Thread: Equations 2

is it the gradient of the slope at x=5 or something that weve to find?

The value of the slope at x = 5. Rise over run.

no idea what that means so im going to do what i think it is

ok, im pretty darn sure i iether did the wrong thing, or i made a mistake somewhere, coz im shit at maths, but this is what i got.

-435/20 thats it as a fraction, but as a number its -21.75

id love to know how your supposed to do it though

Simple quotient rule, I believe. vdu/dx is 0 at x=5. The denominator tends to 0 as x tends to 5, and the numerator, x(9x3 +x2 -10x +25) has no (x-5) factor so we have an asymptote?

Here's something a bit less formulaic:

(3x2 +1)1/2 -2x1/2 +x -1 = 0

Indicies are on the right, so (3x2 +1)1/2 is root(3xsquared +1), etcetera.

Originally Posted by Xei

Simple quotient rule, I believe. vdu/dx is 0 at x=5. The denominator tends to 0 as x tends to 5, and the numerator, x(9x3 +x2 -10x +25) has no (x-5) factor so we have an asymptote?

Here's something a bit less formulaic:

(3x2 +1)1/2 -2x1/2 +x -1 = 0

Indicies are on the right, so (3x2 +1)1/2 is root(3xsquared +1), etcetera.

bloody hell, is vdu/dx the derivative of y(what i knew to be dy/dx) coz i got dy/dx to be what ever it was i said earlier

You've got one polynomial, u = 9x3 +x2 -10x +25, divided by another, v = x -5.

I'm guessing you just divided the differential of the top by the differential of the bottom?

Unfortunately you can't do that. You have to use the quotient rule, which says that d(u/v)/dx = (vdu/dx - udv/dx) / v2.

y = (9x^3 + x^2 - 10x + 25) / (x - 5)

x can't be 5, you'd be dividing by zero [5-5 = 0]. Therefore, there is no slope or a graph for that matter.

Or am I meddling in some advanced, hi-tech, difficult-to-understand calculus concept here?

no, i brougt the bottom of the right hand side over to the left(to make things easier for me) then seperately brough x over to the right then -5 over or something like that, it was a bit more complicated but. theni simplified it a bit then diferentiated it, then substituted x = 5 in and i got that answer.

but i was finding the gradient at the point x=5, i dont know if that was what i was supposed to do anyway

Originally Posted by ThreeLetterSyndrom

y = (9x^3 + x^2 - 10x + 25) / (x - 5)

x can't be 5, you'd be dividing by zero [5-5 = 0]. Therefore, there is no slope or a graph for that matter.

Or am I meddling in some advanced, hi-tech, difficult-to-understand calculus concept here?

holy shit he is right, its so god dam obvious i didnt even think about that

no, i brougt the bottom of the right hand side over to the left(to make things easier for me) then seperately brough x over to the right then -5 over or something like that, it was a bit more complicated but. theni simplified it a bit then diferentiated it, then substituted x = 5 in and i got that answer.

but i was finding the gradient at the point x=5, i dont know if that was what i was supposed to do anyway

Yeah, but unfortunately the only way to do it is how I said. You can't multiply out the fraction and then differentiate both sides; the thing on the left would then be y(x-5) and I'm guessing you don't know how to differentiate that.

y = (9x^3 + x^2 - 10x + 25) / (x - 5)

x can't be 5, you'd be dividing by zero [5-5 = 0]. Therefore, there is no slope or a graph for that matter.

Or am I meddling in some advanced, hi-tech, difficult-to-understand calculus concept here?

Not bad actually. Just one thing; you have to check that the top doesn't divide by (x-5). For example,

x2 -8x +15 / x -5

=

(x -5)(x -3) / x -5

= (x -3)

Which doesn't go off to infinity at 5, or any number for that matter.

You can check the polynomial doesn't divide by x-5 because if it did, if you put x=5 in, it would = 0, which it doesn't.

haha I did this stuff senior year of hs Yes even if you take the derivitive it is divided by 0...unless this is another one of those rules I have forgotten!

Originally Posted by ThreeLetterSyndrom

y = (9x^3 + x^2 - 10x + 25) / (x - 5)

x can't be 5, you'd be dividing by zero [5-5 = 0]. Therefore, there is no slope or a graph for that matter.

Or am I meddling in some advanced, hi-tech, difficult-to-understand calculus concept here?

Uhm... that part isn't calculus, it's discrete algebra. It still has a slope.

Xei, you don't need the quotient rule, you can simplify it a lot before going into any calculus at all. The only calc required is to find the actual slope.

Oh, crap. I accidentally made it more complicated than it was supposed to be. I forgot to multiply something when I created it. Let's simplify it instead of dealing with all those fractions.

Solve for slope at x = 5

Code:

y = (x^2 - 10x + 25) / (x - 5)

lulz ninja. now its much easier! its just derivative of x-5 = 1 so the slope is 1 everywhere

...you do need the quotient rule for the fudged version.

Obviously the above is just x -5 which has a gradient of 1.

plz do mein nau.

Originally Posted by Xei

Yeah, but unfortunately the only way to do it is how I said. You can't multiply out the fraction and then differentiate both sides; the thing on the left would then be y(x-5) and I'm guessing you don't know how to differentiate that.

yes it did come to y(x-5), but that is also yx-5y, so then i brought over 5, then i brought over x. i know i didnt have to make it as long winded as that, but i only did it becasue it was easier to think about it that way. it doesnt matter anyway, becasue it was the qrong answer and i cant do this new one either coz its pretty much the same so im not going to attempt it, im not going to further embarres myself

Yay, right. Sorry I screwed up the math


Here's one I got asked by a 10 year old when I was teaching robotics this summer. He wanted to know how a certain mathematical constant was derived. This is how, don't just recognize what the formula is, actually show how you did it Anyone who recognizes the formula can tell us the answer.

Solve.

Code:

4(∫(0, 1) (x^2 + y^2 = 1))

x^2 + y^2 dx/dy = 1

-x^2 dx = y^2 dy

-x/3 ^ 3 = y/3^3

y = -x 0,1

= -1?

I dont recognize the formula so I have no way of checking myself.

ninja are you gonna do mine or not?

It just requires a bit of thinking outside the box. Proper maths.

And eh... why are you trying to integrate an equation?

Originally Posted by Xei

ninja are you gonna do mine or not?

It just requires a bit of thinking outside the box. Proper maths.

And eh... why are you trying to integrate an equation?

isnt that what you are supppoosed to do, to get y

You integrate functions, not equations.

I think ninja is thinking of 4S(0,1)(1-x2)1/2dx

sorry, its just that im used to intergrating whenever i see that S thing

yeah I thought the s meant integrate...

It does. And you can't integrate an equation. Think about the kind of things you integrate; they're expressions.