Question

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

Answer 1

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

Answer 2

You could also say derivation in direction of some dimension.

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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...

Answer 9

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)

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

10sqrt(1) + C = 40

Answer 19

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

Answer 20

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)

Answer 21

correct

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

poke: yes..

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

which equals 40sqrt(t)

Answer 31

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

Answer 32

its just reversing the steps of taking the derivative

Answer 33

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?

Answer 34

thats correct

Answer 35

but we have to finish our problem here.....

tell me, do these curves have the same derivative:

y= 2x+4 and y= 2x-17

Answer 36

no

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

o... both would be 2

Answer 41

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

Answer 42

dy/dx = 2

y = 2x + C right?

Answer 43

where C can be any constant, any number

Answer 44

right

Answer 45

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

Answer 46

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

Answer 47

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?

Answer 48

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

Answer 49

40 = 40sqrt(1) + C

40 = 40 + C

0 = C .... so yes, in this probelm C would be zero

Answer 50

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

Answer 51

1/2?

Answer 52

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?

Answer 53

i mean -20

Answer 54

ya. i divided it instead

Answer 55

:) happens to me all the time lol

Answer 56

does this make better sense to you now?

Answer 57

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!

Answer 58

good luck, Ciao! :)