Calc I

[Hyudra]

Nothing to see here.

 

[isekii]

is this your homework ?

 

[NorthRiver]

Math is the DEVIL!!!!!!!

 

[Dudd]

I'd help you, but that would be 10 more questions than I've done for my own calc class this year, so I'll pass.

 

[agnitrate]

Pathetic. Did you even bother trying these before you blindly copied them off of your homework sheet?

NO I WILL NOT DO YOUR HOMEWORK!

-silver

 

[isekii]

heh i did all of these 3 semesters ago~

dont remember much~

 

[Hyudra]

hah yea, kinda getting some homework done, tho this feels more like a test im doing.

 

[Hyudra]

Originally posted by: agnitrate
Pathetic. Did you even bother trying these before you blindly copied them off of your homework sheet?

NO I WILL NOT DO YOUR HOMEWORK!

-silver

well let me see this is like 10 I need from 24 i went through, I have no clue how to approach these since the answers given are very weird.

 

[Legendary]

9. ( 1/(2 * sqrt(x-sqrt(x))) ) * (1 - (1 / (2 * sqrt(x))))

There's your answer - try to read it.
By the way, the only way to learn how to do calc is to learn it.

This is just a reminder - parallel lines have the same slope *cough* DERIVATIVE *cough*
Remember that.

 

[agnitrate]

Originally posted by: Hyudra

Originally posted by: agnitrate
Pathetic. Did you even bother trying these before you blindly copied them off of your homework sheet?

NO I WILL NOT DO YOUR HOMEWORK!

-silver

well let me see this is like 10 I need from 24 i went through, I have no clue how to approach these since the answers given are very weird.

Sorry, I just find it annoying that people always help those by giving them answers instead of showing them how to do it. For all we know, you didn't try these at all. Add this to the fact that you didn't say anything about how you even tried to approach them or answers that you got for any of them, and it comes off as unconvincing.

Especially where the questions are just finding the derivative of a function, you don't even state what you got when you tried.

Forgive me for being a little less than helpful.

-silver

 

[Hyudra]

Originally posted by: Legendary
9. ( 1/(2 * sqrt(x-sqrt(x))) ) * (1 - (1 / (2 * sqrt(x))))

There's your answer - try to read it.
By the way, the only way to learn how to do calc is to learn it.

This is just a reminder - parallel lines have the same slope *cough* DERIVATIVE *cough*
Remember that.

alright i got what you got, but heres what I have to choose from..

#9
a) 1/ (2sqr(x)sqr(x-sqr(x))) b) sqr(x)/(2sqr(x)sqr(x-sqr(x))) c) 2sqr(x)-1/ 4sqr(x)sqr(x-sqr(x)) d) 1 / sqr(x-sqr(x)) e) sqr(x-sqr(x)) / 2(2sqr(x)-1)

 

[dighn]

Originally posted by: Hyudra
1. Find the point or points on the graph of y = x^3 + 3x where the tangent to the graph is parallel to 6x - y = 5.
a) (1,0) or (-1,0) B) (1,4) C) (3,0) D) (-1,-4) or (1,4) e) (1,1)

ok i'm not gonna do you homework to you but 1) is very easy

parallell = same slope

slope of 6x-y=5 = 6

find where the slope ie dy/dx of the first euqation is 6 also. i'll leave it at that

 

[Hyudra]

Originally posted by: dighn

Originally posted by: Hyudra
1. Find the point or points on the graph of y = x^3 + 3x where the tangent to the graph is parallel to 6x - y = 5.
a) (1,0) or (-1,0) B) (1,4) C) (3,0) D) (-1,-4) or (1,4) e) (1,1)

ok i'm not gonna do you homework to you but 1) is very easy

parallell = same slope

slope of 6x-y=5 = 6

find where the slope ie dy/dx of the first euqation is 6 also. i'll leave it at that

What? how do you know what to do with the points?

 

[dighn]

Originally posted by: Hyudra

Originally posted by: dighn

Originally posted by: Hyudra
1. Find the point or points on the graph of y = x^3 + 3x where the tangent to the graph is parallel to 6x - y = 5.
a) (1,0) or (-1,0) B) (1,4) C) (3,0) D) (-1,-4) or (1,4) e) (1,1)

ok i'm not gonna do you homework to you but 1) is very easy

parallell = same slope

slope of 6x-y=5 = 6

find where the slope ie dy/dx of the first euqation is 6 also. i'll leave it at that

What? how do you know what to do with the points?

eh? the equation is y = x^3 + 3x. find x where dy/dx is 6, then sub x into original equation to find (x,y)

 

[Hyudra]

Originally posted by: dighn

Originally posted by: Hyudra

Originally posted by: dighn

Originally posted by: Hyudra
1. Find the point or points on the graph of y = x^3 + 3x where the tangent to the graph is parallel to 6x - y = 5.
a) (1,0) or (-1,0) B) (1,4) C) (3,0) D) (-1,-4) or (1,4) e) (1,1)

ok i'm not gonna do you homework to you but 1) is very easy

parallell = same slope

slope of 6x-y=5 = 6

find where the slope ie dy/dx of the first euqation is 6 also. i'll leave it at that

What? how do you know what to do with the points?

eh? the equation is y = x^3 + 3x. find x where dy/dx is 6, then sub x into original equation to find (x,y)

AHHH I'm freakin stupid as hell, those (x,y) coordinates confused the hell outta me. I got it now

 

[XZeroII]

If I wanted to do calculus homework, I would be sitting in front of MY book using MY pencil and paper doing MY homework.

 

[XZeroII]

BTW: If you don't even know how to take a derivative:

8. find the derivative of the function f(x) = (x^2)/3 - 3/(x^2). - no clue
a) (4x^2 - 9) / 6x b) (2x^4+18)/(3x^3) c) (x^3-9)/3x d) (2x^4-18)/(2x^3) e) (18-2x^4)/(2x^3)

then you need to talk to the teacher/professor. You should be able to do that in less than one minute. Use the quotient rule (f'g - fg')/g^2 on the two items separated by the minus sign.

edit: I put in the wrong quotient rule. It's correct now. Sorry.

 

[dighn]

Originally posted by: XZeroII
BTW: If you don't even know how to take a derivative:

8. find the derivative of the function f(x) = (x^2)/3 - 3/(x^2). - no clue
a) (4x^2 - 9) / 6x b) (2x^4+18)/(3x^3) c) (x^3-9)/3x d) (2x^4-18)/(2x^3) e) (18-2x^4)/(2x^3)

then you need to talk to the teacher/professor. You should be able to do that in less than one minute. Use the quotient rule (f'g - fg')/2 on the two items separated by the minus sign.

hmm i didn't notice that. seems like you need to read the book more or something

 

[Krk3561]

Ummm, we're you paying any attention in class or read your textbook? These are really easy (and I'm a HS sophmore, hehe). Here are the first 2, i dont feel like doing anymore:


1. dy/dx= 3x^2 +3 parrallel: y= 6x - 5 ------ plug in x values (1, -1, or 3 (answer choices) now and find the ones that make em equal then put it back into the first equation and pick the choice

2. You want to find the normal of the curve. Find dy/dx at x=2, gives you slope of the curve. Take the opposite reciprocal of the slope and that is the slope of the normal. Then plug x=2 into y to find the value of the normal at x=2. Once you have all that, figure out the equation, easy to do, you already have the value at x=2 and the slope.

 

[Hyudra]

Originally posted by: dighn

Originally posted by: XZeroII
BTW: If you don't even know how to take a derivative:

8. find the derivative of the function f(x) = (x^2)/3 - 3/(x^2). - no clue
a) (4x^2 - 9) / 6x b) (2x^4+18)/(3x^3) c) (x^3-9)/3x d) (2x^4-18)/(2x^3) e) (18-2x^4)/(2x^3)

then you need to talk to the teacher/professor. You should be able to do that in less than one minute. Use the quotient rule (f'g - fg')/2 on the two items separated by the minus sign.

hmm i didn't notice that. seems like you need to read the book more or something

OMG do you guys have to concentrate on that one easy problem. thats like the only one I didn't feel like doing.

 

[XZeroII]

5. if f(x) = x^(5/3) - x^(2/3), for what values of x will f(x) have horizontal tangents (I tried and I got ZERO, tho i cant choose a zero)
a) 3 b) 2 c) 2/3 d) 3/5 e) 2/5


Find the derivative and set the derivative equal to zero.

 

[VBboy]

LOL LOL I love this sh*t !!! I had A's in Calc Too bad this won't help you learn anything, only to get a good grade on this homework...

1. In order for a tangent line of a graph to be parallel to a straight line, the first derivative of the graph must be equal to the slope of the straight line.

f(x) = x^3 + 3x. f'(x) = 3x^2 + 3.

The line specified was 6x - y = 5, aka y = 6x - 5. Here, the slope is 6.
Now, solve the equation f'(x) = 6, i.e. 3x^2 + 3 = 6. Find the x's, and those will be the points where the tangent line of the f(x) is parallel to the specified straight line.


2. Equation of the line perpendicular to something. There, find the PARALLEL line first. Then, compute the slope of the perpendicular line as: -1/(slope_of_parallel_line). E.g. is the line is y = 2x - 3, then the slop of the line perpendicular to it will be -1/2 = - 0.5


3. Boring.

4. Boring.

5. Horisontal tangent is when the slope of the tangent line is 0. An example would be the tangent line to the graph f(x)=x^2 at x = 0. There, the X-axis happens to be the tangent line, with the slope = 0.
To solve, find the derivative and make it equal to 1. Solve for x.

6. Same as before.

7. Simple partial differentiation. Read the formula. Split the function and differentiate in parts.

8. Same thing, read on how to differentiate a complex function (not "complex" as in the square root of -1).

9. Same thing.

 

[bigshooter]

Number 9 is the chain rule or composite function rule or whtaever you want to call it. You take the derivative of the inside, times teh derivative of the outside as many times as you need to.