anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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amistre64
  • amistre64
they voth mean basically the same theing, it is a notation that reads: the derivative with respect to "variable".
nowhereman
  • nowhereman
You could also say derivation in direction of some dimension.
amistre64
  • amistre64
dy/dt tends to mean: the derivative of y with respect to t (for time)

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anonymous
  • anonymous
so what if it's dx/dt = 20/sqrt(t)? then it asks to find the antiderivative given x(1) = 40
amistre64
  • 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
  • amistre64
initial conditions helpt to anchor the family of curves that is produced to a single curve.
anonymous
  • anonymous
dy/d(something) just means the rate of change of 'y'(here) with respect to 'something'
amistre64
  • 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
  • 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
  • amistre64
take out the constant, and change the radical to an appropriate exponent
anonymous
  • anonymous
so is dx/dt the derivative of a function called x? which in this case would be 20/sqrt(t)?
nowhereman
  • nowhereman
You should also know, that 1/sqrt(t) is the the derivative of 2\sqrt(t)
amistre64
  • amistre64
20 times the integral of 1/t^(1/2)
amistre64
  • amistre64
20 times the integral of t^(-1/2)
amistre64
  • amistre64
20 (1/2)(sqrt(t) = 10sqrt(t)
anonymous
  • anonymous
my book says the answer is 40*sqrt(t)
amistre64
  • 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
  • amistre64
10sqrt(1) + C = 40
amistre64
  • amistre64
if we derive 10sqrt(t) we get -> 10/2 sqrt(t)..right? so yeah, my mistake :)
amistre64
  • 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
  • anonymous
correct
amistre64
  • amistre64
now this X(t) function is floating around....it needs an anchor...it needs a +C to hold it in place.
amistre64
  • 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
  • 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
  • amistre64
poke: yes..
anonymous
  • anonymous
so why isnt the final answer 40/sqrt(t)??
amistre64
  • 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
  • amistre64
40t^(1/2) = 40sqrt(t)
amistre64
  • amistre64
t^(-1/2 + 2/2) 20t^(1/2) 20 * ---------------- = --------- = 40t^(1/2) -1/2 + 2/2 1/2
amistre64
  • amistre64
which equals 40sqrt(t)
anonymous
  • anonymous
ok... you changed the 1/sqrt(t) to 2/sqrt(t) and i got lost at that step. it makes sense now.
amistre64
  • amistre64
its just reversing the steps of taking the derivative
anonymous
  • 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
  • amistre64
thats correct
amistre64
  • 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
  • anonymous
no
amistre64
  • amistre64
take the derivative of both of them...what do you get?
anonymous
  • anonymous
one is x^2 +4x and the other is x^2-17x
amistre64
  • amistre64
not the integral....the derivative...
anonymous
  • anonymous
o... both would be 2
amistre64
  • amistre64
thats right.... then how do we know which curve is our answer when we suit them up with the integrals?
amistre64
  • amistre64
dy/dx = 2 y = 2x + C right?
amistre64
  • amistre64
where C can be any constant, any number
anonymous
  • anonymous
right
amistre64
  • amistre64
our integral is floating around..it needs to be anchored to a specific point in order to help us out
anonymous
  • anonymous
in the example i originally gave the C would be 0. right?
amistre64
  • 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
  • anonymous
u would do 40sqrt(1) + C = 40
amistre64
  • amistre64
40 = 40sqrt(1) + C 40 = 40 + C 0 = C .... so yes, in this probelm C would be zero
amistre64
  • amistre64
what would our answer be if we wanted X(1) = 20?
anonymous
  • anonymous
1/2?
amistre64
  • 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
  • anonymous
i mean -20
anonymous
  • anonymous
ya. i divided it instead
amistre64
  • amistre64
:) happens to me all the time lol
amistre64
  • amistre64
does this make better sense to you now?
anonymous
  • 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
  • amistre64
good luck, Ciao! :)

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