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anonymous
 5 years ago
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
 5 years ago
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

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0they voth mean basically the same theing, it is a notation that reads: the derivative with respect to "variable".

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0You could also say derivation in direction of some dimension.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0dy/dt tends to mean: the derivative of y with respect to t (for time)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so what if it's dx/dt = 20/sqrt(t)? then it asks to find the antiderivative given x(1) = 40

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0all 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
 5 years ago
Best ResponseYou've already chosen the best response.0initial conditions helpt to anchor the family of curves that is produced to a single curve.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dy/d(something) just means the rate of change of 'y'(here) with respect to 'something'

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x' = 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
 5 years ago
Best ResponseYou've already chosen the best response.0Yes 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
 5 years ago
Best ResponseYou've already chosen the best response.0take out the constant, and change the radical to an appropriate exponent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so is dx/dt the derivative of a function called x? which in this case would be 20/sqrt(t)?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0You should also know, that 1/sqrt(t) is the the derivative of 2\sqrt(t)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.020 times the integral of 1/t^(1/2)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.020 times the integral of t^(1/2)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.020 (1/2)(sqrt(t) = 10sqrt(t)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my book says the answer is 40*sqrt(t)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0poke: yes, dx/dt is the derivative of a funtion X(t) find X(t) and input your initial condition that x(1) = 40

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if we derive 10sqrt(t) we get > 10/2 sqrt(t)..right? so yeah, my mistake :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ahhh.... i see my mistake :) i did it in me head for starters. the reciprocal of 1/2 = 2..... 20(2) sqrt(t) = 40sqrt(t)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0now this X(t) function is floating around....it needs an anchor...it needs a +C to hold it in place.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.040sqrt(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
 5 years ago
Best ResponseYou've already chosen the best response.0so when you pull the 20 out originally you have 20*(1/sqrt(t))... then the antideriv of that is 20*(2/sqrt(t))?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so why isnt the final answer 40/sqrt(t)??

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.040t^(1/2) = 40sqrt(t)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0t^(1/2 + 2/2) 20t^(1/2) 20 *  =  = 40t^(1/2) 1/2 + 2/2 1/2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0which equals 40sqrt(t)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok... you changed the 1/sqrt(t) to 2/sqrt(t) and i got lost at that step. it makes sense now.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its just reversing the steps of taking the derivative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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 = 5t4?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0but we have to finish our problem here..... tell me, do these curves have the same derivative: y= 2x+4 and y= 2x17

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0take the derivative of both of them...what do you get?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one is x^2 +4x and the other is x^217x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0not the integral....the derivative...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0thats right.... then how do we know which curve is our answer when we suit them up with the integrals?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0dy/dx = 2 y = 2x + C right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0where C can be any constant, any number

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0our integral is floating around..it needs to be anchored to a specific point in order to help us out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in the example i originally gave the C would be 0. right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we 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
 5 years ago
Best ResponseYou've already chosen the best response.0u would do 40sqrt(1) + C = 40

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.040 = 40sqrt(1) + C 40 = 40 + C 0 = C .... so yes, in this probelm C would be zero

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0what would our answer be if we wanted X(1) = 20?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.020 = 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
 5 years ago
Best ResponseYou've already chosen the best response.0ya. i divided it instead

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0:) happens to me all the time lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0does this make better sense to you now?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ya. 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!
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