Calculus Help (and by help I mean homework)

[corbin]

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)© = (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. 

[448191]

Isn't there some sort of rule against helping with homework?

[PugJr]

Isn't there some sort of rule against helping with homework?

 

Only if caught.  And honestly, what type of stalker/person who makes sure he does his homework, would check an internet forum? And that being a php forum. I think its pretty safe.

[Daniel0]

Right, so we know that the volume of a cone is given by [imath]V=\frac{1}{3}\pi r^2 h[/imath] where [imath]h[/imath] is the height. You were right about substituting in something (though you did it incorrectly). Seeing as [imath]r=\frac{3h}{2}[/imath] for this particular cone, we can substitute that in to only have one variable:

 

[math]

\begin{split}

V&=\frac{1}{3}\pi r^2 h \\

&=\frac{1}{3}\pi \left( \frac{3h}{2} \right)^2 h \\

&=\frac{1}{3}\pi \frac{9h^2}{4} h \\

&=\frac{1}{3} \cdot \frac{9h^3\cdot\pi}{4} \\

&=\frac{3\pi}{4} h^3

\end{split}

[/math]

 

If we solve for [imath]h[/imath] then we get:

 

[math]\begin{split}

V=\frac{3\pi}{4} h^3 \Rightarrow \frac{4V}{3\pi} = h^3 \Rightarrow h = \sqrt[3]{\frac{4V}{3\pi}}

\end{split}[/math]

 

As you also said, we know that [imath]\frac{dV}{dt} = 10t[/imath] where [imath]t[/imath] is the time. Using the above we can also find [imath]\frac{dh}{dV}[/imath], but what we're looking for is [imath]\frac{dh}{dt}[/imath].

 

That is where the chain rule comes in, because [imath]\frac{dh}{dt} = \frac{dh}{dV} \cdot \frac{dV}{dt}[/imath].

 

Your turn

 

Isn't there some sort of rule against helping with homework?

 

I don't mind that as long as the poster makes it clear that it's homework.

[corbin]

Hrmmm, how would I solve that last equation?  Or is that not the one to be solved?  IT doesn't seem like there's a way to plug stuff in without just getting dh/dt = dh/dt.

 

 

(Also, I see where I should have been plugging in for r earlier instead of h.)

 

As for the step before that:

 

h = (4V/3pi)^(1/3)

 

dh/dt = (1/3)((4/3pi)V)^(-2/3)(4/3pi)(dV/dt)

 

dh/dt = ((4/3pi)V)^(-2/3)(4/9pi)(10)

 

But then I don't know V?

 

 

 

Bleh, got to head to school now, but I do plan on actually figuring the rest of this out later (I think later I'll try finding the derivative at the V = (3pi/4)h^3 step).  (In other words, ignore this post until later ;p.)

 

 

 

Isn't there some sort of rule against helping with homework?

 

 

I actually have never checked the rules in regards to homework x.x.

 

But I lucked out:

 

Users will not post their homework questions expecting to get their homework coded for them. If you have a question about part of the assignment that you do not understand, please post asking for an explanation rather than the code.

 

 

Do not post your homework to be done.

 

[Daniel0]

Hmm... I'm not actually sure that [imath]\frac{dV}{dt}=10t[/imath]. The assignment says "sand is falling off of a conveyor and onto a conical pile at a rate of 10 cubic feet per minute". I take this to mean "each minute, 10 ft³ are added to the file, so this means that [imath]V_1(t)=10t[/imath]. We'll say that [imath]V_2(h) = \frac{3\pi}{4} \cdot h^3[/imath], so we know that when the file is 15 ft high, then the volume must be [imath]V_2(15) = \frac{10125\cdot\pi}{4} \approx 7952.16[/imath]. Thus [math]V_1(t) = \frac{10125\cdot\pi}{4} \Rightarrow t = \frac{2025\cdot\pi}{8} \approx 795.216[/math]

 

This means that the pile must be 15 ft high after [imath]t = \frac{2025\cdot\pi}{8}[/imath] minutes.

 

If we have [imath]h=\sqrt[3]{\frac{4V}{3\pi}}[/imath] then we can just plug in our [imath]V_1[/imath] function to express the height by the time.

 

[math]h(t) = \sqrt[3]{\frac{4\cdot V_1(t)}{3\pi}} = \sqrt[3]{\frac{4 \cdot 10t}{3\pi}} = \sqrt[3]{\frac{40t}{3\pi}}[/math]

 

If we differentiate that we get:

 

[math]h'(t) = \frac{2\cdot 5^{\frac{1}{3}} \cdot 3^{\frac{2}{3}}}{9\cdot \pi^{\frac{1}{3}} \cdot t^{\frac{2}{3}}}[/math]

 

So now we just plug in the time at which the height is 15:

 

[math]h'\left( \frac{2025\cdot\pi}{8} \right) = \frac{8}{405\cdot\pi} \approx 0.006288[/math]

 

Hmm... is that right? I suppose you can go over it and check if I made any errors. It's your homework after all

[corbin]

Ahhh, I guess that would be where I was getting confused this morning (well, yesterday morning now). 

 

Your math does make sense, and I checked a second ago, and that's what the book says is the answer.

 

I think my class probably would have worked it a little differently though (although really you way seems like it would be easier down the road if problems grow more complex).  We haven't had to find a rate based off of a rate yet.

 

To be honest, I was actually reviewing for a test* last night.  I had 90% of the chapter down, but the last section threw me off a bit.  I randomly picked questions to see if I could work them, and I'm wondering now if I picked one she didn't expect us to know how to do.  On the test there was a related rate problem, but it was much simpler (it was like the planes one).

 

 

*Hopefully it doesn't bug you that it was technically test review and not homework.  In a way I would think it's better since it wasn't work we technically had to do.

 

 

 

Anyway, thanks for your help!  It seems like the farther I get in math classes, the worse I realize I am at math.  I think I get this problem now, but now I need to do a few others to make sure I really get it hehe.

[Daniel0]

You're welcome. How much calculus have you gone through? After a while I'm sure it'll click. Depending on what you need to do, it might help        thinking about a derivative in different ways. E.g. as a tangent line, the rate of change in a given point, or thinking as differentiation as the opposite of integration.

[Zane]

Depending on what you need to do, it might help        thinking about a derivative in different ways. E.g. as a tangent line, the rate of change in a given point, or thinking as differentiation as the opposite of integration.

 

I either don't know shit about calculus or that sentence doesn't make any sense whatsoever.  (Most likely I don't know shit, because I don't)

[Daniel0]

How doesn't it make sense?

 

E.g. as a tangent line,

 

This was of course supposed to have been the slope of a tangent line. Actually calculating the tangent line's equation is trivial though.

[corbin]

You're welcome. How much calculus have you gone through? After a while I'm sure it'll click. Depending on what you need to do, it might help        thinking about a derivative in different ways. E.g. as a tangent line, the rate of change in a given point, or thinking as differentiation as the opposite of integration.

 

We really haven't gone through much.  Finding out if something is differentialable at points, finding derivatives with the difference quotient, the power rule, chain rule, multiplication rule (I don't think it's called "multiplication rule": f(x) = ab; f'(x) = ab' + a'b) and a few other things.  Oh, and we have not done any integration yet (although we have gotten very slightly into it in chemistry).

 

I typically try to think of it in different ways.  The first thing I always think of is the slope of a tangent line, then I think of the change of something compared to the change of something else.  When ever I come across word problems like those, I try to think of it as the change of something compared to the change of something else, (for example, when the time changes 1 minute in that problem, the volume changes 10 cubic ft).  Chemistry is what made the rate of change thing really set in.  It didn't occur to me until about a week ago that delta A/delta B in chemistry is the same as dA/dB in math...  So then I realized (although I already knew to some extent) that taking the derivative of something in respect to x is just finding the change in something as compared to the change in x.

 

I think I have a fairly good grasp of everything that we've done so far in calculus.  It was just these word problems threw me I guess since there were so many steps, and I'm terrible at finding relations between two things.

[Daniel0]

It didn't occur to me until about a week ago that delta A/delta B in chemistry is the same as dA/dB in math...

 

That's not entirely true. Δx typically means "some change in x" (though Big Delta means other things in other contexts) whereas dx is "an infinitesimal change in x". This is also why you write dy/dx as the derivative. Because y is dependent on x, we have that when Δx -> 0 then Δy/Δx -> dy/dx, which is also the typical way of proving that something is a derivative of something else.

[corbin]

Hrmmm, I guess that makes sense since you're trying to find the change at one specific point, hence the smaller the bottom is, the more accurate the top would be.  I think?

 

 

I guess I shouldn't have assumed they were the same.  It seems like they are in some situations though, but I guess not always hehe.

[Daniel0]

Right, Δy/Δx describes the average change over a given time period. As you decrease that time period, you'll get closer to the change in a particular point in time, and as you make the time period infinitely small you get infinitely close, which is "close enough" because nothing is smaller than infinitely small (per definition).

[corbin]

So I guess that's why the difference quotient works since it's a limit approaching 0.  Nifty.

[corbin]

Ahhh, back again....

 

 

I had hoped not to have to ever have a calculus question I couldn't figure out myself again, but not even a week later I'm back.

 

 

This time it's about trig functions though.

 

 

So anyway, hopefully someone can help me with this and will want to:

 

What is known:

 

y = x^3

dx/dt = 3 cm/sec

the x and y scales are the same.

 

Derivable from the given information:

 

dy/dt = 3x^2 * dx/dt

dy/dt = 9x^2

 

Question:

 

What is the rate of change o the angle of inclination of the line connecting the origin to (x,y) when x = 2?

 

 

So, this is what I tried:

 

 

tan (a) = y/x

 

Seems simple enough, right?

 

Eventually it comes down to:

 

dtan a/dt = 12

 

 

But, then there's the problem of the whole knowing the rate change of tan a, not of a?

 

 

How in the world do I get da/dt out of that?

 

I tried to get alone in the beginning but failed to do so:

 

 

a = arctan(y/x)

 

So then obviously:

 

da/dt = (d arctan(y/x))/dt

 

But what in the world is the derivative of arctan y/x in respect to t?

 

 

I found via google that the derivative of arctan x in respect to x is:

 

1/(1+x^2)

 

So then I thought maybe I could figure out where dx/dx would be and just replace it with dx/dt, but as I expected, that failed miserably.

 

 

So basically where in the world do I go with this?  Do I go the d tan a/dt route or the arctan route?  Or, do I find the length of the connecting line (sqrt(x^2 + y^2)) then find the derivative of that and go that route?  I tried that and didn't get very far either.

 

 

Anyway, if someone could just hint at which path would be the easiest and perhaps tell me how to get around the part I'm getting stuck on, that would be wonderful.

 

 

 

 

 

(Side note:  Blerh, I should find a math forum, or start posting on Yahoo Answers or something when ever I have a math problem.... lol.)

[Daniel0]

If [imath]y=x^3[/imath] then the angle between the x-axis and the line between the origin and the point (x,y) must be given by:

 

[math]A(x) = \arctan \left( \frac{y}{x} \right) = \arctan \left( \frac{x^3}{x} \right) = \arctan \left( x^2 \right)[/math]

 

Then when we find the derivative:

 

[math]A'(x) = \frac{1}{\left(x^2\right)^2 + 1} \cdot 2x = \frac{2x}{x^4 + 1}[/math]

 

So when [imath]x=2[/imath] the rate of change of the angle is [imath]A'(2) = \frac{4}{17} \approx 0.235294[/imath].

[corbin]

Ahhh, so that was the derivative?  That would be the rate of change of the angle with respect to x, right?  So to get it with respect to time would take a bit more, right?

 

 

Someone pointed out to me today that d/dt * tan a is just sec(a)^2 * da/dt....  For some reason I had stuck in my head that it wouldn't be possible to get a alone.

 

Once someone pointed that out to me, it was easy:

 

sec^2(a) * (da/dt) = 2x (dx/dt)

 

then to skip a few steps (plug all the knowable stuff in):

 

12/sec^2(arctan(4)) = 12/17 rad/sec

 

 

 

 

Once again, thanks so much Daniel....  Hopefully I'm done with poorly thought out calculus questions now .

[Daniel0]

Oh, I didn't notice there was an element of time as well.

[corbin]

Ahhh, ok.  Wasn't sure if your answer was with respect to x or t.  Guess it was x .