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Year 13 Maths Help Thread

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Original post by kiiten
Kinda. Please could you remind me of that core 2 example you were talking about (ive forgotten most of it :3 )


Here's an example from Core 2. Solve: log2(x)=3\displaystyle \log_2(x)=3

This of course reads as "what is the number I get, x, when I raise the base, 2, to the third power?"

From here you would fully go with cancelling the logarithm by making both sides an index of a certain number, that number being the base of the logarithm:

2log2(x)=23\displaystyle 2^{\log_2(x)}=2^3

and you should know that aloga(b)=b\displaystyle a^{\log_a(b)}=b therefore the equation becomes:

x=23\displaystyle x=2^3

Now you can apply the same idea to your equation without moving the exponent of x down.

Original post by kiiten

Ok so if there are no -ve values why is the answet + or - ?? When it cant be a -ve? (I realised im over complicating it - sorry).


No, there ARE negative 'x' values but you are getting rid off them with the way you are working through your answer by moving the exponent of x down as then you have ln(x)\ln(x) and x cannot be negative for this expression, while x can be negative in ln(x2)\ln(x^2) as negative numbers squared become positive, so you'd be taking the log of a positive number which is valid hence negative x values are valid.
(edited 7 years ago)
Reply 661
How can I start to find r=1nr10r\sum^n_{r=1} r10^{-r} in terms of nn?

That is, just drop me a hint and not the full solution.
Original post by Palette
How can I start to find r=1nr10r\sum^n_{r=1} r10^{-r} in terms of nn?

That is, just drop me a hint and not the full solution.


Differentiate the geometric series.
Reply 663
Original post by Zacken
Differentiate the geometric series.


The derivative of 10x10^{-x} gives (ln10)10x-(\ln 10)10^{-x} but I don't know how it will help.

Would this differentiation approach work for any r=1nf(r)g(r)\sum^n_{r=1} f(r)g(r) where g(r) is a geometric series?
Original post by Palette
The derivative of 10x10^{-x} gives (ln10)10x-(\ln 10)10^{-x} but I don't know how it will help.

Would this differentiation approach work for any r=1nf(r)g(r)\sum^n_{r=1} f(r)g(r) where g(r) is a geometric series?


Not what I meant.

f(k)=r=0nkrdfdk=r=1nrkr1=k1r=1nrkr\displaystyle f(k) = \sum_{r=0}^n k^r \Rightarrow \frac{df}{dk} = \sum_{r=1}^n rk^{r-1}= k^{-1}\sum_{r=1}^n rk^r or along these lines w/e

and no to your last question
Reply 665
Original post by Zacken
Not what I meant.

f(k)=r=0nkrdfdk=r=1nrkr1=k1r=1nrkr\displaystyle f(k) = \sum_{r=0}^n k^r \Rightarrow \frac{df}{dk} = \sum_{r=1}^n rk^{r-1}= k^{-1}\sum_{r=1}^n rk^r or along these lines w/e

and no to your last question


I don't want to sound stupid, but why can we differentiate f(k)f(k) when rr can only take positive integer values between [0 and nn?
Original post by Palette
I don't want to sound stupid, but why can we differentiate f(k)f(k) when rr can only take positive integer values between [0 and nn?


Simplified, your question is isomorphic to: "Why can we differentiate x2x^2 when 22 is a positive integer?"

I explicitly pointed out that we're differentiating with respect to kk (a variable that can vary over the reals) by writing df/dkdf/dk and not f(k)f'(k). We're not differentiating with respect to rr, it's just a constant.
Reply 667
Original post by Zacken
Simplified, your question is isomorphic to: "Why can we differentiate x2x^2 when 22 is a positive integer?"

I explicitly pointed out that we're differentiating with respect to kk (a variable that can vary over the reals) by writing df/dkdf/dk and not f(k)f'(k). We're not differentiating with respect to rr, it's just a constant.


I get it now- thanks for your help!
Reply 668
Original post by Zacken
Simplified, your question is isomorphic to: "Why can we differentiate x2x^2 when 22 is a positive integer?"

I explicitly pointed out that we're differentiating with respect to kk (a variable that can vary over the reals) by writing df/dkdf/dk and not f(k)f'(k). We're not differentiating with respect to rr, it's just a constant.


Having looked at my question again, I realized how silly it was.
(edited 7 years ago)
Original post by Palette
Having looked at my question again, I realized how silly it was.


The original one or the "r constant" one? :tongue:

By the way, even if rr could take on positive integer values, you could still take on rr as a real variable, differentiate with respect to it and then send it back to the integer space in nice enough conditions and/or sufficient handwaving.
Original post by Palette
Having looked at my question again, I realized how silly it was.


Same - i only realise when its pointed out to me. Im just glad im anonymous on this site aha :biggrin:

- thanks to the people who helped me on that ln question. I think i understand now :smile:
Original post by k.russell
now I am a future biologist, so maybe not a leading authority on engineering - but I do think I've worked it out.
It makes sense the block will reach terminal velocity, right, when the forces acting forward = those acting backward.
So you set up the diagram, mg sin(a) down the slope and uN + kV up the slope, as we are adding in the air resistance term (kV). N = mg cos (a) , so you end up with this;
mg sin(a) = u mg cos(a) + kV
and you rearrange it get V out, because that V is gonna be the terminal velocity. At lesser values of V, the forces down slope would be greater and the object would accelerate. Do you agree with that?? Hope you can understand it lol (I had to use a for alpha and u for mu)


Yeah that makes sense thanks a lot, forgot to include a=0 when it's at constant velocity ie terminal velocity
Original post by metrize
Yeah that makes sense thanks a lot, forgot to include a=0 when it's at constant velocity ie terminal velocity



lol np, I am glad (and surprised) I could help out :smile: I am guessing you are studying some M units, how many have you done so far?
Sorry to intrude, but just out of interest, I'd like to ask those who do both physics and further maths which is the more difficult? (I do further maths, not physics)


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Original post by RedSquirrels
Sorry to intrude, but just out of interest, I'd like to ask those who do both physics and further maths which is the more difficult? (I do further maths, not physics)


Posted from TSR Mobile

Further maths can be very easy if you're interested in maths.
Original post by RedSquirrels
Sorry to intrude, but just out of interest, I'd like to ask those who do both physics and further maths which is the more difficult? (I do further maths, not physics)


Posted from TSR Mobile


I think Physics. You can choose fairly 'easy' (as in work hard, practice a lot, gain understanding and ace the exam) modules and it can be almost... straightforward.

Physics doesn't test your mathematical abilities as much but there's a lot of theory, used to have oracticals as well which was a nightnare, and like a lot of science subjects the exams/markschemes are quite annoying about things.
Hi, I was wondering if anyone could help me with this C3 Trigonometry question?image.jpeg
Original post by Leah3852
Hi, I was wondering if anyone could help me with this C3 Trigonometry question?image.jpeg


Well, that's the same thing as saying x=sinkx = \sin k for 0<k<π20 < k < \frac{\pi}{2}, what's the range of sin k as k moves from 0 to pi/2. That is, what portion of the y-axis does sin k map to when you look at a graph of sin k?

For b(i) you want to do cosk=cos(arcsinx)\cos k = \cos (\arcsin x) and for b(ii) you want tan(arcsinx)\tan(\arcsin x). Try drawing a right-angled triangle and finding the angle arcsinx\arcsin x. then using cos (arcsin x) = adjacent/hypo, etc... let me know hoe you get on. Prove your attempt and I can see whether you're going along the right lines and what hints you need.
How do I do this question?

Find the value of λ\lambda for which λx2ex\lambda x^2e^x is a particular integral for d2ydx22dydx+y=ex\frac{d^2y}{dx^2} -2 \frac{dy}{dx} + y =e^x
Original post by NotNotBatman
How do I do this question?

Find the value of λ\lambda for which λx2ex\lambda x^2e^x is a particular integral for d2ydx22dydx+y=ex\frac{d^2y}{dx^2} -2 \frac{dy}{dx} + y =e^x


Find y' and y'' and then just plug it into the equation and compare coefficients.

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