Pre-Calc 1 exam question - stumped past 5 days on it :( help

[Queenisabella87]

The walls of an A-frame cottage make a 60'(degree) angle with the ground. A 12 foot ladder leans against the wall with the bottom of the ladder touching the wall.

If the bottom of the ladder is pulled away from the cottage at a rate 1 ft / second.


How was is the top sliding down the wall when the ladder is starting to move.

:shock: Thank you again so very much, this is one of the 2 questions that stumped me for almost a week.

I CANT THANK YOU ENOUGH

~Isabella

 

[tkhunny]

Re: Pre-Calc 1 exam question - stumped past 5 days on it

#1 Which angle are we talking about? With your A-Frame house, I suppose it's rather obvious, but please try to be clear in the definition and fail to require your reader to guess what you mean.

As you move the ladder, you may wish to consder the obtuse triangle:

A = Base of ladder.
B = Base of house where it touches the ground.
C = Top of ladder, where it touches the house.

The Law of cosines suggests: \(\displaystyle AC^{2} = AB^{2} + BC^{2} - 2(AB)(BC)\cdot\cos(\angle ABC)\)

Since we know \(\displaystyle \angle ABC = 120^{\circ}\), we have \(\displaystyle AC^{2} = AB^{2} + BC^{2} - (AB)(BC)\)

Since we know \(\displaystyle AC = 12\;ft\), we have \(\displaystyle 144\; ft^{2} = AB^{2} + BC^{2} - (AB)(BC)\)

We assume now that BC and AB are separate functions of time (BC = f(t) and AB = g(t)) and employ the implicit derivative.

\(\displaystyle 0 = 2\cdot AB \frac{dAB}{dt} + 2\cdot BC \frac{dBC}{dt} - AB\cdot\frac{dBC}{dt} + BC\cdot \frac{dAB}{dt}\)

At the start of the process, we know that AC = BC and AB = 0 ft.

It's only algebra at this point. Just crawl through the whole thing one step at a time.

 

[Queenisabella87]

Re: Pre-Calc 1 exam question - stumped past 5 days on it

thank you so much Tkhunny

I'll do it now and see where it gets me.

 

[Queenisabella87]

Re: Pre-Calc 1 exam question - stumped past 5 days on it

Hey TK, how do I find out AC ?

 

[Mrspi]

Re: Pre-Calc 1 exam question - stumped past 5 days on it

Hey TK, how do I find out AC ?

I'm not TK and I'm sure he's happy about that! BUT.....if you have READ TK's response, you should see that A is the foot of the ladder and C is the top of the ladder. So AC is the length of the ladder...and that is GIVEN.

 

[Queenisabella87]

Re: Pre-Calc 1 exam question - stumped past 5 days on it

Thank you oh Wise one.

so AC = A x C right? just makin sure haha. My brains almost mushy!

 

[Mrspi]

Re: Pre-Calc 1 exam question - stumped past 5 days on it

Thank you oh Wise one.

so AC = A x C right? just makin sure haha. My brains almost mushy!

Well, no.....

AC definitely does NOT mean A*C.

The geometric interpretation of AC is "the distance between points whose names are 'A' and 'C'"

 

[Queenisabella87]

Re: Pre-Calc 1 exam question - stumped past 5 days on it

No I know im sorry, lol my heads somewhere else.


SO you see where he gave me that last equation where 0 = ...

Knowing that AC = BC and AB = 1ft at the first movements


Where do I go about solving for the speed from there? Thats where im stuckk

 

[Deleted member 4993]

Re: Pre-Calc 1 exam question - stumped past 5 days on it

Please re-read TK's response.

Please understand the terms he has used - very carefully.

AB is not equal to 1ft. It is the rate of change of AB that 1 ft/sec.

You should see all the terms that you need to derive.