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Reply 120
Original post by HR--
I'm just about to go into 2nd year at university and I've chosen an outside course where you're supposed to have done Advanced Higher Maths. I left at the end of 5th year at school so only did higher.

So basically, I'm trying to learn as much of AH maths as I can within the next 2 weeks for the start of the course. :giggle:

Any tips?

Oh and thanks for the resources in the OP- very helpful!


What's the course on? Do you know what kind of maths is involved?

Khanacademy has loads of great stuff for all kinds of maths, and there are various maths videos on YouTube for just about everything.
Reply 121
Original post by JordanR
What's the course on? Do you know what kind of maths is involved?

Khanacademy
has loads of great stuff for all kinds of maths, and there are various maths videos on YouTube for just about everything.

It's a course on Problem Solving in Physics. It's supposed to be a course for direct entry Physics students into 2nd year so they're expected to have AH maths and physics. I emailed the course organizer and she said it's basically applying Higher and Advanced Higher maths to solving physics problems. It's pretty maths-based. I did physics courses in my 1st year at uni but no maths.

Thanks for the link. I'll take a look. I guess I'll just have to try to quickly work through the AH course notes.
Reply 122
Original post by HR--
It's a course on Problem Solving in Physics. It's supposed to be a course for direct entry Physics students into 2nd year so they're expected to have AH maths and physics. I emailed the course organizer and she said it's basically applying Higher and Advanced Higher maths to solving physics problems. It's pretty maths-based. I did physics courses in my 1st year at uni but no maths.

Thanks for the link. I'll take a look. I guess I'll just have to try to quickly work through the AH course notes.


Ah.

In that case, I'd focus on your vector work, your trig work and your calculus work. Although that's just what I'd imagine would be in there.
Reply 123
You should try PatrickJMT as well
http://adf.ly/2c7MH
Reply 124
Original post by Evan7788
You should try PatrickJMT as well
http://adf.ly/2c7MH



Original post by JordanR
Ah.

In that case, I'd focus on your vector work, your trig work and your calculus work. Although that's just what I'd imagine would be in there.


Brilliant! Thanks very much for your help. Much appreciated.

Right, I better get to it. I bet I'm going to hate this after a week and a half of only looking at maths. :tongue:
KhanAcademy is brilliant for extra help but PatrickJMT absolutely nailed it for me. I remember spending late nights just looking at his videos and absorbing everything. Make sure you pause the video during the example so you can do it for yourself!
really stuck. Can you guys please help me on a couple of questions:

1) y= e^2t. show that d^2y/dt^2 - 2t dy/dt - 2y = 0


2) if y=x^2 lnx, show that dy/dx = x ln(ex^2)

thanks
Reply 127
Original post by adam_johnson
really stuck. Can you guys please help me on a couple of questions:

1) y= e^2t. show that d^2y/dt^2 - 2t dy/dt - 2y = 0


2) if y=x^2 lnx, show that dy/dx = x ln(ex^2)

thanks


That's a second-order differential equation. Not sure if you've learned those yet, since they're not until the end of the second unit. I haven't learned them completely, so I wouldn't want to make an arse of it. :tongue:

2.)

y=x2lnxy=x^2lnx

Use the product rule:

f(x)g(x)+f(x)g(x)f'(x)g(x) + f(x)g'(x)

2xlnx+x21x2x*lnx + x^2*\frac{1}{x}

x(2lnx+1)x(2lnx+1)

Which is equal to what you have to show it's equal to, but only for positive values of x.
(edited 12 years ago)
Original post by JordanR
That's a second-order differential equation. Not sure if you've learned those yet, since they're not until the end of the second unit. I haven't learned them completely, so I wouldn't want to make an arse of it. :tongue:

2.)

y=x2lnxy=x^2lnx

Use the product rule:

f(x)g(x)+f(x)g(x)f'(x)g(x) + f(x)g'(x)

2xlnx+x21x2x*lnx + x^2*\frac{1}{x}

x(lnx+1)x(lnx+1)

Which is equal to what you have to show it's equal to, but only for positive values of x.


Sorry, I've been away from school for a few years and I'm lost. Where does the 2 disappear from the second to last to last line?
Reply 129
Original post by flown_muse
Sorry, I've been away from school for a few years and I'm lost. Where does the 2 disappear from the second to last to last line?


Sorry, typo. :tongue: Fixed now.
Original post by JordanR
Sorry, typo. :tongue: Fixed now.


Aw good :smile: This is starting to ring bells again, and then I got to that bit and thought I'd missed something. Thanks :smile:
Reply 131
Original post by flown_muse
Aw good :smile: This is starting to ring bells again, and then I got to that bit and thought I'd missed something. Thanks :smile:


Does maths come back to you quickly? I don't want to study something that doesn't require much maths, and then forget everything I know. :c
Original post by JordanR
Does maths come back to you quickly? I don't want to study something that doesn't require much maths, and then forget everything I know. :c


I seem to have a weird memory, I soak stuff up and it only takes a wee bit to remember it. I hope I remember fast cause my course starts on Wednesday! :eek:

Can you remind me what the rule is for differentiating lnx again? :redface:
(edited 12 years ago)
Reply 133
Original post by flown_muse
I seem to have a weird memroy, I soak stuff up and it only takes a wee bit to remeber it. I hope I remember fast cause my course starts on Wednesday! :eek:

Can you remind me what the rule is for differentiating lnx again? :redface:


ddxlnx=1/x \frac{d}{dx} lnx = 1/x

Although I remember that because I remember that:

f(x)f(x)=lnx+c\int \frac{f'(x)}{f(x)} = lnx + c

And obviously differentiation is just the opposite of integration. c:
Original post by JordanR
ddxlnx=1/x \frac{d}{dx} lnx = 1/x

Although I remember that because I remember that:

f(x)f(x)=lnx+c\int \frac{f'(x)}{f(x)} = lnx + c

And obviously differentiation is just the opposite of integration. c:


Wah. Where does the e come from them? (Sorry I'm being so thick)
Original post by flown_muse
Wah. Where does the e come from them? (Sorry I'm being so thick)


Remember that ln e = 1. Using log rules, you have 2lnx + 1 = ln(x^2) + 1 = ln(x^2) + lne = ln(ex^2)
Reply 136
Original post by ukdragon37
Remember that ln e = 1. Using log rules, you have 2lnx + 1 = ln(x^2) + 1 = ln(x^2) + lne = ln(ex^2)


Yep. I was going to write this, but I had to rush to McDonald's before it closed. :tongue:
Original post by ukdragon37
Remember that ln e = 1. Using log rules, you have 2lnx + 1 = ln(x^2) + 1 = ln(x^2) + lne = ln(ex^2)


I'm getting there, thanks! :smile:
I'm doing it this year.

I took it as preparation for university.

Plan on studying civil engineering.

Therefore , I'll take what I get , but I'll still do my best :smile:
Original post by ukdragon37
Remember that ln e = 1. Using log rules, you have 2lnx + 1 = ln(x^2) + 1 = ln(x^2) + lne = ln(ex^2)


how does ln(x^2) + lne go to ln(ex^2) ?

oh sorry I just got it. log rules
(edited 12 years ago)

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