trying to help. Maybe you don't know how hard it is to un-fvck someone's brain after they learn something the wrong way in a subject as difficult as Caculus, but I do.

For someone so bitchy about correctness, your solution is completely wrong X=12 and dX/dt = 2, then t = 6. If t =6 and dY/dt = -4.8, then the ladder ends up underground Also, care to explain how you got 2X*dX/dt and 2Y*dY/dt? Do you remember your partial derivatives?