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C1- Integration help!

Can someone help me with the following two questions please :smile:

dxdt=3t2−2t+1\frac{dx}{dt}=3t^2-2t+1 and that x=2x=2 when t=1t=1, find the value of xx when t=2t=2

Right so the question comes under the topic of integration but I don't really understand what its asking me. If I was doing this in the exam I wouldn't know that I would have to integrate. So once I have integrated this. I got

t3−t2+x+C t^3-t^2+x+C What do I do next?

Next one is dxdt=(t+1)2\frac{dx}{dt}=(t+1)^2 and the x=0x=0 when t=2t=2, find the value of xx when t=3t=3. I think if I knew how to do the first question I should be able to do the second one. Any help would be appreciated. Thanks :smile: P.S how would I know that I would have to integrate this?
(edited 10 years ago)
Reply 1
Avatar for the bear
the bear
20
when you integrate you must include a constant c
Reply 2
Avatar for tmorrall
tmorrall
You need to recognise the question gives you dx/dt which is the the differential with respect to t of y.
So you integrate dx/dt with respect to t and you should acquire t^3+t^2+t+k. You have t^3+t^2+x which shows you do not know what the actual process of integration is doing. Rewrite dx/dt if it make it more familiar to have say dy/dx. If you don't have limits to integrate between then you must assume there is some constant (k) which can be zero. Since x had been differentiated any constant will be lost in this process when integrating we must assume there was one there.


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Reply 3
Avatar for Super199
Super199
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Original post by the bear
when you integrate you must include a constant c

DAMN IT! I completely forgot
Reply 4
Avatar for Super199
Super199
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Original post by tmorrall
You need to recognise the question gives you dx/dt which is the the differential with respect to t of y.
So you integrate dx/dt with respect to t and you should acquire t^3+t^2+t+k. You have t^3+t^2+x which shows you do not know what the actual process of integration is doing. Rewrite dx/dt if it make it more familiar to have say dy/dx. If you don't have limits to integrate between then you must assume there is some constant (k) which can be zero. Since x had been differentiated any constant will be lost in this process when integrating we must assume there was one there.


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Right I understood that part, but what is it asking me to do when it gives values of x and t?
Reply 5
Avatar for tmorrall
tmorrall
Original post by Super199
Right I understood that part, but what is it asking me to do when it gives values of x and t?


So if you integrate dx/dt with respect to t you get the function of x so now you know x=t^3+t^2+t+k. You know a value for t when x is whatever. Sub them both in solve for your k so now you know the full equation since you know the constant k now. You can then do the next part. Note also that you have integrated with respect to t so the when you integrate the 1 it becomes t not x. If you had dy/dx is 10x+1 integrate with respect to x you get 5x^2+x+k not 5x^2+y+k :smile:


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(edited 10 years ago)
Reply 6
Avatar for Super199
Super199
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Original post by tmorrall
So if you integrate dx/dt with respect to t you get the function of x so now you know x=t^3+t^2+t+k. You know a value for t when x is whatever. Sub them both in solve for your k so now you know the full equation since you know the constant k now. You can then do the next part. Note also that you have integrated with respect to t so the when you integrate the 1 it becomes t not x. If you had dy/dx is 10x+1 integrate with respect to x you get 5x^2+x+k not 5x^2+y+k :smile:


Posted from TSR Mobile

So if I sub 1 for t. 13−12+1+c?1^3-1^2+1+c? What do I do about the x? Where would I sub x =2?
Reply 7
Avatar for tmorrall
tmorrall
Original post by Super199
So if I sub 1 for t. 13−12+1+c?1^3-1^2+1+c? What do I do about the x? Where would I sub x =2?


You know that x=t^3-t^2+t+c at the point where t=1 AND x=2 so replace both your t and your x with their corresponding values and you get 2=1^3-1^2+1+c
2=1-1+1+c
c=1
So now you know the constant you have an equation which you know everything about: x=t^3-t^2+t+1

I hope I am not coming across too patronising. Trying to be as thorough as possible :smile:


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Reply 8
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Super199
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Original post by tmorrall
You know that x=t^3-t^2+t+c at the point where t=1 AND x=2 so replace both your t and your x with their corresponding values and you get 2=1^3-1^2+1+c
2=1-1+1+c
c=1
So now you know the constant you have an equation which you know everything about: x=t^3-t^2+t+1

I hope I am not coming across too patronising. Trying to be as thorough as possible :smile:


Posted from TSR Mobile

Oh wow you my friend are a lad!!!! I understand wow thanks. I got x = 7 as my answer :smile:. Thanks man! :smile:
Reply 9
Avatar for tmorrall
tmorrall
Original post by Super199
Oh wow you my friend are a lad!!!! I understand wow thanks. I got x = 7 as my answer :smile:. Thanks man! :smile:


Yeah thats it. The more you do calculus the greater your understanding will become. Arguably most applicable maths to real life this stuff!
Haha yeah no problem


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Reply 10
Avatar for Super199
Super199
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Original post by majmuh24
Integration is the inverse of differentiation, so if you have a derivative dxdt\dfrac{dx}{dt} and want to find a function for x in terms of t, you have to integrate it with respect to t.

To find C, substitute the values that you have been given so x=2 and t=1, so just put these values into the equation you have from integrating the derivative.

Help me with this one please :smile:
y12=x13+3y^\frac{1}{2}=x^\frac{1}{3}+3
a). Show that y=x23+Ax13+By=x^\frac{2}{3}+Ax^\frac{1}{3}+B where A and B are constants to be found
B). Hench find the integral of ydx.

How do I start?
Original post by Super199
Help me with this one please :smile:
y12=x13+3y^\frac{1}{2}=x^\frac{1}{3}+3
a). Show that y=x23+Ax13+By=x^\frac{2}{3}+Ax^\frac{1}{3}+B where A and B are constants to be found
B). Hench find the integral of ydx.

How do I start?


Well, what would you have to do to turn y12y^{\frac{1}{2}} into yy
Reply 12
Avatar for Super199
Super199
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Original post by TenOfThem
Well, what would you have to do to turn y12y^{\frac{1}{2}} into yy

square it?
(x13)2=x23(x^\frac{1}{3})^2=x^\frac{2}{3} laws of indices. Then 32=93^2=9 Where is the Ax13Ax^\frac{1}{3} from?
(edited 10 years ago)
Reply 13
Avatar for davros
davros
16
Original post by Super199
square it?
(x13)2=x23(x^\frac{1}{3})^2=x^\frac{2}{3} laws of indices. Then 32=93^2=9 Where is the Ax13Ax^\frac{1}{3} from?


Well what do you get when you expand (a+b)2(a+b)^2 ?
Original post by Super199
square it?
(x13)2=x23(x^\frac{1}{3})^2=x^\frac{2}{3} laws of indices. Then 32=93^2=9 Where is the Ax13Ax^\frac{1}{3} from?


square the RHS too
Reply 15
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Super199
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Original post by davros
Well what do you get when you expand (a+b)2(a+b)^2 ?

Right got it! Cheers :smile:

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