222 replies to this topic

[Secret Igshar]

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...

That's actually the fundamentals behind the chain rule.

I wonder if anyone can figure this out without tearing their hair out. Find the prime factorization of 2¹⁰-1

[Xylox]

That's actually the fundamentals behind the chain rule.

I wonder if anyone can figure this out without tearing their hair out. Find the prime factorization of 2¹⁰-1

3 x 11 x 31
:D

[Waker]

Here's a relatively simple derivative problem I'm being stupid with:

Find the equation(s) of the tangent lines to y=cosx that are horizontal.

So uh y'=0 at intervals of pi. How do you express this in Taylor form?

[Secret Igshar]

Here's a relatively simple derivative problem I'm being stupid with:

Find the equation(s) of the tangent lines to y=cosx that are horizontal.

So uh y'=0 at intervals of pi. How do you express this in Taylor form?

Um. I would just write it out as y'=0 when x = π*n, n∈Z

Z is the set of all integers, meaning {..., -2, -1, 0, 1, 2, ...}

That's just me, though. That's how I was told to express it in my AP Calc class in High School. :0 For Sine it's more fun since it's π/2 + π*n. lolz.

[Waker]

No Taylor form, huh? Kay.

'Nother problem.

Find the equation of the normal line to the graph of y=(4-x^2)^.5 at the point (2,0).

So I find the tangent and get the normal from that. The tangent is the derivative of the original equation~ y'= .5(4-x^2)^(-.5)*2x = x/(4-x^2)^.5 and... now what?

it equates to x*sqrt(4-x^2)/(4-x^2). When you plug in 2 for x, both numerator and denominator become zero. Can I apply L'Hopital's rule in this case? :l

[Secret Igshar]

No Taylor form, huh? Kay.

'Nother problem.

Find the equation of the normal line to the graph of y=(4-x^2)^.5 at the point (2,0).

So I find the tangent and get the normal from that. The tangent is the derivative of the original equation~ y'= .5(4-x^2)^(-.5)*2x = x/(4-x^2)^.5 and... now what?

it equates to x*sqrt(4-x^2)/(4-x^2). When you plug in 2 for x, both numerator and denominator become zero. Can I apply L'Hopital's rule in this case? :l

You're gonna hate me for saying this. When you take the derivative, any value which gets spit out by an x coordinate simply gives you the slope of the line tangent to the graph at that point.

The equation of the tangent line is:

y = (dy/dx)x + b

Where dy/dx is evaluated at the point you are given as the point you want. If you wish to find a line perpendicular to the tangent line, you adjust the slope and b changes accordingly. How do you change b? Opposite reciprocal. Once you get to the point where you have:

dy/dx = -x/sqrt(4-x²)

You could simply switch it to:

m = sqrt(4-x²)/x

Which yields a slope of 0.

But why does this work?

You're finding the slope, right? Your slope is undefined, yes? What line has an undefined slope? A vertical line. You're being given a vertical tangent line, so you're looking for a horizontal normal line to the graph. You can do this all graphically, because you're looking at the top half of a circle of radius 2. If you don't know this, SHAME ON YOU.


The answer is just y=0. >_>



To answer your direct question, YES. That is how you would solve it with Calculus rather than graphical interpretation. You would set up a limit to attempt to find the value of the derivative at that point.

Spoiler alert: If you attempt to use L'Hopital's rule on that, you'll be stuck right back in an undefined value for the derivative, due to the nature of the derivative of the square root function. It will stuff that radical right back into the denominator and cancel things left and right to return you exactly back to where you were before you multiplied by 1.

Why doesn't this help in this case? Simple. Your initial function was undefined at the value, not indeterminate. You can ONLY use L'Hopital's rule to evaluate a limit of a rational function in which the numerator and denominator of the most reduced form both evaluate to zero. That implies that the value of that ratio could be ANY number.

side math: 0/0 = any number because 0/0 = k implies 0 = k0, thus k can be any number.

You have 2/0 = k, which implies 2 = k0, which is just impossible.



tl;dr, You thought waaaaay too hard about a simple semicircle problem.

[Waker]

tl;dr, You thought waaaaay too hard about a simple semicircle problem.

:<

>> The function has an inverse. Without solving for the inverse, find the derivative of the inverse function at x=a and graph the inverse function.

f(x)=x³+4x-1, a=-1

[Secret Igshar]

:<

>> The function has an inverse. Without solving for the inverse, find the derivative of the inverse function at x=a and graph the inverse function.

f(x)=x³+4x-1, a=-1

The graph is easy. Take the whole function's graph and just reflect it across the line y=x.

f'(x) = 3x² + 4
f'(-1) = 3(1) + 4 = 7

For the derivative of the inverse, all you do is 1/f'(x) so you get 1/7 as your answer.

[Waker]

The graph is easy. Take the whole function's graph and just reflect it across the line y=x.

f'(x) = 3x² + 4
f'(-1) = 3(1) + 4 = 7

For the derivative of the inverse, all you do is 1/f'(x) so you get 1/7 as your answer.

The answer at the back of the book is 1/f'(0) which = 1/4 ._.

So apparently f^-1(x)=0...

Are you using 1/f'(f^-1(x))?

[Secret Igshar]

:<

>> The function has an inverse. Without solving for the inverse, find the derivative of the inverse function at x=a and graph the inverse function.

f(x)=x³+4x-1, a=-1

The answer at the back of the book is 1/f'(0) which = 1/4 ._.

So apparently f^-1(x)=0...

Are you using 1/f'(f^-1(x))?

Nice try, kid. You're gonna have to try harder than that to trip me up. Taking the derivative at 0 is silly, considering you're looking at a, not 0. Derp.

The formula is d(f^-1(y))/dy = 1/f'(f^-1(y))

f^-1(y) = x

so.

dx/dy = 1/(dy/dx)

That's how it works. ._.

[Waker]

Nice try, kid. You're gonna have to try harder than that to trip me up. Taking the derivative at 0 is silly, considering you're looking at a, not 0. Derp.

The formula is d(f^-1(y))/dy = 1/f'(f^-1(y))

f^-1(y) = x

so.

dx/dy = 1/(dy/dx)

That's how it works. ._.

Umm. Here's how I did it:

Note - Used d/dx[f^-1(x)]=1/f'(f^-1(x))

Found inverse of original equation at a=-1.

-1=y³+4y-1

0=y³+4y ==> 0=y(y²+4) so y=0

Then f'(0)=2x²+4 evaluates to 4. So uh. 1/f'(0)=1/4


EDIT: UUUUHHH. About the f^-1(x) in there. Isn't it implied that the x and y will be switched?

Edited by Waker, 04 October 2010 - 11:07 PM.

[Dion]

Simple question;

How do you factor a third degree polynomial such as

x^3 - x^2 + x - 2 = 4

Without going through extraordinary measures and taking cube roots and square roots everywhere?

[Waker]

Simple question;

How do you factor a third degree polynomial such as

x^3 - x^2 + x - 2 = 4

Without going through extraordinary measures and taking cube roots and square roots everywhere?

Get zero on one side. Then split factor~ That is, split the side opposite 0 into two parts, factor those. For example, with

x³-x²-2x+2 = 0

x²(x-1)-2(x-1) = 0 easier to factor --> (x²-2)(x-1).

[Dion]

Get zero on one side. Then split factor~ That is, split the side opposite 0 into two parts, factor those. You'll usually have something like x²(x-1)-2(x-1) which is easier to factor --> (x²-2)(x-1).

Your math is wrong, obviously.

[Dion]

Get zero on one side. Then split factor~ That is, split the side opposite 0 into two parts, factor those. For example, with

x³-x²-2x+2 = 0

x²(x-1)-2(x-1) = 0 easier to factor --> (x²-2)(x-1).

The problem is that the equation is x³-x²-2x+2 is not the equation in question. :v

Edited by Dion, 06 October 2010 - 07:55 PM.

[Secret Igshar]

Simple question;

How do you factor a third degree polynomial such as

x^3 - x^2 + x - 2 = 4

Without going through extraordinary measures and taking cube roots and square roots everywhere?

Derp derp. You just gotta use your p's over q's.

Teach Me Mogster Lesson π: p's and q's

A p is any factor of the constant term, in this case, -6.
A q is any factor of the lead coefficient, in this case, 1.

You put your p's over q's and then try out EVERY one of those in trial division until you find one which results in zero.

Factors of p: {-1, -2, -3, -6, 1, 2, 3, 6}
Factors of q: {-1, 1}

Teach Me Mogster Lesson φ: Trial Division:

x³ - x² + x - 6 = 0

Change this to look like so:

_| 1 -1 1 -6

In the upper left box, you put in the test number, say 2.

[U]2[/U]| 1 -1  1 -6
   [U]- +2 +2 +6[/U]
   1  1  3  0

[spoiler=More spelled out for explanatory purposes]
Drop the first number down to below the line.

[U]2[/U]| 1 -1  1 -6
   [U]-         [/U]
   1

Multiply the most recent number below the line by your trial number (2) and add it to the NEXT number in line. (Note: 1*2=2) Write the result below the line.

[COLOR=RoyalBlue][U]2[/U][/COLOR]| 1 [COLOR=Yellow]-1[/COLOR]  1 -6
   [U]- +[COLOR=DarkOrchid]2[/COLOR]      [/U]
   [COLOR=Red]1[/COLOR]  [COLOR=White]1[/COLOR]

Repeat this process for the next number in line.

 [COLOR=RoyalBlue][U]2[/U][/COLOR]| 1 -1  [COLOR=Yellow]1[/COLOR] -6
    [U]- +2 +[COLOR=DarkOrchid]2[/COLOR]   [/U]
    1  [COLOR=Red]1[/COLOR]  [COLOR=White]3[/COLOR]

Repeat for the next number in line.

  [COLOR=RoyalBlue][U]2[/U][/COLOR]| 1 -1  1 [COLOR=Yellow]-6[/COLOR]
     [U]- +2 +2 +[COLOR=DarkOrchid]6[/COLOR] [/U]
     1  1  [COLOR=Red]3[/COLOR]  [COLOR=White]0[/COLOR]

Because the result of the FINAL step is 0, you found a root of the equation.
[/spoiler]

Uh, tbh, I actually just picked one and it worked. I guess that's a Math Major for you. Anyway~

With this, you now know that x=2 is a zero of the function, so now you have a new function:

(x-2)(x² + x + 3) = 0

Then you would continue and do another trial division on the second bit (or just factor it), to find that it is a prime polynomial. HOW I KNOW THIS BY LOOKING!?

1² - 4(1)(3) [Discriminant]

is a negative value, meaning that the polynomial is never equal to zero (has imaginary roots). Another way to see this is that you are adding x² + x. No matter how largely negative x is, x² will always be larger and positive, thus the quantity x² + x is always positive. Add to that a positive integer (3) and you will always have a positive value, never zero.


So, in summation, the factored form of

x³ - x² + x - 6 = 0

is

(x-2)(x² + x + 3) = 0

QED, bitches.

[Waker]

Uff. HEY GUYS. Why is when xy=4 have dy/dt=-5/8 when x=8 and you are given dx/dt=10?

When I derive it (see: xy=4) I get (dx/dt)(dy/dt)=0 which equates to 10(dy/dt)=0. In this case dy/dt ought to be 0...

[Secret Igshar]

Uff. HEY GUYS. Why is when xy=4 have dy/dt=-5/8 when x=8 and you are given dx/dt=10?

When I derive it (see: xy=4) I get (dx/dt)(dy/dt)=0 which equates to 10(dy/dt)=0. In this case dy/dt ought to be 0...

Product rule, fool.

d/dt (xy = 4) =

y(dx/dt) + x(dy/dt) = 0

x=8 so y=0.5
dx/dt=10 so~

(0.5)(10) + 8(dy/dt) = 0
5 = -8(dy/dt)
dy/dt = -5/8

:>

[Waker]

Product rule, fool.

d/dt (xy = 4) =

y(dx/dt) + x(dy/dt) = 0

x=8 so y=0.5
dx/dt=10 so~

(0.5)(10) + 8(dy/dt) = 0
5 = -8(dy/dt)
dy/dt = -5/8

:>

O RITE, product rule. Forgot they were two different variables. Gorram I am going to almost fail Ap Carucurus if I keep forgetting these kinds of things.

[Dion]

In the upper left box, you put in the test number, say 2.

Beautifully explained. Better than my teacher. But one issue. How do we randomly decide to use 2? (It being the only solution that works) .-.