Rates of Change (1 Viewer)

michaeljennings · Jun 7, 2015

Has anyone got a worked solution to Chapter 7E, Q11 from the Cambridge Year 12 Textbook?

Cheers

InteGrand · Jun 7, 2015

michaeljennings said:
Has anyone got a worked solution to Chapter 7E, Q11 from the Cambridge Year 12 Textbook?

Cheers

$11. (a) Note that the angular velocity is $\omega = \frac{v}{r}=\frac{2}{1}=2\text{ rad/s}$. Since the motion is circular, $x=r\cos (\omega t)$ and $y=r\sin (\omega t)$ (assuming the motion starts at $(0,r)$; it doesn't matter if we assume this, since we're only asked questions about rates of change of $x$, which is unaffected by the initial position), where $r=1$. Note that $\frac{\mathrm{d}x}{\mathrm{d}t}<0$ when the particle is above the $x$-axis, since the motion is anticlockwise.$

$So $\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}(r\cos (\omega t)) = -r\omega \sin (\omega t) = -\omega y = -\omega \sqrt{r^2 - x^2} = -2\sqrt{1-x^2}$, by substituting in known values and since $y \geq 0$ (so we took the positive square root).$

$So the answer is $\dot{x} = -2\sqrt{1-x^2}$ (with units of m/s).$

$(b) So at the point $P(0,1)$, $\dot{x} = -2\sqrt{1-x^2} = -2 \sqrt{1-0^2}=-2$ (with units of m/s). This should have been obvious without the formula, because at the instant the particle reaches the point $P(0,1)$, it is moving \emph{parallel to the }$x$-\emph{axis} (and in the \emph{negative} direction, since the motion is anticlockwise) with a speed of 2 m/s, so its instantaneous rate of change of $x$-value would be $-2$ m/s.$

mreditor16 · Jun 7, 2015

InteGrand said:
$11. (a) Note that the angular velocity is $\omega = \frac{v}{r}=\frac{2}{1}=2\text{ rad/s}$. Since the motion is circular, $x=r\cos (\omega t)$ and $y=r\sin (\omega t)$ (assuming the motion starts at (0,r)), where $r=1$. Note that $\frac{\mathrm{d}x}{\mathrm{d}t}<0$ when the particle is above the $x$-axis, since the motion is anticlockwise.$

$So $\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}(r\cos (\omega t)) = -r\omega \sin (\omega t) = -\omega y = -\omega \sqrt{r^2 - x^2} = -2\sqrt{1-x^2}$, by substituting in known values and since $y \geq 0$ (so we took the positive square root).$

$So the answer is $\dot{x} = -2\sqrt{1-x^2}$ (with units of m/s).$

$(b) So at the point $P(0,1)$, $\dot{x} = -2\sqrt{1-x^2} = -2 \sqrt{1-0^2}=-2$ (with units of m/s). This should have been obvious without the formula, because at the instant the particle reaches the point $P(0,1)$, it is moving \emph{parallel to the }$x$-\emph{axis} (and in the \emph{negative} direction, since the motion is anticlockwise) with a speed of 2 m/s, so its instantaneous rate of change of $x$-value would be $-2$ m/s.$

Integrand, your solution can't be used, because this is a 3U question and you have used a 4U approach.

InteGrand · Jun 7, 2015

mreditor16 said:
Integrand, your solution can't be used, because this is a 3U question and you have used a 4U approach.

$What did I do that is disallowed? You mean the use of $v=r\omega$? This is just a simple manipulation of the definition of angular velocity: $\omega \equiv \frac{2\pi}{T}=\frac{2\pi r}{T}\times \frac{1}{r}=\frac{v}{r},$ where $T$ is the period of motion and $v$ is the tangential speed. (Plus, I thought the HSC accepts 4U methods in 3U, and I don't see what was so advanced about the given solution in the first place.)$

mreditor16 · Jun 7, 2015

InteGrand said:
$What did I do that is disallowed? You mean the use of $v=r\omega$? This is just a simple manipulation of the definition of angular velocity: $\omega \equiv \frac{2\pi}{T}=\frac{2\pi r}{T}\times \frac{1}{r}=\frac{v}{r},$ where $T$ is the period of motion and $v$ is the tangential speed. (Plus, I thought the HSC accepts 4U methods in 3U, and I don't see what was so advanced about the given solution in the first place.)$

Well, put it this way, all questions in a 3U paper (and more broadly, 3U textbooks) are (or maybe I should say "should be" to stay safe on this issue) designed such that the questions can be done using only 3U (and below) concepts. So you're essentially assuming 3U students can pluck the 4U concept of angular velocity out of thin air when faced with this question. I'm not saying your method won't be accepted, it's just the OP seems to be wanting a method appropriate to show to 3U students.

InteGrand · Jun 7, 2015

$If we're not allowed to make reference to angular velocity, then we could do the following: $x^2 + y^2 = 1 \Rightarrow 2x \dot{x} + 2y\dot{y} = 0$ (*) by implicit differentiation with respect to $t$. Use the facts that $y = +\sqrt{1-x^2}$ and $\dot{y} = \frac{\mathrm{d}y}{\mathrm{d}x}\cdot \dot{x}$ (chain rule), with $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-x}{\sqrt{1-x^2}}$, and substitute these into Equation (*), then solve for $\dot{x}$.$

Edit: sorry, this ends up with 0 = 0

mreditor16 · Jun 7, 2015

It's like giving an integral that can only be solved by integration by parts. Under your logic, you can argue it can be it in the 3U test, since 3U students could derive the method of integration by parts on the spot and then utilise it in the question they face.

mreditor16 · Jun 7, 2015

InteGrand said:
$If we're not allowed to make reference to angular velocity, then we could do the following: $x^2 + y^2 = 1 \Rightarrow 2x \dot{x} + 2y\dot{y} = 0$ (*) by implicit differentiation with respect to $t$. Use the facts that $y = +\sqrt{1-x^2}$ and $\dot{y} = \frac{\mathrm{d}y}{\mathrm{d}x}\cdot \dot{x}$ (chain rule), with $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-x}{\sqrt{1-x^2}}$, and substitute these into Equation (*), then solve for $\dot{x}$.$

Again, I don't like saying this, but if I am informed correctly, implicit differentiation is a 4U concept.

InteGrand · Jun 7, 2015

Well the tangential speed needs to be used since that affects the answer. And tangential speed is inherently linked to angular velocity. And circular motion isn't explicitly a 3U concept, is it (rather, 4U I think)? But we still need to use the tangential speed, which would theoretically be unheard of for 3U students then, assuming circular motion is 4U but not 3U.

InteGrand · Jun 7, 2015

mreditor16 said:
Again, I don't like saying this, but if I am informed correctly, implicit differentiation is a 4U concept.

And implicit differentiation is just the chain rule essentially. Whenever we use 'implicit differentiation', a 3U student could theoretically use the chain rule – just replace y with y(x) and differentiate using product rules etc.

mreditor16 · Jun 7, 2015

Well, technically, you could say everything in 4U is in 3U, because we do use 3U concepts to derive 4U concepts. So this sort of discussion will only go in circles.

InteGrand · Jun 7, 2015

mreditor16 said:
Well, technically, you could say everything in 4U is in 3U, because we do use 3U concepts to derive 4U concepts. So this sort of discussion will only go in circles.

Well, at least circles is 3U

mreditor16 · Jun 7, 2015

InteGrand said:
Well, at least circles is 3U

But not circular motion

Crisium · Jun 7, 2015

Ekman · Jun 7, 2015

mreditor16 said:
Again, I don't like saying this, but if I am informed correctly, implicit differentiation is a 4U concept.

Well I remember doing implicit differentiation in 3u before learning it in 4u

Carrotsticks · Jun 7, 2015

Use of implicit differentiation is allowed in the HSC (ie: 2014 MX1 Question 13 Rates of Change, an implicit differentiation proof was accepted).

The syllabus is mucky and not 100% clear cut defined, so arguing technicalities is pointless.

Technically, students are expected to be able to identify forms f'(x)/f(x) when integrating. So things like x/(1+x^2) are to be done by inspection by 2U students.

However, would we then expect them to be able to solve integrals such as the integral of 1/(x+root(x)), which is technically in that form? How about sec(x)?

Whilst we're all talking about the validity of methods in the HSC, how about first asking whether the question would even be there? In this case, probably not, without a decent amount of hand holding.

Crisium · Jun 7, 2015

Does anyone know how to do it using an Extension 1 method?

Carrotsticks · Jun 7, 2015

Crisium said:
Does anyone know how to do it using an Extension 1 method?

Well the students are given the tangential velocity to be 2, which may be converted to angular velocity using basic arithmetic.

If you see the conversion from tangential to angular velocity as something to be deduced on the spot, I am sure it is within the scope of 3U difficulty (as they've been expected to spot much more difficult things in the past).

If you see the conversion as the application of a formula, then I guess one could say it's out of the syllabus, though barely.

Crisium · Jun 7, 2015

I managed to do it without angular velocity, etc.

OP If you look at the worked examples above you will notice that the theory applied to the ladder one can sort of be related to this question

braintic · Jun 7, 2015

Using parametric coordinates (cosθ, sinθ),
and using l=rθ, ie. l=θ, so dl/dθ = 1

dx/dt = dx/dθ × dθ/dl × dl/dt

= -sinθ × 1 × 2
= -2y
= -2√(1-x²)

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