Math Problem....HELP!?!?!?!

[Mooka]

It's all jeero's and vone's with me. Atleast I can understand an Indian accent.

I have trouble with that.

My main problem with Indian professors is their math backrounds are ridiculously strong. They breeze over it so fast, sometimes you get the concepts confused.

[SkinsMaster88]

You da man!!

Alright, this is for your eyes, and your eyes only. It's making it's ES debut.

http://www.youtube.com/watch?v=nTegy6sBQVA

If you like me, you better put a ring on me.

No one else watch this.

You've got great dancing skills...

[SonOfWashington]

STB21, in this situation, your parabola is opening downward since you have a y-intercept lower than your vertex, so there will be a symmetrical point across the line x=3 at (6,-6)

If you have a graphing calculator (I have a TI-83 Plus, so I'll use that), you can use it to figure it out this way:

1. Go to STAT > EDIT.
2. Enter your points (3,3), (0,-6), and as I mentioned above, (6,-6).
3. Hit STAT again, go right to CALC, choose QuadReg.
4. Hit enter twice and you have your solution.
TO CHECK
1. Go to your y-editor (y=). Arrow up to highlight Plot 1, and hit enter. In the graph you should see the three points.
2. Plug in your quadratic equation and if it goes through the points, winner winner chicken dinner.

If you don't have the TI-83 and want an on-screen one, PM me and I can email it to anyone who wants it.

~SOW

[Cherick]

I tried.......

[ixcuincle]

I'm just glad I quit math after Calc II. I've seen what's after Calc II and it's not pleasant.

The worst thing about Math and Physics were the tests, you were given a three page test, there are like 5 problems. However each problem takes you like an hour to do lol...

I had a good Physics teacher back in the day but neither him or the TA could explain the complicated workings of basic freshman physics. Once you bring in any of the 6 trig functions, I'm immediately lost. Same with logs and ln, they would scribble partial derivatives and logs on the white board and I had no idea what was going on.

[mjah]

Where do you go to school and what are you studying? Triple integrals were kind of annoying especially when you have to think in freaking 4 dimensions. They are even more annoying when trying to set up a probability problem

Getting multiple integrals involved with probability problems was pretty much my favorite part of math, ever.

Probability starts out so discrete, and then all of a sudden whooooaaaaBAM you're smearing it into distributions and finding a good everyday use for all that Calculus. So cool.

[MLSKINS]

ES, the new homework center. llam.

I would help you but I am done with math, FOREVER!!! When I see that I just blackout.

I thought I was done with math forever, but I switched to IT and had to take two more math classes.....

Anyway, I need help with these two questions

Given the functionL y^2-x^3+4=0

Use implicit differentiation to find dy/dx
and
Evaluate dy/dx at the point (-2, 2)

For the life of me, I cannot solve this dumb problem.

[S.T.real,lights,out]

Nevermind that isn't right..

http://www.math.ucdavis.edu/~kouba/C...licitDiff.html


Maybe this would help...

[MLSKINS]

The messed up part is that I took notes, some good notes too, and I still can't get this stupid question right. I hate math....

[Corcaigh]

It's been > 25 years since I thought about this but isn't it as simple as:

As they are asking you to use implicit differentiation, differentiate term by term
Use the chain rule and create an equation for y'
Substitute for x and y

[MLSKINS]

It's been > 25 years since I thought about this but isn't it as simple as:

As they are asking you to use implicit differentiation, differentiate term by term
Use the chain rule and create an equation for y'
Substitute for x and y

I forgot about the chain rule. I am about to try it out now..

[ixcuincle]

[DCSaints_fan]

I thought I was done with math forever, but I switched to IT and had to take two more math classes.....

Anyway, I need help with these two questions

Given the functionL y^2-x^3+4=0

Use implicit differentiation to find dy/dx
and
Evaluate dy/dx at the point (-2, 2)

For the life of me, I cannot solve this dumb problem.

Differentiate the whole side with respect to x
1) d/dx (y^2 -x^3 + 4) = d/dx(0)
2) d/dx (y^2) - d/dx(x^3)+ d/dx(4) = 0 - Deriviate of sum is equal to the sum of the deriviatives
3) 2y(dy/dx) - 3x^2 + 0 = 4 - You use the chain rule on the first term, second term is just usual polynomial differentiation
dy/dx = (4-3x^2)/2y

[ixcuincle]

Was rambling in chat about how if I ever returned to an elementary college physics, calculus, multivar, linear algebra, or differential equations course, I'd fail it...and I have over 9 years of college experience

I would seriously struggle and I've been in college for over 9 years

That's how hard that stuff is

[MLSKINS]

It's been > 25 years since I thought about this but isn't it as simple as:

As they are asking you to use implicit differentiation, differentiate term by term
Use the chain rule and create an equation for y'
Substitute for x and y

Differentiate the whole side with respect to x
1) d/dx (y^2 -x^3 + 4) = d/dx(0)
2) d/dx (y^2) - d/dx(x^3)+ d/dx(4) = 0 - Deriviate of sum is equal to the sum of the deriviatives
3) 2y(dy/dx) - 3x^2 + 0 = 4 - You use the chain rule on the first term, second term is just usual polynomial differentiation
dy/dx = (4-3x^2)/2y

Thanks. Once I remebered how to do the chain rule I was set....

Hopefully this time I am done with math for real. I can't take this stuff anymore...

http://i.imgur.com/sVAzu.jpg

I have seen everything now.