I've recently come across some equations for velocity and acceleration and they have at the end of them: dx/dt etc. and i was wondering if anyone could explain to me what they mean, because unfortunately i have no calculus knowledge and i have come to understand that its within the language of calculus. To do with derivatives i believe..

The Wikipedia entry on derivative is quite informative. I suggest you start there.

 

In short, the derivative of [math]x(t)[/math] with respect to [math]t[/math] is a measure of the rate at which [math]x[/math] changes with respect to [math]t[/math]. The Wikipedia entry should make this clear to you.

 

Come back to us if you have any specific questions.

The derivative is the slope of the line (i.e. the ratio ajb mentions), the integral is the area under a curve.

thanks for the link, it is helpful. Also, what does this mean: a^-1 = 1/a2 or just a..?

thanks for the link, it is helpful. Also, what does this mean: a^-1 = 1/a2 or just a..?

 

[imath]a^{-n} = \frac{1}{a^n}[/imath] so [imath]a^{-1} = \frac{1}{a}[/imath]

but why?, i still don't understand why dividing 1 into a = a^-1...

So, anything over itself(aside from zero or infinity) is equal to one. We can use that to cancel out things in fractions. Let us consider a²/a. We can factor out an a in the numerator to get [imath]\frac{aa}{a}[/imath] which, given the trick learned above, reduces to [imath]\frac{a}{1}=a[/imath]. So, a² divided by a¹ is a¹. It turns out that this generalizes and dividing like bases is simply a matter of subtracting the power of the base in the denominator by the power of the like base in the numerator which yields the remaining power of base after simplification. Since any number(besides zero or infinity) taken to the 0 power is 1, we can see why 1/a is the same as a^-1. Since a⁰=1, we can do the division with a⁰. 0-1=-1. That means the power of the base is -1.

Edited February 27, 201114 yr by ydoaPs

ok cheers.

Sorry, but in dx/dt, where does the d come from and what does it mean ?

Sorry, but in dx/dt, where does the d come from and what does it mean ?

 

Really it is just notation and is defined by the definition of the derivative.

 

Informally you can think of [math]dx[/math] as a "small change" in [math]x[/math] and similarly for [math]dt[/math]. Intuitively, the derivative the the ratio of these "small changes".

Ok, so d is anything to worry about then...it's just representing the derivative of the function or something like that.

 

And also how exactly do unit vectors fit into the equation so to speak..?

Ok, so d is anything to worry about then...it's just representing the derivative of the function or something like that.

 

And also how exactly do unit vectors fit into the equation so to speak..?

They are vectors of magnitude 1 in each direction of the co-ordinate system. If you have a three dimensional co-ordinate system, there are three unit vectors; one for x, y, and z axes. Any vector can be written as a sum of the unit vectors multiplied by three constants.

 

[math]\vec{v}=a\vec{i}+b\vec{j}+c\vec{k}[/math]

 

where v is a vector; a, b, and c are constants; and i, j, and k are unit vectors.