222 replies to this topic

[Deadly-Dreamer-X]

I really need help in this.

Find the inverse using elementary transformation.

[9 7 3]
.5 -1 4.
[6 8 2]

I can't even start with it!

[Last Tsurugi]

I'm no good at math D: I just had a Trigonometry test and I failed it horribly. I got a 46, which was far lower than I anticipated, and I studied too. I guess rationalizing the Denominator and long division did me in.

That being said, can anyone help me with rationalizing the denominator in the different instanced it is used in?

[Dion]

Ok, a simple trigonometry problem, I suppose for anyone in calculus;

sinx + cosx = 2(sinx)^2 + 1/(sinx - cosx)

Verify please? I've been at this for a while now.

[Lexaeus]

Anyway, find me the line parallel to:

x = 2 + 3t
y = 4 + 6t
z = 1 + 2t

that passes through the point (3,5,7)

Obviously it's in 3D, not 2. :> ENJOY.

Shoot. I need to continuing studying Calculus this summer if I want to make this large jump. This seems adaptable to do from 2d parametric, but bleh I need to refresh.

[Secret Igshar]

I really need help in this.

Find the inverse using elementary transformation.

[9 7 3]
.5 -1 4.
[6 8 2]

I can't even start with it!

What do you mean by inverse? >_>; All I know of that would be an "inverse" for a matrix would be something like this:

|9 5 6|
|7 -1 8|
|3 8 2|

I'm no good at math D: I just had a Trigonometry test and I failed it horribly. I got a 46, which was far lower than I anticipated, and I studied too. I guess rationalizing the Denominator and long division did me in.

That being said, can anyone help me with rationalizing the denominator in the different instanced it is used in?

Rationalizing the denominator is actually rather simple. You just multiply the fraction by one. The form of one that you use just needs to be "denominator over denominator"

For example, if you have the fraction (1 / sqrt(2) ) you would multiply the whole fraction by ( sqrt(2) / sqrt(2) ) and end up with ( sqrt(2) / 2 ), thus making the denominator into a rational number rather than an irrational number. If you meant something other than this, I've no clue what you mean. >_>;

Ok, a simple trigonometry problem, I suppose for anyone in calculus;

sinx + cosx = 2(sinx)^2 + 1/(sinx - cosx)

Verify please? I've been at this for a while now.

I don't remember enough of my Trig identities to do this thing... D: This is super old, though, so I hope you actually managed to figure it out. >>;

Shoot. I need to continuing studying Calculus this summer if I want to make this large jump. This seems adaptable to do from 2d parametric, but bleh I need to refresh.

You need to understand what it means to be parallel in terms of parametric equations before you can attempt to solve it. :> It's actually incredibly simple once you know how it works.

[Deadly-Dreamer-X]

What do you mean by inverse? >_>; All I know of that would be an "inverse" for a matrix would be something like this:

|9 5 6|
|7 -1 8|
|3 8 2|

Hey, teach. Wow, that was QUICK! ; Anyway, welcome back.

I think that's the transpose. Transpose are denoted by A'(A Dash) and Inverse as A^-1(A inverse).

There is an incomplete note. But I will share it with you:

Invertible matrices:

If A is a square matrix of order m if there exists some another square matrix B of the same order M such that AB = BD = I then B is called the inverse of A and is denoted by A^-1

1) By using elementary operation find the inverse of the matrix
A=

|1 2|
|2-1|

Ans:

write A= IA

i.e 
|1 2| = |1 0| A (row operation (2-2)(i)-1-2(2)=5[sup0[/sup]) 
|2 -1|    |0 1|

R[sub]1[/sub] |1  2| = |1 0|        R[sub]2[/sub] -> R[sub]2[/sub] - 2 R[sub]2[/sub]; 0-2(1)=-2
R[sub]2[/sub] |0 -5|   |-2  1|      1-2(0)=1              

And the rest of the note...

And don't ask me what those are, and perhaps that's why I am came to you, teach. A small recap (about it)will definitely suffice, and will make my day. Thank you.

[Waker]

I'm horrible at factoring.

How do you get x^3 + x as the derivative of

lim x->3 of f(x) = (x^3 + x - 30)/(x - 3)?

Some attempts:

(x + 3)(x - 5)(x + 2)

(x + 3)(x^2 - 3x - 10)

(x - 3)(x + 5)(x + 2)

(x - 3)(x^2 + 7x + 10)

[Lexaeus]

I'm horrible at factoring.

How do you get x^3 + x as the derivative of

lim x->3 of f(x) = (x^3 + x - 30)/(x - 3)?

Some attempts:

(x + 3)(x - 5)(x + 2)

(x + 3)(x^2 - 3x - 10)

(x - 3)(x + 5)(x + 2)

(x - 3)(x^2 + 7x + 10)

that limit is not describing the change in x, what are you doing?

[Secret Igshar]

I'm horrible at factoring.

How do you get x^3 + x as the derivative of

lim x->3 of f(x) = (x^3 + x - 30)/(x - 3)?

Some attempts:

(x + 3)(x - 5)(x + 2)

(x + 3)(x^2 - 3x - 10)

(x - 3)(x + 5)(x + 2)

(x - 3)(x^2 + 7x + 10)

Um. Not sure what the limit has to do with the whole notion of taking the derivative, but there's two ways to go about it.

You can use the Rational Root Theorem to get your p's and q's and therefore get your potential factors to factor the upper portion or just go brute force and use the quotient rule.

d(u/v)/dx = (vdu - udv)/v²

v = x-3
dv = 1
u = x³ + x - 30
du = 3x² + 1

Substitute~

( (x-3)(3x² + 1) - (x³ + x - 30)(1) ) / (x-3)²

(3x³ - 9x² + x - 3 - x³ - x + 30) / (x-3)²

(2x³ - 9x² +27) / (x-3)²

From there, you use the Rational Root Theorem. (probably should have done that before differentiating, in hindsight).

For that you put +/- p/q where p is a factor of 27 and q is a factor of 2 (in this case)

This gives you the following as possible roots:

+/- { 1, 1/2, 3, 3/2 } ; Judging from the function we're working with, I'll start by testing 3. I hope for two 3's, so I attempt to divide out a second one. (denominator has (x-3)² in it)

[U]3[/U]| 2  -9   0   27
   [U]-  +6  -9  -27[/U]
[U]3[/U]| 2  -3  -9   0
   [U]-  +6  +9[/U]
   2   3   0

At the end of all this bullshit, you end up with:

(2x + 3)(x - 3)² / (x - 3)²

or simply 2x + 3. :> Not sure where x³ + x could come from, either. >_>;

Unless I didn't do what you were trying to do...

OH WAIT I SHOULD DO THE LIMIT, TOO.

lim_x->3f(x)

f(x) = (x³ + x - 30) / (x - 3)

To do the limit, you just RRT on the original function, choosing to test 3.

[U]3[/U]| 1  0  1 -30
   [U]- +3 +9 +30[/U]
   1  3 10  0

(x² + 3x + 10)(x - 3) / (x - 3) in your limit.

x² + 3x + 10 evaluated at 3 gives you:

9 + 9 + 10 = 28. QED.

[Waker]

._.

I wrote/read the problem wrong.

> In problems 27-36, the given limit is a derivative, but of what function and at what point?

30. lim x->3 (x^3 + x - 30)/(x - 3)

So uh. I take this to be the same as [(x+h)^3+(x+h)-30]/h and the function is x^3 + x and with value x as 3, you get the point (3,30) ._.

Edited by Waker, 09 September 2010 - 09:15 AM.

[Secret Igshar]

._.

I wrote/read the problem wrong.

> In problems 27-36, the given limit is a derivative, but of what function and at what point?

30. lim x->3 (x^3 + x - 30)/(x - 3)

So uh. I take this to be the same as [(x+h)^3+(x+h)-30]/h and the function is x^3 + x and with value x as 3, you get the point (3,30) ._.

Well if you wanna do it that way, the definition of the derivative is one of these:

lim_h->0[ f(x+h) - f(x) ] / h
lim_s->x[ f(s) - f(x) ] / [s - x]

I would take this as the latter. Basically what we're looking at is this:

lim_x->3[ f(x) - f(3) ] / [x - 3]

f(3) is a constant, so it can be assumed that at least a part of the -30 is your -f(3). I'm not sure on a solid means to reverse this whole thing, but with a bit of trial and error, you'll find that 30 = f(3) and x³ + x = f(x).

What I'd probably do is plug 3 in on top and find that it zeroes out (30-30) which actually puts you in 0/0, which allows usage of L'Hopital's rule if you actually wanted to solve the Limit. REGARDLESS, you'll see that if x³ + x is to be your f(x), you'll get f(3) = 30. When you plug f(x) and f(3) into that equation where I swapped x and s for 3 and x, you'll find yourself right back into the original limit.

Not sure if it can be done any way other than trial and error. Maybe someone else would know. >__>

[Waker]

Oh god I'm having problems trying to derive

sqrt(x^2+4)

I don't get how it derives to x(x^2+4)^(-1/2)

Using the chain rule I get .5x^(-1/2)*2x...

Since f(x)=x^(1/2) and g(x)=x^2+4

so f^1(x)=(1/2)x^(-1/2) and g^1(x)=2x...

f^1[g(x)]*g^1(x) = (1/2)(x^2+4)^(-1/2) * 2x which = x(x^2+4)^(-1/2) oh.

Never mind. Well I guess I'll keep this here since it explains the process of the chain rule a little.

[Secret Igshar]

Oh god I'm having problems trying to derive

sqrt(x^2+4)

I don't get how it derives to x(x^2+4)^(-1/2)

Using the chain rule I get .5x^(-1/2)*2x...

Those two expressions are identical, good sir, if you simply had kept the x²+4 under the radical rather than somehow transforming it into just x.

[Waker]

Those two expressions are identical, good sir, if you simply had kept the x²+4 under the radical rather than somehow transforming it into just x.

What.

is a radical.


Anyway, I don't know how to determine the slope of the graph of 2x^2 - 3xy + y^3 = -1 at the point (2, -3). Rather, I am having trouble moving the ys to one side :<

[Secret Igshar]

What.

is a radical.

sqrt(x) is the same as radical x.

Anyway, I don't know how to determine the slope of the graph of 2x^2 - 3xy + y^3 = -1 at the point (2, -3). Rather, I am having trouble moving the ys to one side :<

2x² - 3xy + y³ = -1

The way to do this one is just implicit differentiation, that is, you just apply the operator d/dx to the full expression like so (y' = dy/dx)

4x - (3y + 3xy') + 3y²y' = 0
4x - 3y = (3x - 3y²)y'
(4x - 3y) / (3x - 3y²) = y'

Evaluated at the point (x,y) = (2,-3)

(4*2 - 3*(-3)) / (3*2 - 3*(-3)²) = y'
(8 + 9) / (6 - 27) = y'
-17/21 = y'

[Waker]

sqrt(x) is the same as radical x.


2x² - 3xy + y³ = -1

The way to do this one is just implicit differentiation, that is, you just apply the operator d/dx to the full expression like so (y' = dy/dx)

4x - (3y + 3xy') + 3y²y' = 0
4x - 3y = (3x - 3y²)y'
(4x - 3y) / (3x - 3y²) = y'

Evaluated at the point (x,y) = (2,-3)

(4*2 - 3*(-3)) / (3*2 - 3*(-3)²) = y'
(8 + 9) / (6 - 27) = y'
-17/21 = y'

How'd you get y' in 3y²y'? Methought y^3 derived to 3y^2

Edited by Waker, 20 September 2010 - 07:41 PM.

[Secret Igshar]

How'd you get y' in 3y²y'?

You're differentiating y³ with respect to x. The chain rule states that if you have a function of the form f(x), df/dx is equivalent to f'(x)dx/dx. dx/dx is 1, derp. In this case, f(x) = x³ and the "x" is y, so the chain rule makes it f'(y)*dy/dx which, in this case, comes out to 3y²y'.

That was probably very confusing. ._.

[Waker]

You're differentiating y³ with respect to x. The chain rule states that if you have a function of the form f(x), df/dx is equivalent to f'(x)dx/dx. dx/dx is 1, derp. In this case, f(x) = x³ and the "x" is y, so the chain rule makes it f'(y)*dy/dx which, in this case, comes out to 3y²y'.

That was probably very confusing. ._.

That was confusing. I don't get how differentiating x³ is different from y³. I'll just ask what dy and dx are in f'(y)*dy/dx...

Edited by Waker, 20 September 2010 - 08:03 PM.

[Secret Igshar]

That was confusing. I don't get how differentiating x³ is different from y³. I'll just ask what dy and dx are in f'(y)*dy/dx...

When you differentiate x³ with respect to x, you get 2x²*dx/dx. dx/dx (the rate of change of x with respect to x) is always 1, so it never shows up.

If you take a function y(x) and differentiate it with respect to x, you get its derivative, dy/dx (the rate of change of y with respect to x), which is NOT always 1. So if you differentiate y³ with respect to x, you get 3y²*dy/dx. The d's can be thought of as delta signs, since that's essentially what they are. They're sometimes replaced by deltas in Physics.

[Waker]

When you differentiate x³ with respect to x, you get 2x²*dx/dx. dx/dx (the rate of change of x with respect to x) is always 1, so it never shows up.

If you take a function y(x) and differentiate it with respect to x, you get its derivative, dy/dx (the rate of change of y with respect to x), which is NOT always 1. So if you differentiate y³ with respect to x, you get 3y²*dy/dx. The d's can be thought of as delta signs, since that's essentially what they are. They're sometimes replaced by deltas in Physics.

Oh~ I see now. dy/dx = y' I see, I see. I never knew that there were byproducts to deriving or that dx/dx shows up in all derivations of x...