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they voth mean basically the same theing, it is a notation that reads: the derivative with respect to "variable".
You could also say derivation in direction of some dimension.
dy/dt tends to mean: the derivative of y with respect to t (for time)
so what if it's dx/dt = 20/sqrt(t)? then it asks to find the antiderivative given x(1) = 40
all it means is that the amount of change in the top part depends on the amount of change that occurs to the variable inthe bottom part
initial conditions helpt to anchor the family of curves that is produced to a single curve.
dy/d(something) just means the rate of change of 'y'(here) with respect to 'something'
x' = 20/sqrt(t) means that there is a function X(t) that this came from.....and they want you to know where it came from...
Yes but as integration is the inverse operation of differentiation he is right, that in order to find x(t) he has to integrate dx/dt which in that example is 20/sqrt(t)
take out the constant, and change the radical to an appropriate exponent
so is dx/dt the derivative of a function called x? which in this case would be 20/sqrt(t)?
You should also know, that 1/sqrt(t) is the the derivative of 2\sqrt(t)
20 times the integral of 1/t^(1/2)
20 times the integral of t^(-1/2)
20 (1/2)(sqrt(t) = 10sqrt(t)
my book says the answer is 40*sqrt(t)
poke: yes, dx/dt is the derivative of a funtion X(t) find X(t) and input your initial condition that x(1) = 40
10sqrt(1) + C = 40
if we derive 10sqrt(t) we get -> 10/2 sqrt(t)..right? so yeah, my mistake :)
ahhh.... i see my mistake :) i did it in me head for starters. the reciprocal of 1/2 = 2..... 20(2) sqrt(t) = 40sqrt(t)
now this X(t) function is floating around....it needs an anchor...it needs a +C to hold it in place.
40sqrt(t) + C is what you want.... plug in your initial condition that X(1) = 40 and solve for "C" then you have the right answer
so when you pull the 20 out originally you have 20*(1/sqrt(t))... then the antideriv of that is 20*(2/sqrt(t))?
so why isnt the final answer 40/sqrt(t)??
you add one to the exponent, then put that number as the denominator and also use that number as the new exponent. -1/2 + 2/2 = 1/2
40t^(1/2) = 40sqrt(t)
t^(-1/2 + 2/2) 20t^(1/2) 20 * ---------------- = --------- = 40t^(1/2) -1/2 + 2/2 1/2
which equals 40sqrt(t)
ok... you changed the 1/sqrt(t) to 2/sqrt(t) and i got lost at that step. it makes sense now.
its just reversing the steps of taking the derivative
so basically when it says like dy/dt that means there is a function called y and the variable used will be t? such as dy/dt = 5t-4?
but we have to finish our problem here..... tell me, do these curves have the same derivative: y= 2x+4 and y= 2x-17
take the derivative of both of them...what do you get?
one is x^2 +4x and the other is x^2-17x
not the integral....the derivative...
o... both would be 2
thats right.... then how do we know which curve is our answer when we suit them up with the integrals?
dy/dx = 2 y = 2x + C right?
where C can be any constant, any number
our integral is floating around..it needs to be anchored to a specific point in order to help us out
in the example i originally gave the C would be 0. right?
we have your problem.. X = 40sqrt(t) + C ....... how do we find the right equation of the curve? when we are given the X(1) = 40?
u would do 40sqrt(1) + C = 40
40 = 40sqrt(1) + C 40 = 40 + C 0 = C .... so yes, in this probelm C would be zero
what would our answer be if we wanted X(1) = 20?
20 = 40sqrt(1) + C 20 = 40 + C 20 -40 = C -20 = C X = 40sqrt(t) - 20 is the curve wed want right? you see how I did that?
i mean -20
ya. i divided it instead
:) happens to me all the time lol
does this make better sense to you now?
ya. thanks for your help. i have to run off to class and take a quiz over this stuff. i should be fine now. thanks again!
good luck, Ciao! :)