Super199
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Can someone help me with the following two questions please

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

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

Next one is \frac{dx}{dt}=(t+1)^2 and the x=0 when t=2, find the value of x when t=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 P.S how would I know that I would have to integrate this?
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the bear
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when you integrate you must include a constant c
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tmorrall
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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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Super199
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(Original post by the bear)
when you integrate you must include a constant c
DAMN IT! I completely forgot
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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?
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tmorrall
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(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


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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


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So if I sub 1 for t. 1^3-1^2+1+c? What do I do about the x? Where would I sub x =2?
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tmorrall
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(Original post by Super199)
So if I sub 1 for t. 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


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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


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Oh wow you my friend are a lad!!!! I understand wow thanks. I got x = 7 as my answer . Thanks man!
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tmorrall
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(Original post by Super199)
Oh wow you my friend are a lad!!!! I understand wow thanks. I got x = 7 as my answer . Thanks man!
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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Super199
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(Original post by majmuh24)
Integration is the inverse of differentiation, so if you have a derivative \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
y^\frac{1}{2}=x^\frac{1}{3}+3
a). Show that y=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?
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TenOfThem
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(Original post by Super199)
Help me with this one please
y^\frac{1}{2}=x^\frac{1}{3}+3
a). Show that y=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 y^{\frac{1}{2}} into y
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Super199
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(Original post by TenOfThem)
Well, what would you have to do to turn y^{\frac{1}{2}} into y
square it?
(x^\frac{1}{3})^2=x^\frac{2}{3} laws of indices. Then 3^2=9 Where is the Ax^\frac{1}{3} from?
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davros
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(Original post by Super199)
square it?
(x^\frac{1}{3})^2=x^\frac{2}{3} laws of indices. Then 3^2=9 Where is the Ax^\frac{1}{3} from?
Well what do you get when you expand (a+b)^2 ?
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TenOfThem
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(Original post by Super199)
square it?
(x^\frac{1}{3})^2=x^\frac{2}{3} laws of indices. Then 3^2=9 Where is the Ax^\frac{1}{3} from?
square the RHS too
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Super199
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(Original post by davros)
Well what do you get when you expand (a+b)^2 ?
Right got it! Cheers
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