Thread: Fancy Maths Stuff!

Originally Posted by Xei

Invader: It pops up more when you integrate a separable differential equation or do integration by substituting the differential at the end of the integral.

And nah, more like )(ei. That's how pretty much everyone writes maths xs, otherwise they look like multiplication. I use . for multiplication though so it's fine.

Also do you do lines through your 7s? :V

Hrmmm ok. I've seen them drawn like that sometimes depending on the teacher. We only use x to represent multiplication in elementary school, but once the variable is introduced we use the dot for multiplication the way you do.
And yes, I draw lines through my 7's, though it's not common for people to do that here. I don't do it to my z's. My z's are so angled that they can't possibly be confused for 2s. Evar.

Just a comment on the discussion of the word gradient: as far as I know, gradient is a term used in multivariable calculus -- slope is used when there is only 1 variable.

More specifically, the "value" of the gradient at a given point is a vector whose components are the partial derivatives with respect to the first variable, the second, etc, so more generally the gradient is a vector field (as Wikipedia says). According to what I've been told and what I can tell, the term gradient doesn't really make sense in 1D (you would get a 1-component vector), though I guess it's pretty frequent to use the term "gradient" in such a situation to mean slope.

Originally Posted by Wildman

Just a comment on the discussion of the word gradient: as far as I know, gradient is a term used in multivariable calculus -- slope is used when there is only 1 variable.

More specifically, the "value" of the gradient at a given point is a vector whose components are the partial derivatives with respect to the first variable, the second, etc, so more generally the gradient is a vector field (as Wikipedia says). According to what I've been told and what I can tell, the term gradient doesn't really make sense in 1D (you would get a 1-component vector), though I guess it's pretty frequent to use the term "gradient" in such a situation to mean slope.

Yea, it's introduced with vector analysis as part of Calculus III, at least in my experience.

Over here we tend to call that 'grad'. Gradient is usually used to refer to the one dimensional concept. I've never heard slope used for anything.

I'm impressed that you do vector calculus at school.

Originally Posted by Xei

I'm impressed that you do vector calculus at school.

It's not too bad, we only ever did calculations concerning vector fields, which is fairly elementary to vector analysis. It's useful for things like determining the rotation of a hurricane over a certain period of time as it moved across some distance or figuring out why they rotate counterclockwise in the northern hemisphere and move through the ocean clockwise and vice versa for the southern hemisphere. Other common applications are things like velocity fields, gravitational fields, and electric force fields. There's a lot of integrals involved. The definition of a gradient in two variables goes something like this:

Let z = f(x,y) be a function of x and y such that f_x and f_y exist. Then the gradient of f, denoted by 𝛁f(x,y), is the vector

𝛁f(x,y) = f_x(x,y)i + f_y(x,y)j

𝛁f is read "del f" and yea, another notation for gradient is "grad f(x,y)."

Huh, wow. We only met vector calculus at the end of our first year of university, and this is in a mathematics course.

We covered multiple integrals, line integrals, surface integrals, and the main theorems.

I enjoy math a lot, but as I've not yet taken a calculus course, I won't be able to contribute to this thread. However, I will say I put lines through my z's and 7's!

To be honest the question was explicitly not about calculus.

I don't think a basic college-level algebra mind could have handled that problem. That's how I feel about it anyway, I'm not sure where to start or what I should end up with.

BTW I love math.

Originally Posted by Phion

(b-m)^2 - 4(1)(c-y_0-mx_0) = 0

Just found this thread. In case anybody wants to follow along with Phion's solution (which I recommend, it's quite elegant), he made a typo when taking the discriminant.

It ends up being

(b - m)² - 4a(c + mx₀ - y₀).

Solving for m in that equation gives m = 2ax₀ + b which is correct. Phion's typo leaves it missing the factor of a and (I'm pretty sure, haven't checked it) with a complex gradient.

I'm confused?

How is this more complicated than x = (-b +- sqrt(b^2 - 4a(c - y)) / 2a?

Where is the information pertaining to the slope contained in that equation?

EDIT:
Never mind. I see it. You can do

2ax + b = sqrt(b² - 4a(c-y))

but that leaves with with an ugly radical to deal with and how would you have known, without already knowing, that 2ax + b is the slope. And why should sqrt(b² - 4a(c-y)) be the slope too?

edit:
How did you get the equation. It looks like you just did:

ax² + bx + c = y
ax² + bx + (c - y) = 0

and then plugged that into the quadratic equation. So yeah, it's pure luck that you can even get the equation for slope out of that (even if it is set to some ridiculous radical).