The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Solving Differential Equations by substitution

(FP2 AQA)

question5.PNG

Have I messed up yet? I'm repeating this question where I got it wrong (there is also no answer for this on the solutions)
Attachment not found
answer5.jpg485.1KB
What you've done doesn't look right, although I'm not entirely clear where one line starts and another begins at points.

I think you would make life easier for yourself if you wrote dt/dx in terms of 'x', not t; then when you differentiate the 2nd time (w.r.t. x), you just have −1x2dydt\dfrac{-1}{x^2} \dfrac{dy}{dt}; differentiating 1/x^2 is easy, and differentiating dy/dt (considered as a function of t) w.r.t x is just like differentiating y w.r.t. x, so it's essentially just doing what you did to relate dy/dx to dy/dt in the previous step.

It should only be about 5 lines of working in total if you make no mistakes.
Reply 2
Avatar for ckfeister
ckfeister
OP
19
Original post by DFranklin
What you've done doesn't look right, although I'm not entirely clear where one line starts and another begins at points.

I think you would make life easier for yourself if you wrote dt/dx in terms of 'x', not t; then when you differentiate the 2nd time (w.r.t. x), you just have −1x2dydt\dfrac{-1}{x^2} \dfrac{dy}{dt}; differentiating 1/x^2 is easy, and differentiating dy/dt (considered as a function of t) w.r.t x is just like differentiating y w.r.t. x, so it's essentially just doing what you did to relate dy/dx to dy/dt in the previous step.

It should only be about 5 lines of working in total if you make no mistakes.


The method your doing is different from the homestudy provider, have you got a video on this?
Original post by ckfeister
The method your doing is different from the homestudy provider, have you got a video on this?
No, why would I have a video? If you want to do it keeping things in terms of t, that's fine too - I just think it's a little more confusing in this case.
Reply 4
Avatar for ckfeister
ckfeister
OP
19
Original post by DFranklin
No, why would I have a video? If you want to do it keeping things in terms of t, that's fine too - I just think it's a little more confusing in this case.


How is this confusing... I also meant youtube.............................
Original post by ckfeister
How is this confusing... I also meant youtube.............................
I understood you. I just think it's a rather sad situation when a student's first response, made within minutes of a post, is "have you got a video on that?".
Reply 6
Avatar for ckfeister
ckfeister
OP
19
Original post by DFranklin
I understood you. I just think it's a rather sad situation when a student's first response, made within minutes of a post, is "have you got a video on that?".


Ok, as you would like to be brutal, I will be brutal. Nothing you said made sense, very sad isn't it :smile:
FWIW, as far as I can see, your first error is in the line starting (fg)' (it's a sign error). There are other errors later.
Reply 8
Avatar for ckfeister
ckfeister
OP
19
Original post by DFranklin
I understood you. I just think it's a rather sad situation when a student's first response, made within minutes of a post, is "have you got a video on that?".


As it is so " sad " would you also justify why this video proves my way was actually correct? d/dx means to differentiate, you can't differentiate them so you use the product rule.
https://www.youtube.com/watch?v=bTx3xegXgA4
(edited 6 years ago)
Original post by ckfeister
As it is so " sad " would you also justify why this video proves my way was actually correct? d/dx means to differentiate, you can't differentiate them so you use the product rule.Are you seriously lecturing a maths graduate on what d/dx means? Really?

I also explicitly said that your way was "fine", so I'm really not sure what point you're trying to prove here. I certainly can't be bothered to watch a video on the topic.
Reply 10
Avatar for Notnek
Notnek
21
Original post by ckfeister
As it is so " sad " would you also justify why this video proves my way was actually correct? d/dx means to differentiate, you can't differentiate them so you use the product rule.
https://www.youtube.com/watch?v=bTx3xegXgA4

Your/the video's method is correct but it's not the most efficient. Normally the ExamSolutions videos are excellent but I don't like his method for this question:

dtdx=2t12\frac{dt}{dx} = 2t^\frac{1}{2}

And since x=t12x=t^\frac{1}{2}, we have that dtdx=2x\frac{dt}{dx} = 2x.

Writing the derivative in terms of x makes the rest of the working simpler with less room for errors.

DFranklin has shown you the mistake you made using your method. It's possible you wouldn't have made the mistake if you used a "better" method but who knows.
Reply 11
Avatar for ckfeister
ckfeister
OP
19
Original post by Notnek
Your/the video's method is correct but it's not the most efficient. Normally the ExamSolutions videos are excellent but I don't like his method for this question:

dtdx=2t12\frac{dt}{dx} = 2t^\frac{1}{2}

And since x=t12x=t^\frac{1}{2}, we have that dtdx=2x\frac{dt}{dx} = 2x.

Writing the derivative in terms of x makes the rest of the working simpler with less room for errors.

DFranklin has shown you the mistake you made using your method. It's possible you wouldn't have made the mistake if you used a "better" method but who knows.


I was offering to but he didn't explain I asked for video if any quickly on youtube but apparently, it's sad... I also got this information on cloud learn the study provider and I could only find this on youtube that was relevent.. if you could put me in the right direction to find on youtube it would be great.
(edited 6 years ago)
Reply 12
Avatar for Notnek
Notnek
21
Original post by ckfeister
I was offering to but he didn't explain I asked for video if any quickly on youtube but apparently, it's sad... I also got this information on cloud learn the study provider and I could only find this on youtube that was relevent.. if you could put me in the right direction to find on youtube it would be great.

It's not easy to find a video explaining a style of tackling a question. But I don't think you need a video and you can work this out yourself. For your question:

x=t−1x=t^{-1} then x2=t−2x^2 = t^{-2}

So you had dtdx=−t−2\frac{dt}{dx}=-t^{-2} which can be written as dtdx=−x2\frac{dt}{dx}=-x^2. Does this make sense?

Now try continuing from here so the next step is to find dydx\frac{dy}{dx} and then d2ydx2\frac{d^2y}{dx^2}.


Or you can continue using your method - that isn't the biggest problem unless you find that you are making a lot of mistakes. You have been given the mistake you made using your method so try to correct that.
(edited 6 years ago)
Reply 13
Avatar for ckfeister
ckfeister
OP
19
Original post by Notnek
It's not easy to find a video explaining a style of tackling a question. But I don't think you need a video and you can work this out yourself. For your question:

x=t−1x=t^{-1} then x2=t−2x^2 = t^{-2}

So you had dtdx=−t−2\frac{dt}{dx}=-t^{-2} which can be written as dtdx=−x2\frac{dt}{dx}=-x^2. Does this make sense?

Now try continuing from here so the next step is to find dydx\frac{dy}{dx} and then d2ydx2\frac{d^2y}{dx^2}.


Or you can continue using your method - that isn't the biggest problem unless you find that you are making a lot of mistakes. You have been given the mistake you made using your method so try to correct that.


Ohh now I get it, thanks.
Reply 14
Avatar for simon0
simon0
13
Your solution could be clearer but my recommendations are:

- Check your answer for d2ydx2, \dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}, (you have missed out some terms),

-Your attempt to simplify d2y/dx2, \mathrm{d}^{2}y / \mathrm{d}x^{2}, (which you did not use in the end) is not correct and is not a step I would take.
(edited 6 years ago)
Please look at the part a answer

Original post by ckfeister
(FP2 AQA)

question5.PNG

Have I messed up yet? I'm repeating this question where I got it wrong (there is also no answer for this on the solutions)
Attachment not found
Reply 16
Avatar for ckfeister
ckfeister
OP
19
Original post by Notnek
It's not easy to find a video explaining a style of tackling a question. But I don't think you need a video and you can work this out yourself. For your question:

x=t−1x=t^{-1} then x2=t−2x^2 = t^{-2}

So you had dtdx=−t−2\frac{dt}{dx}=-t^{-2} which can be written as dtdx=−x2\frac{dt}{dx}=-x^2. Does this make sense?

Now try continuing from here so the next step is to find dydx\frac{dy}{dx} and then d2ydx2\frac{d^2y}{dx^2}.


Or you can continue using your method - that isn't the biggest problem unless you find that you are making a lot of mistakes. You have been given the mistake you made using your method so try to correct that.



For
dtdx=−x2\frac{dt}{dx}=-x^2.

I get
Unparseable latex formula:

\frac{dt}{dx}=-x^-^2

. as dxdt=−x2\frac{dx}{dt}=-x^2. Correct or wrong?
Reply 17
Avatar for ckfeister
ckfeister
OP
19
Original post by gurungg48
Please look at the part a answer

QUOTE=ckfeister;74195320](FP2 AQA)

question5.PNG

Have I messed up yet? I'm repeating this question where I got it wrong (there is also no answer for this on the solutions)
Attachment not found


Where does the d2ydt2 \frac{d^2y}{dt^2} t part come from I thought it was
Unparseable latex formula:

t^-^2

Reply 18
Avatar for Notnek
Notnek
21
Original post by ckfeister
For
dtdx=−x2\frac{dt}{dx}=-x^2.

I get
Unparseable latex formula:

\frac{dt}{dx}=-x^-^2

. as dxdt=−x2\frac{dx}{dt}=-x^2. Correct or wrong?

Correct.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you