Calc II Help

[Soccerman06]

Have a TI-83 do it for you. Anyways, I could have answered this 2 months ago, for some reason my calc 1 class got farther than this. I forgot this because my teacher was a douche and didnt feel like teaching this how to do anything (yes a college prof sucked).

 

[MrDudeMan]

Originally posted by: Random Variable

A monkey.

The magnitude of the vector in the ?-direction = x*sin?cosf + y*sin?sinf + z*cos?
= ?sin?cosfsin?cosf + ?sin?sinfsin?sinf + ?cos?cos?
= ?sin²?(cos²f + sin²f) + ?cos²?
= ?sin²? + pcos²? = ?

The magnitude of the vector in the ?-direction = x*cos?cosf + y*cos?sinf -z*sin?
=?sin?cosfcos?cosf + ?sin?sinfcos?sinf - ?cos?sin?
=psin?cos?(cos²f + sin²f) - pcos?sin? = 0

The magnitude of the vector in the f-direction = -x*sinf + y*cosf + z*0
= -?sin?cosfsinf + ?sin?sinfcosf = 0

Therefore vector function (x,y,z) in cartesian coordinates is the same as vector function (?,0,0) in spherical coordinates.

nice. thanks for sharing, even though it was like pulling teeth 😉

edit: removed nested quote action

 

[manlymatt83]

x^3/3 + x^2/2 + 2x is the anti-derivative of ( x^2 + 2 ) - ( - x ) with respect to x. Since you're doing a definite integral (the 0 to 1 thing), you have to evaluate your integral at both 0 and 1 and subtract them. That is what "[ x^3/3 + x^2/2 + 2x ] BAR 1 - 0" means. So you need to do (1^3/3 + 1^2/2 + 2*1) - (0^3/3 + 0^2/2 + 2*0) and you will get your answer.

-Tom

EDIT: Didn't bother checking the integration, but Bigsm00th is right, should be 2x instead of x.

How is it the anti derivaive? I see x^2 + 2 + x .... so I see where they get x^3/3 ... but shouldn't the anti-derivate of 2 be 2x and the anti derivative of x be x^2/2 ?
OHHHH

did they SWITCH THEM on me?

 

[Mo0o]

They simply found the antiderivative and the bar followed by the two numbers imply those are the bounds of your integral. Did you bother reading the book?

 

[Random Variable]

Originally posted by: Bigsm00th

Originally posted by: Random Variable

A monkey.

The magnitude of the vector in the ?-direction = x*sin?cosf + y*sin?sinf + z*cos?
= ?sin?cosfsin?cosf + ?sin?sinfsin?sinf + ?cos?cos?
= ?sin²?(cos²f + sin²f) + ?cos²?
= ?sin²? + pcos²? = ?

The magnitude of the vector in the ?-direction = x*cos?cosf + y*cos?sinf -z*sin?
=?sin?cosfcos?cosf + ?sin?sinfcos?sinf - ?cos?sin?
=psin?cos?(cos²f + sin²f) - pcos?sin? = 0

The magnitude of the vector in the f-direction = -x*sinf + y*cosf + z*0
= -?sin?cosfsinf + ?sin?sinfcosf = 0

Therefore vector function (x,y,z) in cartesian coordinates is the same as vector function (?,0,0) in spherical coordinates.

nice. thanks for sharing, even though it was like pulling teeth 😉

edit: removed nested quote action

It's another proof in itself to show that I'm using the correct transformations.

 

[MrDudeMan]

Originally posted by: mjuszczak

x^3/3 + x^2/2 + 2x is the anti-derivative of ( x^2 + 2 ) - ( - x ) with respect to x. Since you're doing a definite integral (the 0 to 1 thing), you have to evaluate your integral at both 0 and 1 and subtract them. That is what "[ x^3/3 + x^2/2 + 2x ] BAR 1 - 0" means. So you need to do (1^3/3 + 1^2/2 + 2*1) - (0^3/3 + 0^2/2 + 2*0) and you will get your answer.

-Tom

EDIT: Didn't bother checking the integration, but Bigsm00th is right, should be 2x instead of x.

How is it the anti derivaive? I see x^2 + 2 + x .... so I see where they get x^3/3 ... but shouldn't the anti-derivate of 2 be 2x and the anti derivative of x be x^2/2 ?
OHHHH

did they SWITCH THEM on me?

read my post about using integral and anti-derivative. they arent exactly the same thing.

 

[Soccer55]

Originally posted by: mjuszczak

x^3/3 + x^2/2 + 2x is the anti-derivative of ( x^2 + 2 ) - ( - x ) with respect to x. Since you're doing a definite integral (the 0 to 1 thing), you have to evaluate your integral at both 0 and 1 and subtract them. That is what "[ x^3/3 + x^2/2 + 2x ] BAR 1 - 0" means. So you need to do (1^3/3 + 1^2/2 + 2*1) - (0^3/3 + 0^2/2 + 2*0) and you will get your answer.

-Tom

EDIT: Didn't bother checking the integration, but Bigsm00th is right, should be 2x instead of x.

How is it the anti derivaive? I see x^2 + 2 + x .... so I see where they get x^3/3 ... but shouldn't the anti-derivate of 2 be 2x and the anti derivative of x be x^2/2 ?
OHHHH

did they SWITCH THEM on me?

Yes, addition is commutative. x^2 + 2 + x is the same as x^2 + x + 2.

-Tom

EDIT: I should also point out that my use of the word "anti-derivative" was technically incorrect as Bigsm00th pointed out. x^3/3 + x^2/2 + 2x + C is actually the anti-derivative of ( x^2 + 2 ) - ( - x ) with respect to x. Anti-derivatives always have that "+ C" term tacked onto the end.

 

[MrDudeMan]

Originally posted by: Random Variable

Originally posted by: Bigsm00th

Originally posted by: Random Variable

A monkey.

The magnitude of the vector in the ?-direction = x*sin?cosf + y*sin?sinf + z*cos?
= ?sin?cosfsin?cosf + ?sin?sinfsin?sinf + ?cos?cos?
= ?sin²?(cos²f + sin²f) + ?cos²?
= ?sin²? + pcos²? = ?

The magnitude of the vector in the ?-direction = x*cos?cosf + y*cos?sinf -z*sin?
=?sin?cosfcos?cosf + ?sin?sinfcos?sinf - ?cos?sin?
=psin?cos?(cos²f + sin²f) - pcos?sin? = 0

The magnitude of the vector in the f-direction = -x*sinf + y*cosf + z*0
= -?sin?cosfsinf + ?sin?sinfcosf = 0

Therefore vector function (x,y,z) in cartesian coordinates is the same as vector function (?,0,0) in spherical coordinates.

nice. thanks for sharing, even though it was like pulling teeth 😉

edit: removed nested quote action

It's another proof in itself to show that I'm using the correct transformations.

to be honest i dont completely remember the conversions and rules but it looks familiar enough regardless. its been a while since ive done that.

 

[Ionizer86]

Originally posted by: mjuszczak
I'm working on a problem where the example goes from:

The integral from 0 - 1 of: [ ( x^2 + 2 ) - ( - x ) ] dx

to

= [ x^3/3 + x^2/2 + x ] BAR 1 - 0


What are they doing here? The integral sign is dissapearing and all of a sudden a bar appears at the end from 1 to 0... are they getting the anti derivatives? Combining? What are they doing? Its the only step I don't understand.

Don't you mean
= [ x^3/3 + x^2/2 + 2x ] BAR 1 - 0 ?

They integrated it, that's all. Have they taught you how to do integrals by definition, using lim h--> 0 [ f(x - h ) - f(x) ]/ h ?

I'd hope so, because it takes a few exercises with that equation for you to at least understand the math. Then you can jump from that integral to the equation with the bar, as you put it.

Edit: wow, when I was writing this there were only like 3 replies! A little bit late 🙁

 

[manlymatt83]

Originally posted by: Soccer55

Originally posted by: mjuszczak

x^3/3 + x^2/2 + 2x is the anti-derivative of ( x^2 + 2 ) - ( - x ) with respect to x. Since you're doing a definite integral (the 0 to 1 thing), you have to evaluate your integral at both 0 and 1 and subtract them. That is what "[ x^3/3 + x^2/2 + 2x ] BAR 1 - 0" means. So you need to do (1^3/3 + 1^2/2 + 2*1) - (0^3/3 + 0^2/2 + 2*0) and you will get your answer.

-Tom

EDIT: Didn't bother checking the integration, but Bigsm00th is right, should be 2x instead of x.

How is it the anti derivaive? I see x^2 + 2 + x .... so I see where they get x^3/3 ... but shouldn't the anti-derivate of 2 be 2x and the anti derivative of x be x^2/2 ?
OHHHH

did they SWITCH THEM on me?

Yes, addition is commutative. x^2 + 2 + x is the same as x^2 + x + 2.

-Tom

EDIT: I should also point out that my use of the word "anti-derivative" was technically incorrect as Bigsm00th pointed out. x^3/3 + x^2/2 + 2x + C is actually the anti-derivative of ( x^2 + 2 ) - ( - x ) with respect to x. Anti-derivatives always have that "+ C" term tacked onto the end.

+C because they are indefinite right?

 

[MrDudeMan]

Originally posted by: mjuszczak

Originally posted by: Soccer55

Originally posted by: mjuszczak

x^3/3 + x^2/2 + 2x is the anti-derivative of ( x^2 + 2 ) - ( - x ) with respect to x. Since you're doing a definite integral (the 0 to 1 thing), you have to evaluate your integral at both 0 and 1 and subtract them. That is what "[ x^3/3 + x^2/2 + 2x ] BAR 1 - 0" means. So you need to do (1^3/3 + 1^2/2 + 2*1) - (0^3/3 + 0^2/2 + 2*0) and you will get your answer.

-Tom

EDIT: Didn't bother checking the integration, but Bigsm00th is right, should be 2x instead of x.

How is it the anti derivaive? I see x^2 + 2 + x .... so I see where they get x^3/3 ... but shouldn't the anti-derivate of 2 be 2x and the anti derivative of x be x^2/2 ?
OHHHH

did they SWITCH THEM on me?

Yes, addition is commutative. x^2 + 2 + x is the same as x^2 + x + 2.

-Tom

EDIT: I should also point out that my use of the word "anti-derivative" was technically incorrect as Bigsm00th pointed out. x^3/3 + x^2/2 + 2x + C is actually the anti-derivative of ( x^2 + 2 ) - ( - x ) with respect to x. Anti-derivatives always have that "+ C" term tacked onto the end.

+C because they are indefinite right?

yes. this one isnt indefinite though, so you can leave that off.

 

[PhaZe]

http://www.jonathanarias.com/atot.doc

hope that helps

 

[manlymatt83]

perfect!! thanks!!!!