Calculus Help (and by help I mean homework)

Posted on 16th Feb 2014 by admin

Hrmmm, I hate posting about math homework, partly because it's homework, and partly because I hate whenever I can't figure out math myself.


Yes, I realize this is an extremely long post. Feel free to just tell me how to do it instead of reading my rambling for 3 pages about what I've tried.


Anyway, in class we've been doing stuff with related rates... In most cases it's fairly simple, but there are a couple of things that have me stumped. (By a couple I mean two.)

The first one:

"An air traffic controller spots two planes at the same altitude converging on a point as they fly at right angles to each other. One plane is 150 miles from the point moving at 450 miles per hour, and the other plane is 200 miles from the point moving at 600 miles per hour. At what rate is the distance between the planes decreasing?" (See attachment for a drawing.)

Proof that I have tried:

So, I'll call the distance between the plane 150 miles away a, and the other distance b.

The position of the plane 1 (in other words a):
a = 150 - 450t
For plane 2: b = 200 - 600t

(One of the other questions was how long until they collide. That was easy since it was just 200-600h = 150-450h then h = 1/3 [or you could set one of them equal to 0 since the collision is when the distance from the point is 0, but they would have to be equal to prove that anyway].)

So, the distance (I'll call that c) between the planes should be c = a^2 + b^2. Simple enough... So the change in distance with respect to time should be: (Edit: I must be tired... Just realized that I got the freakin' distance formula wrong earlier... Wonder if it would've worked out if I hadn't.)

(d/dt)(c) = (d/dt)(a^2) + (d/dt)(b^2)
dc/dt) = (d/dt)(a^2 + b^2)
dc/dt = (d/dt)(150^2 - 300(450t) + 450^2t^2 + 200^2 - 400(600t) + 600^2t^2)
Then distribute the d/dt:
dc/dt = 0 - (300)(450)(dt/dt) + 2(450^2)(t)(dt/dt) + 0 - 400(600)(dt/dt) + 2(600^2)(t)(dt/dt)
dc/dt = 135000 + 405000t - 240000 + 720000t
dc/dt = 1125000t - 105000


The reason I know that is wrong is because the book says -750 mph and that is definitely not equal... But, I also know that the units won't be correct. I should end up with miles/hours. It has hours, but it's just miles, not miles/hours.... So at some point I managed to take out a <something>/hours that I should not have. Which makes me wonder what information I was given or can figure out (and it be a constant) that would relate <something> to hours with hours dividing it....


I guess I know that the derivative of a is 450 miles/hour, and I know that b' is 600 miles/hour. Oh wow! I think I just figured it out! No, just worked it out on paper, and I get some weird answer.

Well, that means that going back to c = sqrt(a^2 + b^2) when taking the derivative of a and b, those can be substituted (I think?). But that still leaves a problem, since it comes up with:

c = .5(a^2 + b^2)^-.5(2a + 2b)

That doesn't make sense though. I don't think the chain rule would be correct there.


Anyway, I keep running in circles with that problem.

Edit:

Wow... Apparently I should've googled. http://www.cliffsnotes.com/WileyCDA/CliffsReviewTopic/Related-Rates-of-Change.topicArticleId-39909,articleId-39897.html deals with a different problem, but it did make me realize that I was going about it wrong. Apparently I need to take the derivative and then substitute (which makes more sense actually). (Also, I got the distance forumla wrong earlier, so not sure if it would have worked out had I not. Guessing it would've still been wrong because I didn't need to find the position dynamically since the derivative of the distance would be the same at any two positions.)

Answer:

c^2 = a^2 + b^c

2c(dc/dt) = 2a(da/dt) + 2b(db/dt)

Then just plug in c (250), a (150), b (200) and da/dt (450) and db/dt (600) and solve for dc/dt.... And 750


The other problem (well type of problem) that confuses me is as follows:


"At a sand and gravel plant, sand is falling off of a conveyor and onto a conical pile at a rate of 10 cubic feet per minute. The diameter of the base of the cone is approximately 3 times the altitude. At what rate is the height of the pile changing when the pile is 15 feet high?"

What I know:

V = (3pi/4) * hr^2
dV/dt = 10 ft^3/minute
d = 3h
r = 3h/2

What I'm trying to find:
dh/dt when h = 15


So, what I've gotten so far:

V = (3pi/4) * hr^2
dV/dt = (6/4)pi * r (dr/dt)(dh/dt)


That's all fine and good, but I don't know dr/dt and have no way to find it, so I must go back and substitute something.

I've noticed (and the teacher probably told us at some point) that usually it's a good idea to replace in something that gives you the variable that you're solving for the derivative of (or maybe that's just because usually the value of that variable is given hence a constant).

V = (3pi/4) * hr^2
V = (3pi/4) * h(3h/2)^2
V = (3pi/4) * h(9h^2/4)
V = (3pi/4) * (9h^3/4)
V = (27pi/16) * (h^3)

dV/dt = ((3*27pi)/16) * h^2 * dh/dt
10 ft^3/minute = ((3*27pi)/16) * h^2 * dh/dt

Then the craziness:

dh/dt = 10/(((3*27pi)/16) * h^2)
dh/dt = 10/(((3*27pi)/16) * (15)^2)


I thought that maybe I shouldn't use the chain rule when derivating (don't think that's a word) h^3, but using the power rule ends up in a wrong answer as well.

Also, I thought maybe I'm replacing r too early, but I'm not....




If someone will explain to me where I'm going wrong (I've even tried things I know won't work), that would be wonderful! Also,


tl;dr: Find the problems and explain to me how to do them please .

Basically I think what my problem is is determining what is constant and what is not and at what stage in a problem to replace something.