# integration problemWatch

This discussion is closed.
#1
I need to find the value of C from the equation

x = 2t + 5t squared + C

I would be grateful if anyone can give me a hand.
0
13 years ago
#2
when you integrate you got x = 2t - 5t^2 + C i presume is what you mean...so the C is obtrained by pluging in the initial conditions
0
13 years ago
#3
I need to find the value of C from the equation
x = 2t + 5t squared + C

I would be grateful if anyone can give me a hand.
you need one value of x at a certain value of t
it's impossible to find the value of C like this!
0
#4
when you integrate you got x = 2t - 5t^2 + C i presume is what you mean...so the C is obtrained by pluging in the initial conditions

Sorry to be thick but I am still none the wiser!

The original equation was dx/dt = 2 +10t

I have got as far as x = 2t + 5t^2 + C but can't see how to finish.

I would be grateful if you could walk me through it.
0
13 years ago
#5
I think that's enough.

x = 2t + 5t^2 + C

You don't always need to find C, just leave it as that.
0
13 years ago
#6
(Original post by mattj35)
Sorry to be thick but I am still none the wiser!

The original equation was dx/dt = 2 +10t

I have got as far as x = 2t + 5t^2 + C but can't see how to finish.

I would be grateful if you could walk me through it.
You can't do anything else unless the question tells you 'at x=something, t=something'. Remember whenever you differentiate you lose any constants so in order to get it back when you integrate you need a point on the curve. You can't find the exact equation of a curve if you only know its rate of change.
0
13 years ago
#7
C can be anything. All such functions will satisfy your differential equation.

All varying C does is move your graph up and down vertically - that won't change the gradient for a given value of x.

You'd need to know, in order to uniquely determine C, that your graph went through a particular point. This information often comes in the form of an initial condition
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#8
(Original post by RichE)
C can be anything. All such functions will satisfy your differential equation.

All varying C does is move your graph up and down vertically - that won't change the gradient for a given value of x.

You'd need to know, in order to uniquely determine C, that your graph went through a particular point. This information often comes in the form of an initial condition
sorry I forgot to write that x is metres and t = 3
0
#9
(Original post by mattj35)
sorry I forgot to write that x is metres and t = 3

the full question says,

dx/dt = 2 + 10t

where x metres is the distance fallen in time t seconds.

Solve the differential equation and find x when t = 3.
0
13 years ago
#10
(Original post by mattj35)
the full question says,

dx/dt = 2 + 10t

where x metres is the distance fallen in time t seconds.

Solve the differential equation and find x when t = 3.
If x is the distance fallen in t seconds then I presume when t=0, x=0.
0
13 years ago
#11
does the question say: "from rest" or something like that?
0
13 years ago
#12
(Original post by Gaz031)
If x is the distance fallen in t seconds then I presume when t=0, x=0.
Depends if the equation suggests a change in time and distance from 0->t and 0->x respectively, or (what I would think) that its just defining the variables by stating that. So, I dont think that it really tells us anything.

Is this a second part of a question perhaps?
0
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