in calculus, what does the d/dx and d/dt mean? what's the difference? i am starting integration and am confused as to what the problem is asking for

- anonymous

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- amistre64

they voth mean basically the same theing, it is a notation that reads: the derivative with respect to "variable".

- nowhereman

You could also say derivation in direction of some dimension.

- amistre64

dy/dt tends to mean: the derivative of y with respect to t (for time)

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## More answers

- anonymous

so what if it's dx/dt = 20/sqrt(t)? then it asks to find the antiderivative given x(1) = 40

- amistre64

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

- amistre64

initial conditions helpt to anchor the family of curves that is produced to a single curve.

- anonymous

dy/d(something) just means the rate of change of 'y'(here) with respect to 'something'

- amistre64

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...

- nowhereman

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)

- amistre64

take out the constant, and change the radical to an appropriate exponent

- anonymous

so is dx/dt the derivative of a function called x? which in this case would be 20/sqrt(t)?

- nowhereman

You should also know, that 1/sqrt(t) is the the derivative of 2\sqrt(t)

- amistre64

20 times the integral of 1/t^(1/2)

- amistre64

20 times the integral of t^(-1/2)

- amistre64

20 (1/2)(sqrt(t) = 10sqrt(t)

- anonymous

my book says the answer is 40*sqrt(t)

- amistre64

poke: yes, dx/dt is the derivative of a funtion X(t) find X(t) and input your initial condition that x(1) = 40

- amistre64

10sqrt(1) + C = 40

- amistre64

if we derive 10sqrt(t) we get -> 10/2 sqrt(t)..right? so yeah, my mistake :)

- amistre64

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)

- anonymous

correct

- amistre64

now this X(t) function is floating around....it needs an anchor...it needs a +C to hold it in place.

- amistre64

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

- anonymous

so when you pull the 20 out originally you have 20*(1/sqrt(t))... then the antideriv of that is 20*(2/sqrt(t))?

- amistre64

poke: yes..

- anonymous

so why isnt the final answer 40/sqrt(t)??

- amistre64

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

- amistre64

40t^(1/2) = 40sqrt(t)

- amistre64

t^(-1/2 + 2/2) 20t^(1/2)
20 * ---------------- = --------- = 40t^(1/2)
-1/2 + 2/2 1/2

- amistre64

which equals 40sqrt(t)

- anonymous

ok... you changed the 1/sqrt(t) to 2/sqrt(t) and i got lost at that step. it makes sense now.

- amistre64

its just reversing the steps of taking the derivative

- anonymous

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?

- amistre64

thats correct

- amistre64

but we have to finish our problem here.....
tell me, do these curves have the same derivative:
y= 2x+4 and y= 2x-17

- anonymous

no

- amistre64

take the derivative of both of them...what do you get?

- anonymous

one is x^2 +4x and the other is x^2-17x

- amistre64

not the integral....the derivative...

- anonymous

o... both would be 2

- amistre64

thats right.... then how do we know which curve is our answer when we suit them up with the integrals?

- amistre64

dy/dx = 2
y = 2x + C right?

- amistre64

where C can be any constant, any number

- anonymous

right

- amistre64

our integral is floating around..it needs to be anchored to a specific point in order to help us out

- anonymous

in the example i originally gave the C would be 0. right?

- amistre64

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?

- anonymous

u would do 40sqrt(1) + C = 40

- amistre64

40 = 40sqrt(1) + C
40 = 40 + C
0 = C .... so yes, in this probelm C would be zero

- amistre64

what would our answer be if we wanted X(1) = 20?

- anonymous

1/2?

- amistre64

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?

- anonymous

i mean -20

- anonymous

ya. i divided it instead

- amistre64

:) happens to me all the time lol

- amistre64

does this make better sense to you now?

- anonymous

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!

- amistre64

good luck, Ciao! :)

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