AP Calculus Question of the Day

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Discus: SAT/ACT Tests and Test Preparation: May 2003 Archive: AP Calculus Question of the Day
By Heatwave345 (Heatwave345) on Saturday, April 26, 2003 - 12:19 pm: Edit

Hopefully this thread will help all of us out on the AP Calculus Exams.

A 20 ladder slides down a wall at 5 ft/sec. At what spped is the bottom sliding out when the top is 10 feet from the floor?

By Dwayne_Hoover (Dwayne_Hoover) on Saturday, April 26, 2003 - 12:31 pm: Edit

This is a related rates question, just set up an equation and differentiate it,



2(10*root3)ft*-5ft/s + 2(10)*(dy/dt)=0

-100root3ft^2/sec + 20ft*dy/dt=0



dy/dt= 5*root3 ft/second

By Quarky (Quarky) on Saturday, April 26, 2003 - 02:00 pm: Edit

I disagree -- we need to find dx/dt, not dy/dt

x^2 + y^2 = 400
differentiate with respect to t, as dwayne did:
2x(dx/dt) + 2y(dy/dt) = 0
2*sqrt(300)*(dx/dt) + 2*10*(-5) = 0
dx/dt = (2*10*5)/(2sqrt(300)) = 5/3*sqrt(3) = 2.887 ft/sec

By Dwayne_Hoover (Dwayne_Hoover) on Saturday, April 26, 2003 - 02:16 pm: Edit

He is saying how fast is it sliding OUT not DOWN....i used the DY/DT because i set my Y to 10...

By Dwayne_Hoover (Dwayne_Hoover) on Saturday, April 26, 2003 - 02:17 pm: Edit

You are pluggin in the wrong numbers, -5ft/second is the dx/dt of the ladder going DOWN the wall, as the kid stated...we're trying to find out how far the ladder is going OUT...we do not know that, but we do have 10...

I've noticed you on other forums, you think u'r pretty smart huh? hehe

By Parhelia (Parhelia) on Saturday, April 26, 2003 - 02:21 pm: Edit

I agree with Quarky - dx/dt = 5/sqrt3 = 2.887 ft/s.

x^2 + y^2 = 400

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

2(sqrt300)(dx/dt) + 2(10)(-5) = 0

2(sqrt300)(dx/dt) = 100

dx/dt = 100/2sqrt300 = 50/sqrt300 = 5/sqrt3

Forgive me if I made any mistakes.

By Quarky (Quarky) on Saturday, April 26, 2003 - 03:10 pm: Edit

Yeah, I think I'm pretty smart. What about you?

OK, it is clearly stated that the TOP is 10 ft away from the floor. If you set your y to 10 ft, then Y must be the VERTICAL distance, so we already know that the vertical distance is changing at -5 ft/s, and we need to find the HORIZONTAL distance change.

By Dwayne_Hoover (Dwayne_Hoover) on Saturday, April 26, 2003 - 04:11 pm: Edit

Whoops, my bad--i'm slightly dislexic,,so when i see top i see phonetically hear pot--or bottom..sorry

By Heatwave345 (Heatwave345) on Saturday, April 26, 2003 - 05:01 pm: Edit

and the winner is.....
Quarky and you win this smilie face ---->

The next Question:::

limit of 4*x*sin(1/x) as x approches infinity

Do it the old fashioned way. (no calculators)

By Brd (Brd) on Saturday, April 26, 2003 - 05:27 pm: Edit

If you really want to do it the old fashioned way, don't use L'Hopitals rule either -- let's see those epsilons and deltas!

Probably the quickest way to figure this out though (even quicker than L'Hopital's rule), is to recall the first few terms of the power series for sin(1/x) and them multiply that by 4x, which makes it immediately obvious that the limit must be 4 plus a bunch of other terms that all go to zero..

By Dwayne_Hoover (Dwayne_Hoover) on Saturday, April 26, 2003 - 05:52 pm: Edit

that limit is 0.

By Brd (Brd) on Saturday, April 26, 2003 - 05:53 pm: Edit

No, it isn't.

sin(1/x) goes to zero, but x blows up, as x increases without bound. So the question becomes, which grows or decays faster asymptotically -- which term dominates the product in the limit -- or, do they grow and decay at rates that are some fixed multiple of one another? As it turns out, they grow and decay at exactly the same rate. You could also get the idea that this is the case by remembering that sin(y) ~= y for y close to zero, so that sin(1/x) is approximately 1/x when x is large. Then multiplying by the rest of the expression leaves 4.

Or you could use L'Hopitals rule with 4 * 1/(1/x) * sin(1/x).

Or you could prove it with an eps-delta argument.

By Dwayne_Hoover (Dwayne_Hoover) on Saturday, April 26, 2003 - 06:37 pm: Edit

listen dude, rewrite it as,



so cos1/x is getting close to 1, -x*-2 is getting close to 0 and -x^-2 is getting realy close to, so 1/-puny = -infinity.
sorry i misread it at first glance, but by using l'hopitals u get -infinity, not four--please correct me if im wrong..
its real easy to make mistakes when ur just typing on a computer.

By Brd (Brd) on Saturday, April 26, 2003 - 06:52 pm: Edit

The (-x^(-2))/(-x^(-2)) terms cancel (equal one), leaving just cos(1/x), which has limit 1 as x grows without bound.

The limit is 4.

By Dwayne_Hoover (Dwayne_Hoover) on Saturday, April 26, 2003 - 07:49 pm: Edit

Ah--cool thanks, See if i had written this down i would have been able to easily see that they cancel :-) -- thanks!

By Heatwave345 (Heatwave345) on Saturday, April 26, 2003 - 10:26 pm: Edit

Brd is correct, the answer is 4

By Dwayne_Hoover (Dwayne_Hoover) on Saturday, April 26, 2003 - 10:39 pm: Edit

yah i know-- i did it out on paper and gor 4 aas well.

By Heatwave345 (Heatwave345) on Sunday, April 27, 2003 - 03:36 pm: Edit

Time for another one:

Let f and g be functions that are differentiable throughtout their domains and that have the following properties:

(i) f(x+y)= f(x)g(y)+g(x)f(y)
(ii) limit of f(a) as a-->0 is 0
(iii) limit of (g(h)-1)/h as h--> is 0
(iv) f'(0)=1

(a) show that limit of f(a)/a as a-->0 is 1

(b) show that f'(x)=g(x)

(c) Find intergral[g(x)/f(x)dx]

By Brd (Brd) on Monday, April 28, 2003 - 10:30 am: Edit

Well, if no one else is going to bite...

(a) Just using ii), we get a limit of 0/0 as a-->0, so apply L'Hopital's rule using iv) and see that the limit is f'(0)/1 = 1/1 = 1. (*)

(b) Take the total derivative of both sides of (i): (dx + dy)f'(x+y) = dx[f'(x)g(y) + f(y)g'(x)] + dy[f'(y)g(x) + f(x)g'(y)] Looking at just the dy terms we find that f'(x+y)=f'(y)g(x) + f(x)g'(y) for all x and y. In particular, it is true when y=0, so that f'(x) = f'(0)g(x) + f(x)g'(0) = g(x) + f(x)g'(0) (since f'(0)=1 by (iv)). Now, we can pull the same y=0 trick with (i) directly to conclude that g(0)=1. Finally, look at (iii). Using g(0)=1 in (iii) really says that g'(0) = 0. So plug this into the result we obtained above and we are left with f'(x) = g(x) + 0, the desired result.

Looking at this argument, I really see the physicist in me sticking out. Maybe someone can do better.

(c) Having established part (b), we see that Int[g(x)/f(x) dx] = Int[f'(x)/f(x) dx] = Log|f(x)| + C.

(*) I'm not sure iv) is enough without continuity of the derivative also -- to use L'Hopital we really need that lim f'(x) = 1 as x-->0, not just f'(0) = 1.

So, I can think of pretty good candidates for what f and g have to be. Any guesses? :)

By Heatwave345 (Heatwave345) on Monday, April 28, 2003 - 07:36 pm: Edit

nice job Brd!!! You probably won't have any problems gettinf a five on the test.

Here's another one:

Let y=f(x) be the solution to the differential equation dy/dx=arcsin(xy) with the intial condition f(0)=2. Find an approximation for f(1).


BTW Feel free to test me with any questions. i want some practice too!!!

By Heatwave345 (Heatwave345) on Monday, April 28, 2003 - 07:39 pm: Edit


By Brd (Brd) on Monday, April 28, 2003 - 08:11 pm: Edit

Heatwave: Maybe so, but I don't have to take the test, I already have a math degree. :) I just like helping out if no-one chimes in.

Ironically, when I started school, I was a music major -- I clep tested out of college algebra so I would never have to take a math class again. Just goes to show you never know where you will end up...

As for f(x) and g(x) those seem to work fine, and, even better, they are not the ones I had in mind!

By Heatwave345 (Heatwave345) on Tuesday, April 29, 2003 - 07:37 pm: Edit


can anyone answer the arcsin question???

By Stupid_Guy (Stupid_Guy) on Tuesday, April 29, 2003 - 08:16 pm: Edit

i end up with 2...

By Brd (Brd) on Tuesday, April 29, 2003 - 10:28 pm: Edit

Heatwave the only thing I can think of is to use the small angle sin approximation. If sin(x) is linear for small arguments, then so is it's inverse. So, you could look at the approximation dy/dx = xy, which you might happen to reason the solution to: y(x)= 2 e^(x^2/2) (with the initial condition y(0)=2). Evaluated at x=2 you get y(2) = 2 e^2.

I really don't think this is what the problem has in mind (plus I think the approximation I made sucks). When I play with this in Mathematica I quickly get complex solultions (not too surpising since ArcSin is complex-valued for arguments greater than 1) Are you sure you copied it down correctly?

BTW, for the other problem I had in mind f(x) = sin(x) and g(x) = cos(x).

By Heatwave345 (Heatwave345) on Wednesday, April 30, 2003 - 09:56 am: Edit

*cough* Euler *cough*

By Brd (Brd) on Wednesday, April 30, 2003 - 11:04 am: Edit

I didn't like the idea of using Euler with a stepsize as large as 1, and I wouldn't have thought to use Euler with a smaller stepsize without a calculator. :)

By Stupid_Guy (Stupid_Guy) on Wednesday, April 30, 2003 - 02:57 pm: Edit

look, its 2...no matter how small the step is...its gonna be 2. or maybe not

By Heatwave345 (Heatwave345) on Thursday, May 01, 2003 - 08:01 pm: Edit

Brd!!! i need your help

what is

summation((-1)^n/n) from n=(1 to infinity)?

i know it's 1/2, but how???

By Brd (Brd) on Thursday, May 01, 2003 - 09:13 pm: Edit

The series certainly converges, by the alternating series test, but the sum can't be 1/2 -- write it out as:

S = -1 + (1/2 - 1/3) + (1/4 - 1/5) + ...

= -1 + 1/(2*3) + 1/(4*5) + 1/(6*7) + ...

so S = -1 + Sum[ 1/(2n*(2n+1)), n = 1 to infinity]

but look at the new sum on the right,

Sum[ 1/(2n*(2n+1))] = Sum[ 1/(4n^2 + 2n) ]

if we leave of the +2n in the denominator, all we do is make the sum larger, i.e.

Sum[ 1/(2n*(2n+1))] <= (1/4) * Sum[1/n^2] = (1/4)* (Pi^2/6) (Euler proved this 1/n^2 sum)

But Pi^2/24 <= 4^2/24 = 16/24 = 2/3 < 1

So, since S = -1 + (something less than 1), the sum has to be negative, which rules out 1/2.

For what it's worth, Mathematica spits out S= -Log[2] = Log[1/2] = -0.693147 for the sum. How you would prove that, I don't know off the top of my head.

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