Math Help/Tutoring Thread

noxiousraccoon · Post by **noxiousraccoon** » Fri Nov 21, 2008 10:10 pm

Just to keep you guys busy: Compute all first and second partial derivatives of the function: f(x,y) = arctan(x^2 - y^2 - 1)

Was a question on my recent test.

[cc]z@nd! · Post by **[cc]z@nd!** » Sat Nov 22, 2008 12:13 pm

i don't know anything about derivatives, but i'll learn soon. and if i have time this weekend, i might even look it up (although i've got enough work to do as it is).

speaking of tests, though, i just had my third one in precal. the test was actually easier than some of the past ones, except for this little brainteaser of a question. not too hard if you don't overthink it:

Code: Select all

for what real number x will the angle of x radians be the same angle as an angle of x degrees? (as in, for what x will x radians = x degrees)

like i said, this question isn't really hard at all, but depending on how you approach it, it might teach you a lesson about approaching problems from different perspectives. if you get stuck, just take a second to step back and look at it from a different angle.

DeadHamster · Post by **DeadHamster** » Sat Nov 22, 2008 12:30 pm

metkillerjoe · Post by **metkillerjoe** » Sat Nov 22, 2008 12:37 pm

[cc]z@nd! wrote:i don't know anything about derivatives, but i'll learn soon. and if i have time this weekend, i might even look it up (although i've got enough work to do as it is).

He's talking about multivariable calculus (Calc I, Calc II, Calc III?, then Multivariable).

Tonight I may scan in and explain some pages of text for the very basics of find the rate of change of curves.

Altimit01 · Post by **Altimit01** » Sat Nov 22, 2008 12:55 pm

Well since the conversion of rads to degs is 180/pi you have to set up X(180/pi) = X and solve. Then again there's also that looping situation so you have to check against the time when the X in rads = 360n + X in degrees. That should give you the general solution. Ooh I like it.

On the other one, I really don't see the fascination with inverse trig functions. Normal trig yes but inverse gets silly.
Let's see if I can still do this thing.
∂f/∂x = 2x / (( x^2 - y^2 - 1)^2 + 1) = 2x/ (x^4 + y^4 - 2(x^2 * y^2) - 2(x^2) + 2)
∂f/∂y = -2y / (( x^2 - y^2 - 1)^2 + 1) = -2y / (x^4 + y^4 - 2(x^2 * y^2) - 2(x^2) + 2)
The second derivatives just become too tedious without getting out a few pieces of paper and writing this all down.

noxiousraccoon · Post by **noxiousraccoon** » Sat Nov 22, 2008 9:48 pm

Altimit01 wrote:Well since the conversion of rads to degs is 180/pi you have to set up X(180/pi) = X and solve. Then again there's also that looping situation so you have to check against the time when the X in rads = 360n + X in degrees. That should give you the general solution. Ooh I like it.

On the other one, I really don't see the fascination with inverse trig functions. Normal trig yes but inverse gets silly.
Let's see if I can still do this thing.
∂f/∂x = 2x / (( x^2 - y^2 - 1)^2 + 1) = 2x/ (x^4 + y^4 - 2(x^2 * y^2) - 2(x^2) + 2)
∂f/∂y = -2y / (( x^2 - y^2 - 1)^2 + 1) = -2y / (x^4 + y^4 - 2(x^2 * y^2) - 2(x^2) + 2)
The second derivatives just become too tedious without getting out a few pieces of paper and writing this all down.

Restricting the domain perhaps? - pi/2 to pi/2?

[cc]z@nd! · Post by **[cc]z@nd!** » Mon Nov 24, 2008 5:54 pm

DeadHamster wrote:0?

yeah, that's what me and a bunch of other people wrote down, so i'm assuming it's probably right. altimit brought up a good point though. i'm sure after long enough there will be a point where it would be true again, but I didn't bother looking since the test had a time constraint. think you could explain the work behind this?

Altimit01 wrote:X in rads = 360n + X in degrees

Aumaan Anubis · Post by **Aumaan Anubis** » Mon Nov 24, 2008 5:59 pm

[cc]z@nd! wrote:if you get stuck, just take a second to step back and look at it from a different angle.

Sweet pun.

And the only answer I can think, off the top of my head, would be that it would have to be 0.
Seeing as there's 2(pi) radians...
As far as I can tell, there is no degree from 1 - 6.2 that will equal the same radian.

Altimit01 · Post by **Altimit01** » Mon Nov 24, 2008 7:32 pm

[cc]z@nd! wrote:
DeadHamster wrote:0?
yeah, that's what me and a bunch of other people wrote down, so i'm assuming it's probably right. altimit brought up a good point though. i'm sure after long enough there will be a point where it would be true again, but I didn't bother looking since the test had a time constraint. think you could explain the work behind this?

Altimit01 wrote:X in rads = 360n + X in degrees

Well it was basically the same work as before X(180/pi) = X but since the radian progression is so much faster it's possible to loop. I was using the degree side so just add n many revolutions. Final solution would be something like X(180/pi) - 360n = X. In this version just subtract n many revolutions if X(180/pi) is greater than 360 until that value is less the 360. Come to think of it, that's just a modulo operation so if you want to get rid of the n term since it's not important for the solving the problem it's: X(180/pi) modulo 360 = X.

[cc]z@nd! · Post by **[cc]z@nd!** » Tue Dec 09, 2008 9:19 pm

interesting.

well anyways, it's finals week for me, and i've got my pre-cal final this friday at 8am. instead of posting questions here, i'll just keep working on this practice sheet they gave us and check in here occasionally. might even take some pics of the problems and upload them for you guys to take a look at, but i've fallen behind a bit in studying, so i'll take care of that first.

Aumaan Anubis · Post by **Aumaan Anubis** » Mon Dec 15, 2008 7:26 pm

Write an equation for the quadratic function whose graph has a vertex at (-2, 6) and a y-intercept of (0, -4).

Alright. So I gather from this that the slope is (6 - (-4))/(-2-0) = 10/-2 = -5
So the slope is -5.

Then vertex form...

y - 6 = -5(x + 2)^2

But when that happens, the y - intercept is not -4.

So, basically, the correct answer is y - 6 = -2.5(x + 2)^2

But I can't figure out why this doesn't work. We're way past this stuff in our class, but I'm reviewing for my final tomorrow, and I'm utterly confused at why the slope is half of what it should be.

Am I calculating the slope incorrectly?

I know that this should be an incredibly easy problem, but I'm confused.

Cryticfarm · Post by **Cryticfarm** » Mon Dec 22, 2008 5:57 pm

Quadratic not linear.

WaeV · Post by **WaeV** » Mon Dec 22, 2008 6:53 pm

There is bowl shaped like half a sphere, with a radius of 6 in.

If the bowl is being filled with water at 4 cubic in/sec, how quickly is the height of the water level increasing when the height is 2?

Altimit01 · Post by **Altimit01** » Tue Dec 23, 2008 9:44 am

WaeV wrote:There is bowl shaped like half a sphere, with a radius of 6 in.

If the bowl is being filled with water at 4 cubic in/sec, how quickly is the height of the water level increasing when the height is 2?

Just gonna stretch the old brain muscles.

Volume of a sphere = (4/3)(π)r^3
Since the shape filled by the water is always half-spherical it can be described as: (2/3)(π)h^3
Rearrange in terms of h: [(V)(3/2)(1/π)]^(1/3) = h
Separated for simplicity: [(3/2π)^(1/3)] * [V^(1/3)] = h
Derive with respect to time: [(3/2n)^(1/3)](1/3) * [V^(-2/3)] = dh/dt

Which strikes me as a little odd since dV/dt is given which isn't used neither is the max radius. But that's how I'd do it. Solve for the volume for the given h and plug that in to the derivative. Oh well.

WaeV · Post by **WaeV** » Tue Dec 23, 2008 1:54 pm

how did you jump from r^3 to h^3?

Altimit01 · Post by **Altimit01** » Tue Dec 23, 2008 7:14 pm

Well assuming the bowl starts from nothing and goes up to a half sphere, the water is always going to describe a half sphere with a radius of h.

WaeV · Post by **WaeV** » Tue Dec 23, 2008 7:23 pm

h was the variable used for the height of the water. When the water level is only two inches and not six, it doesn't fill a half-sphere.

FleetAdmiralBacon · Post by **FleetAdmiralBacon** » Tue Dec 23, 2008 7:29 pm

Altimit01 wrote:Well assuming the bowl starts from nothing and goes up to a half sphere, the water is always going to describe a half sphere with a radius of h.

No, it isn't. Think about it a bit more.
You can do an integral to find the volume at a given height, from, h=[0,h] using circles of radius sqrt(2rh-h^2), but that doesn't quite help.

Altimit01 · Post by **Altimit01** » Tue Dec 23, 2008 7:43 pm

Ah right. Been awhile since I worked with geometric shapes. Well in that case that solution is wrong.

Bacon, that integral is all that's needed. Just solve that definite integral for the volume in terms of h. Rearrange it so that h is in terms of V. Take the derivative of that.

Am too lazy to actually do that integral right now.

Edit:
Ok, I'm going to put a question out here for a change. This relates to compressing directX graphics but it's a relatively simple problem.

Consider colorspace. 3 dimensions but instead of X, Y and Z you have red green and blue. There are 16 points. I have already computed the line of best fit through them. (If ya'll are curious I can repost the method. It's kind of cool.) The problem, is to find the four best points with which to represent the original 16 points with the following conditions. It's not necessary to use the line of best fit, but it's probably the easiest way.
Conditions:
The four points must be in a line.
The four points must be equidistant from each other.
So in reality all that really matters are determining the two end points since the two points in between are implied by the conditions.

My current method is to treat each of the 16 points as vectors and determine the maximum and minimum projection along the line of best fit and use those as my initial end points. Then I compute the final end points as points at 1/8 along the line and 7/8 along the line to better represent the distribution. Anyone have some better methods?

For those who are curious I can post the full details of this problem in another section.

jks · Post by **jks** » Thu Dec 25, 2008 1:56 am

Took calc semester III last semester, taking differential equations and linear algebra this semester. I'm a math major btw.