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March 30, 2024, 12:22:50 am

Author Topic: 3U Maths Question Thread  (Read 1230693 times)  Share 

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Jefferson

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Re: 3U Maths Question Thread
« Reply #3885 on: February 01, 2019, 04:43:35 pm »
0
Yeah.

The special case of \(y^\prime = 0\) and \( y^{\prime\prime} = 0\) is called a horizontal point of inflexion. This is because the tangent to the curve at that point also has gradient 0, on top of the concavity change. This is what we observe for curves such as \(y=x^3\).

If we want just a (not special) point of inflexion, we just want to eliminate that stationary nature. An example is \(y = \sin x\), at \(x=0\).

Another example is \(y = x(x-1)(x+1)\), at \(x=0\).

Hi RuiAce,
Thanks for responding so quickly, but I think I didn't make my question clear enough and was misinterpreted.
For both of those functions,   y = sin(x)     &     y = x (x-1) (x+1)   , x=0 is an inflexion point, found by setting y"=0.

Sometimes, when y"=0, you don't get an inflextion point, e.g. x4. But this is usually because the first derivative is also equal to 0 (y'=0), making it is a stationary point.

My question was, if we know that y"=0 for a point, but y' ≠ 0 (not equal to), then do we know for sure (just for the sake of the question that it is without a doubt an inflexion pointwithout testing concavity?
« Last Edit: February 01, 2019, 04:51:56 pm by Jefferson »

RuiAce

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Re: 3U Maths Question Thread
« Reply #3886 on: February 01, 2019, 05:07:27 pm »
+2
Hi RuiAce,
Thanks for responding so quickly, but I think I didn't make my question clear enough and was misinterpreted.
For both of those functions,   y = sin(x)     &     y = x (x-1) (x+1)   , x=0 is an inflexion point, found by setting y"=0.

Sometimes, when y"=0, you don't get an inflextion point, e.g. x4. But this is usually because the first derivative is also equal to 0 (y'=0), making it is a stationary point.

My question was, if we know that y"=0 for a point, but y' ≠ 0 (not equal to), then do we know for sure (just for the sake of the question that it is without a doubt an inflexion pointwithout testing concavity?
Nope.

After doing some research, an example on Math Stack Exchange/Wikipedia shows that \(y=x^4-x\) will break here. At 0 we see that \( y^\prime = -1\) and \(y^{\prime\prime} = 0\), but if you look at a Desmos/GeoGebra plot the curve still does not change in concavity at 0.

Edit: If you're interested in researching points where the second derivative vanishes, but we don't end up with an inflexion point, these are called undulation points
« Last Edit: February 01, 2019, 05:12:41 pm by RuiAce »

Jefferson

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Re: 3U Maths Question Thread
« Reply #3887 on: February 01, 2019, 05:16:05 pm »
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Nope.

After doing some research, an example on Math Stack Exchange/Wikipedia shows that \(y=x^4-x\) will break here. At 0 we see that \( y^\prime = -1\) and \(y^{\prime\prime} = 0\), but if you look at a Desmos/GeoGebra plot the curve still does not change in concavity at 0.

Edit: If you're interested in researching points where the second derivative vanishes, but we don't end up with an inflexion point, these are called undulation points

Wow! Thank you for going out of your way once again!
I've spent a decent amount of time on this but couldn't come up with a counter example.
Deeply appreciated.  :)
« Last Edit: February 01, 2019, 05:18:25 pm by Jefferson »

adelaidemurton

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Re: 3U Maths Question Thread
« Reply #3888 on: February 02, 2019, 09:19:13 am »
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Hey I've been struggling with the algebraic manipulation aspects of this question. I would love any help

A solid cylinder, radius r and height h, is to be constructed under the condition that the sum of its height and circumference is S, where S is constant.
   a) if it is given its maximum possible volume, h=S/3
   b) if it is given its maximum possible surface area, r+h = S/2

RuiAce

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Re: 3U Maths Question Thread
« Reply #3889 on: February 02, 2019, 09:55:15 am »
+3
Hey I've been struggling with the algebraic manipulation aspects of this question. I would love any help

A solid cylinder, radius r and height h, is to be constructed under the condition that the sum of its height and circumference is S, where S is constant.
   a) if it is given its maximum possible volume, h=S/3
   b) if it is given its maximum possible surface area, r+h = S/2
\[ \text{Note that the condition we're given is}\\ \boxed{h + 2\pi r = S}.\]
And keep in mind that \(S\) is constant, so we do not need to worry about it until after computing the first derivative.
\[ \text{The volume of the cylinder is }V = \pi r^2 h.\\ \text{Since }h = S - 2\pi r\text{ we have}\\ V = \pi r^2 (S - 2\pi r) \implies \boxed{V = \pi (Sr^2 - 2\pi r^3) } \]
\[ \text{The surface area of the cylinder is }A = 2\pi r^2 + 2\pi r h.\\ \text{Since }h = S - 2\pi r\text{ we have}\\ A = 2\pi r^2 + 2\pi r (S - 2\pi r) \implies \boxed{A = (2\pi - 4\pi^2)r^2 + 2\pi S r} \]
__________________________________________________________________
\[ \text{For each of these, we will need to use }\boxed{h + 2\pi r = S}\text{ a few times.} \]
\[ \text{Setting }\frac{dV}{dr} = 0\text{ gives}\\ \begin{align*} \pi (2Sr - 6\pi r^2) &= 0\\ Sr - 3\pi r^2 &= 0\\ S - 3\pi r &=0 \tag{**}\\ S &= 3\pi r\\ S&= \frac32(S-h)\\ 0 &= \frac12 S - \frac32 h\\ 3h&=S\\ h&=\frac{S}{3} \end{align*} \]
(**) - You actually have \(r=0\) or \(S - 3\pi r = 0\). To do this correctly, you would then look at the second derivative to check that \(r=0\) actually minimises the volume (this makes sense, because if the radius is 0 then the volume should also be 0). This is a much safer way of discarding the solution \(r=0\), than to just divide it out like I did so there.
\[ \text{Setting }\frac{dA}{dr} = 0\text{ gives}\\ \begin{align*} 2(2\pi - 4\pi^2) r + 2\pi S &=0\\ (2\pi - 4\pi^2 r) + \pi (h + 2\pi r) &= 0\\ 2\pi r - 2\pi^2r^2 + \pi h &= 0\\ 2r - 2\pi r + h &= 0\\ 2r + (h-S) + h &= 0\\ 2(r+h) &= S\\ r+h &= \frac{S}{2} \end{align*}\]
Testing for the max with the second derivative is left as your exercise.
« Last Edit: February 02, 2019, 09:58:12 am by RuiAce »

Jefferson

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Re: 3U Maths Question Thread
« Reply #3890 on: February 03, 2019, 12:50:52 pm »
+1
Hi again,This is a question regarding Simple Harmonic Motion.
For equations such as
(acceleration) x.. = -n2x    or
v2 = n2(a2-x2)

Why is it that we always take the positive value for 'n' without any justification?
e.g. x.. = -9x
∴ n = ±3   3.

Also, could you quote the second equation (v2 = n2(a2-x2)) without proving it in the exam? (it's not on the reference sheet)
Thanks.
« Last Edit: February 03, 2019, 01:02:14 pm by Jefferson »

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #3891 on: February 03, 2019, 01:06:31 pm »
+2
Hi again,This is a question regarding Simple Harmonic Motion.
For equations such as
(acceleration) x.. = -n2x    or
v2 = n2(a2-x2)

Why is it that we always take the positive value for 'n' without any justification?
e.g. x.. = -9x
∴ n = ±3   3.

Thanks.

Hey! We actually don't have to necessarily, there are situations where you could take the negative - And indeed I've been caught making this mistake in a lecture before! Say we had this:

An object undergoes SHM according to \(x''=-9x\) and with an amplitude of 2 units. At \(t=0\), it is at the origin. At \(t=\frac{\pi}{18}\), the particle is at \(x=-1\). Find the equation of motion.

In this situation, it might be your intuition to just quickly go \(x=2\sin{3t}\). This meets almost all criteria, except when you substitute:



This is NOT correct, it should be at \(x=-1\). One way to correct this would be to choose the negative \(n\) value, so:



However, it is more common/better practice to instead fix it by putting the negative out the front:



This is why we choose \(n>0\) most of the time, because it's more appropriate to the form of the equation to put the negative out the front. If for some reason your amplitude term out the front needed to be positive (maybe the question sets it up that way?), then you could take the negative value of \(n\) to compensate instead. We just don't do that very often :)
« Last Edit: February 03, 2019, 01:08:49 pm by jamonwindeyer »

Jefferson

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Re: 3U Maths Question Thread
« Reply #3892 on: February 03, 2019, 01:45:55 pm »
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Hey! We actually don't have to necessarily, there are situations where you could take the negative - And indeed I've been caught making this mistake in a lecture before! Say we had this:

An object undergoes SHM according to \(x''=-9x\) and with an amplitude of 2 units. At \(t=0\), it is at the origin. At \(t=\frac{\pi}{18}\), the particle is at \(x=-1\). Find the equation of motion.

In this situation, it might be your intuition to just quickly go \(x=2\sin{3t}\). This meets almost all criteria, except when you substitute:



This is NOT correct, it should be at \(x=-1\). One way to correct this would be to choose the negative \(n\) value, so:



However, it is more common/better practice to instead fix it by putting the negative out the front:



This is why we choose \(n>0\) most of the time, because it's more appropriate to the form of the equation to put the negative out the front. If for some reason your amplitude term out the front needed to be positive (maybe the question sets it up that way?), then you could take the negative value of \(n\) to compensate instead. We just don't do that very often :)

Hi Jamon, thanks for replying. That makes a lot of sense.
Is it okay to quote the equation v2 = n2(a2-x2) without proving it in the exam?

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #3893 on: February 03, 2019, 03:36:31 pm »
+1
Hi Jamon, thanks for replying. That makes a lot of sense.
Is it okay to quote the equation v2 = n2(a2-x2) without proving it in the exam?

Definitely, although it is fairly common to be asked to prove it for marks ;D

david.wang28

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Re: 3U Maths Question Thread
« Reply #3894 on: February 05, 2019, 02:15:58 pm »
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Hello,
I'm stuck on two inverse function questions, i) tan^-1[(8-x)/(1+8x)] and ii) tan^-1[(2x)/(1-x^2)]. I know how to do it by using the standard formula for the differentiation of inverse trig functions, but during my working out, I seemed to have got the wrong answers for both. Can anyone please help me out with these 2 questions? Thanks :)
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RuiAce

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Re: 3U Maths Question Thread
« Reply #3895 on: February 05, 2019, 02:18:50 pm »
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Hello,
I'm stuck on two inverse function questions, i) tan^-1[(8-x)/(1+8x)] and ii) tan^-1[(2x)/(1-x^2)]. I know how to do it by using the standard formula for the differentiation of inverse trig functions, but during my working out, I seemed to have got the wrong answers for both. Can anyone please help me out with these 2 questions? Thanks :)
So to make sure, you would like to differentiate them yes? (You only stated "inverse functions questions" without saying what the questions were but you did say differentiation afterwards.)

david.wang28

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Re: 3U Maths Question Thread
« Reply #3896 on: February 05, 2019, 02:33:17 pm »
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So to make sure, you would like to differentiate them yes? (You only stated "inverse functions questions" without saying what the questions were but you did say differentiation afterwards.)
Yes, I would like to differentiate them please.
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RuiAce

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Re: 3U Maths Question Thread
« Reply #3897 on: February 05, 2019, 02:39:46 pm »
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\[ \text{For example, here's the second one.}\\ \text{Using the quotient rule and chain rule,}\\ \begin{align*} \frac{d}{dx} \tan^{-1} \left( \frac{2x}{1-x^2} \right) &= \frac{2(1-x^2) - 2x(-2x)}{(1-x^2)^2}\cdot \frac{1}{1+\left( \frac{2x}{1-x^2} \right)^2} \\ &= \frac{2-2x^2+4x^2}{(1-x^2)^2 \left[1+ \frac{4x^2}{(1-x^2)^2} \right]}\\ &= \frac{2+2x^2}{(1-x^2)^2 + 4x^2} \tag{expanding}\\ &= \frac{2(1+x^2)}{1-2x^2+x^4+4x^2}\\ &= \frac{2(1+x^2)}{1+2x^2+x^4}\\ &= \frac{2(1+x^2)}{(1+x^2)^2}\\ &= \frac{2}{1+x^2}\end{align*} \]
If you need help with the other one please post some working out

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Re: 3U Maths Question Thread
« Reply #3898 on: February 05, 2019, 03:38:07 pm »
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\[ \text{For example, here's the second one.}\\ \text{Using the quotient rule and chain rule,}\\ \begin{align*} \frac{d}{dx} \tan^{-1} \left( \frac{2x}{1-x^2} \right) &= \frac{2(1-x^2) - 2x(-2x)}{(1-x^2)^2}\cdot \frac{1}{1+\left( \frac{2x}{1-x^2} \right)^2} \\ &= \frac{2-2x^2+4x^2}{(1-x^2)^2 \left[1+ \frac{4x^2}{(1-x^2)^2} \right]}\\ &= \frac{2+2x^2}{(1-x^2)^2 + 4x^2} \tag{expanding}\\ &= \frac{2(1+x^2)}{1-2x^2+x^4+4x^2}\\ &= \frac{2(1+x^2)}{1+2x^2+x^4}\\ &= \frac{2(1+x^2)}{(1+x^2)^2}\\ &= \frac{2}{1+x^2}\end{align*} \]
If you need help with the other one please post some working out
Here is my working out for the first one. Am I on the right track? What else do I need to do?
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Re: 3U Maths Question Thread
« Reply #3899 on: February 05, 2019, 03:40:22 pm »
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Here is my working out for the first one. Am I on the right track? What else do I need to do?
You've done nothing wrong. Now cancel out the \( (1+8x)^2\) in the left fraction's denominator and right fraction's denominator (remember - you're multiplying two fractions) and then expand and simplify.