Frustrating Exam Results Despite Perfect Homework & Quiz Scores

[LT72884]

I don't get it guys. 100% on all the quizes and homework. 42% on the final... Deptartment policy is that you have to get a 60% or above on the final to get a D no matter what your grade is. i understand integration by parts just fine, trig sub just fine, integration with trig functions, parametric equations and taylor series. I even understand the diff EQ and linear algerbra they put on the final for the most part. the only thing i forgot, was how to find slope of a tangent line when just given r=cosa. i know its a long formula and i forgot it. ticks me off that i have to repeat it. i have never had a test so low. I mean, even my trig final i failed was 55%. This sucks that i failed trig and now this class. the integration on the exam was not to bad either

∫1/√x^2+9 dx

∫te^-3t dt

then some bogus initial condition problem that made no sense. don't remember the whole thing but it was something like this

xy`-y=x^2 +5 y(0)=1/2 i think. or something weird like that. i think it is dumb they are adding diff eq to the finals now.

i don't get it. seriously, it should have been an easy exam.

 

[Simon Bridge]

the only thing i forgot, was how to find slope of a tangent line when just given r=cosa
you mean: r'(a) = -sin(a) ?

the integration on the exam was not to bad either

∫1/√x^2+9 dx

∫te^-3t dt
... so you managed these ones OK then?

then some bogus initial condition problem that made no sense. don't remember the whole thing but it was something like this

xy'-y=x^2 +5 y(0)=1/2 i think. or something weird like that. i think it is dumb they are adding diff eq to the finals now. DEs in a calc final - how strange.
Have you figured out how to do those now?

Failing only two questions and getting 42% is a little tricky though.
See if you can get your transcript if you are really puzzled, but with all the quizzes and homework, it sounds like you blanked out in the exam.

 

[LT72884]

I still don't understand the DE's that much. Some of them i understand pretty well. Most of them are not to bad to separate and then bust out, but the one they gave was terrible. haha. plus the professor who wrote the final for the whole school, loves to put hard questions and get lost in the weeds. he LOVES proofs like no other. yeah i can integrate pretty well actually. its the fact that i didnt have all the formulas memorized and we get no notes what so ever. Working the formula and knowing why it works is no problem to me, it was not knowing that was going to be on the final. We were told to study the past finals, but NONE of them last semester or the semester before had a question like that. Oh and my mind did go blank on one of them. could not remember for the life of me with the washer method which varibles to integrate about the y axis. i kept everything solved for y and integrated with dx. oh well.

but still a 42% on a final. that's a kick in thew face man. haha. now i got those feelings of being stuck and weeded out coming on. when in fact i know i know the stuff. i know i can handle DE's but who knows. maybe i need some more practice at calc 2. what's another 3 months... haha

thanks guys. ill try and get a copy of it.

 

[Simon Bridge]

Note: proofs are pretty much what mathematicians so :)

Working the formula and knowing why it works is no problem to me ... could not remember for the life of me with the washer method which variables to integrate about the y axis.

If you knew why it worked, you wouldn't need to remember it.
It sounds like you are too used to applying the formulas.
If you had to find the area between y=x^3 and x=y^3 - would you try something involving a cube-root?

Personally I tend not to memorize formulas if I can help it: too easy to get turned around.
This means I tended to be good at questions like "find the volume of ..." and really bad at questions like "use method xyz to find the volume of..." since I didn't know the name of the method.

but NONE of them last semester or the semester before had a question like that.

... but that's just one question ... missing it is not a fail.
They are allowed to put in questions that have not been asked in a while - it's, kinda, the point.
You cannot expect to know the problems in advance, but if you know how things work, you don't need to.

But I bet there was - if it's this one:

xy'-y=x^2 +5 y(0)=1/2

it's a very standard type of problem. You probably just didn't recognize it when you saw it.

You can do xy'-y=0 then? This one has an extra step.

You have also failed another paper - which suggests there is something wrong with your approach rather than the course or the professors. Unless, of course, very few people passed: I've seen two professors mark students on how he felt about them rather than their performance.

Anyway - the thing to do now is pick yourself up and look hard at where you can adjust your approach to better meet the challenges of the course.
Good luck.

 

[LT72884]

This is the first semester they have had de in calc 2 and we only had 10 homework problems haha. Each question was worth 10 points so if you miss two, you are down to a B. My professor is the best math professor i have ever had. Whats buggin me is i thought i knew what i was doing. Granted i couldn't remember if with washer method if i had to subtract something. I just know its integration of pie r square and if it washer method about y-axis then you have to possibly subtract out the middle. I understand what's going on and why, just so many integration rules for volumes and areas that are simular. Haha. But i hqve all summer to figure it out. Haha

 

[Mark44]

This is the first semester they have had de in calc 2 and we only had 10 homework problems haha. Each question was worth 10 points so if you miss two, you are down to a B. My professor is the best math professor i have ever had. Whats buggin me is i thought i knew what i was doing. Granted i couldn't remember if with washer method if i had to subtract something. I just know its integration of pie r square and if it washer method about y-axis then you have to possibly subtract out the middle. I understand what's going on and why, just so many integration rules for volumes and areas that are simular.

To reiterate what Simon Bridge, it seems that you try to memorize a bunch of formulas without putting much effort into trying to understand where the formulas come from. If you know the area of a circle (I hope you do), the formula for the volume of a disk is just its area times its thickness. A washer is just a disk with a hole (in the shape of a smaller disk) in it.

A shell is really just a thin rectangular slab rolled up into a circular shape. If you understand the basic geometry for these, you don't have to memorize a bunch of formulas.

One of the most important things to do on many of these calculus problems is to draw a sketch of the region, and another of whatever you're using to calculate the area or volume. In my experience, many students resist doing this step, for whatever reason, but without the added insight that a drawing gives, they end up either not being able to do the problem or getting the wrong answer.

Haha. But i hqve all summer to figure it out. Haha

 

[LT72884]

Lol yeah i know where they come from and i know its area times thickness. But what i can't seem to remwmber is which way the disk or washer is facing. The thickness can be either dx or dy and that's where i get hung up. I draw pictures all the time and the problem is, i can draw the shell, disk, or washer in any direction. Ie, use the shell method for y equals sqrt5x about x axis. To me, the hieght of the shell is still in a value of y equals some function. But obviously I am wrong even when i draw the picture, the shell is wrong acording to my professor. Ill draw a pic and show you what i mean. I know the formula is just integration of 2pi(x)f(x)dx because we are dealing with circumfrance now rather than aera of a disk. Thanks guys. I put a lot of time and effort in this class and i did dang good. But it is what it is so now its time to understand it more.

 

[LT72884]

Ok here is a y=x^3, y=8, and x=0. Use shell method to find volume.

∫2pi(x)f(x)dx is what i will use.

when i draw the shell, the height is y=x^3 so i integrate with respect to x

∫2∏(x)(x^3) from 0 to 8

However, i am told that my shell is drawn wrong. i can't see why. to me it looks correct, but i was told the shell should be drawn vertically.

this is why i get hung up. haha.

 

[Simon Bridge]

Ok here is a y=x^3, y=8, and x=0. Use shell method to find volume.

∫2pi(x)f(x)dx is what i will use.

when i draw the shell, the height is y=x^3 so i integrate with respect to x

∫2∏(x)(x^3) from 0 to 8

You are integrating from 0-8 on the x axis, but the range 0-8 on your diagram is on the y axis.
This tells me you do not understand what the formula is doing.

You appear to have picked the x-axis for your integration because that's what you remember of the rule. Instead - ignore the rule, try to work it out.
i.e. if you just had to find the area - what would you do?

(Note: you left out which axis the area should be rotated about. Try sketching the region in 3D.)

 

[LT72884]

opps. picture cut it off. Right hand side of paper was whereit said to rotate around x axis. I have tryied to draw it in 3d many times, haha. however, i have been told that my shell is still drawn in the wrong place and no one ca tell me why. the way i have it drawn on the paper, is the right way to draw it o me. but i know I am wrong and want to know why. haha. height is y=x^3, the radius is some x value and the thickness is in dx because if i roll the shell up, the thickness is a change in x. I am just not seeing where to draw the shell and why. i have been told to draw the shell up and down, not left to right. i can't see why though. up and down does not give me what i need. haha

if i understood correctly, limits of integration tell you what to integrate with respect to right? so if i ignore the rules for a minute and look at the drawing, there is no way to integrate with respect to x because, well, the function continues on in that direction for ever, there is no bounding limit, so therefore it must needs be integrated with y because we know our limits there? Since it will be with y values, we need to solve for x so that we have:

2∏∫(y)(3√y)dy from 0 to 8

Is there any way to integrate with respect to x at all the way i have it? i say no because i don't have a bound like x=5.

thanks

all the other integration stuff is a lot simpler to me, this changing things around drives me nuts. haha.

i do appreciate all your help. time to open the books and study till fall semester. I am burned out from school. haha. i need a break.

 

[Integral]

Look at your drawing again, now think about it. Which way is the thickness of your washer? You have the bottom of the washer at y₁ and the top at y₂ So the thickness is given by Δy which would translate to dy. Now in your integral you are integrating dx? How can that be right?

You cannot do math in autopilot, you have to being thinking about what you are doing at every step of the way.

 

[Simon Bridge]

@Integral - he's trying to use the shell method, even though he appears to have drawn both a shell and a washer - notice that the rotation is about the x axis?

if i understood correctly, limits of integration tell you what to integrate with respect to right? so if i ignore the rules for a minute and look at the drawing, there is no way to integrate with respect to x because, well, the function continues on in that direction for ever, there is no bounding limit

You do actually know a limit in x, but why would you need to - you can integrate with respect to any variable you like ...

so just pick out one shell, radius y, about the x-axis, thickness dy like Integral says.
it has to be inside the volume you are trying to find.
So how wide is it? Look at your diagram - one end is at x=0, the other end is at x=f(y)... so what is the function?

Note: this problem is actually easier to do via the washer method ;)

 

[Mark44]

LT72884,
It's helpful to draw two diagrams for these kinds of problems. One is to show the region that is being revolved, like the one you have. The other should show the solid of revolution and your typical volume element, a shell, disk, or washer.

The 3D drawing doesn't have to be very detailed - a cross-section that shows the back half of the solid and the typical volume element will do.

 

[Simon Bridge]

What he said - when you have a decent idea what the revolved shape turns out to be, a strategy usually presents itself.

Usually it ends up looking like a cylinder that someone has removed bits from.
You can either find the volume of the shape left behind, or get the volume of the cylinder and subtract the volume of the bits that got removed.

Whichever you decide, you either slice the volume like a loaf of bread - which will give you a lot of disks - or you can core it out from the end - in a series of hollow cylinders. One of these will usually be easier than the other.