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

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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! :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.