Author Topic: VCE Specialist 3/4 Question Thread! (Read 2546297 times) Tweet Share

IndefatigableLover · « **Reply #3090 on:** April 25, 2014, 01:50:14 pm »

A particle moves according to a motion law $x=f(t)$ that satisfies the following equation:

$t+x^3=xt$

At t=8s, the particle is 2m away from the origin.
What is the velocity of the particle at t=8s?

keltingmeith · « **Reply #3091 on:** April 25, 2014, 01:57:50 pm »

So, we know that the displacement of a particle is given by $x = f(t)$ , and that it can be written as $t + x^3 = xt$ . At 8 seconds, the particle is 2m away from the origin. So, we know that one solution to this equation is t = 8, x = 2.

We want to find the velocity of the particle, and velocity is the rate of change of displacement. So, let's find $\frac{dx}{dt}$ . Using implict differentiation, you'll get:

$\frac{dx}{dt} = \frac{x-1}{3x^2-t}$

Subbing in for t = 8, x = 2 you will find the velocity of the particle at that point.

IndefatigableLover · « **Reply #3092 on:** April 25, 2014, 02:48:16 pm »

Quote from: EulerFan101 on April 25, 2014, 01:57:50 pm

So, we know that the displacement of a particle is given by $x = f(t)$ , and that it can be written as $t + x^3 = xt$ . At 8 seconds, the particle is 2m away from the origin. So, we know that one solution to this equation is t = 8, x = 2.

We want to find the velocity of the particle, and velocity is the rate of change of displacement. So, let's find $\frac{dx}{dt}$ . Using implict differentiation, you'll get:

$\frac{dx}{dt} = \frac{x-1}{3x^2-t}$

Subbing in for t = 8, x = 2 you will find the velocity of the particle at that point.

By chance could you step me through the implicit differentiation process? Having a mind blank now T_T

keltingmeith · « **Reply #3093 on:** April 25, 2014, 03:00:08 pm »

Sure, no problem.

The easiest way to remember implicit diff is that it's an application of the chain rule.
So, let's say we have a circle $x^2 + y^2 = 1$ , and we want to take the derivative at any point (x, y). Well, when we do this, we'll easily see that you get:

$2x + \frac{d}{dx}(y^2) = 0$

Well, using the chain rule, we can see that:

$\frac{d}{dy}\cdot\frac{dy}{dx}=\frac{d}{dx}$

This means that:
$\frac{d}{dx}(y^2)=\frac{dy}{dx}\cdot\frac{d}{dy}(y^2)=\frac{dy}{dx}\cdot2y$

Using this for the equation in your question:

$\frac{d}{dt}(t)+\frac{d}{dt}(x^3)=\frac{d}{dt}(x\cdot t)$
$1 + \frac{dx}{dt}\cdot\frac{d}{dx}x^3 = x\cdot1 + t\cdot\frac{d}{dt}(x)$
$1 + \frac{dx}{dt}\cdot3x^2 = x + t\cdot\frac{dx}{dt}\cdot1$

Now, we need to collect the differentials together:

$\frac{dx}{dt}\cdot3x^2 - \frac{dx}{dt}\cdot t = x -1$
$\frac{dx}{dt}\cdot(3x^2 - t) = x - 1$
$\frac{dx}{dt} = \frac{x-1}{3x^2-t}$
Note that in the second line I had to apply the product rule, as I was taking the derivative of a product of two variables.

IndefatigableLover · « **Reply #3094 on:** April 25, 2014, 03:30:15 pm »

Quote from: EulerFan101 on April 25, 2014, 03:00:08 pm

Sure, no problem.

The easiest way to remember implicit diff is that it's an application of the chain rule.
So, let's say we have a circle $x^2 + y^2 = 1$ , and we want to take the derivative at any point (x, y). Well, when we do this, we'll easily see that you get:

$2x + \frac{d}{dx}(y^2) = 0$

Well, using the chain rule, we can see that:

$\frac{d}{dy}\cdot\frac{dy}{dx}=\frac{d}{dx}$

This means that:
$\frac{d}{dx}(y^2)=\frac{dy}{dx}\cdot\frac{d}{dy}(y^2)=\frac{dy}{dx}\cdot2y$

Using this for the equation in your question:

$\frac{d}{dt}(t)+\frac{d}{dt}(x^3)=\frac{d}{dt}(x\cdot t)$
$1 + \frac{dx}{dt}\cdot\frac{d}{dx}x^3 = x\cdot1 + t\cdot\frac{d}{dt}(x)$
$1 + \frac{dx}{dt}\cdot3x^2 = x + t\cdot\frac{dx}{dt}\cdot1$

Now, we need to collect the differentials together:

$\frac{dx}{dt}\cdot3x^2 - \frac{dx}{dt}\cdot t = x -1$
$\frac{dx}{dt}\cdot(3x^2 - t) = x - 1$
$\frac{dx}{dt} = \frac{x-1}{3x^2-t}$
Note that in the second line I had to apply the product rule, as I was taking the derivative of a product of two variables.

Awesome EulerFan101! Nice explanation

Cort · « **Reply #3095 on:** April 26, 2014, 08:37:54 pm »

Stupid question time! Yay!

What is the purpose of double differentiation any way? In what context, in relation to the specialist maths course, would it ever be used?

alchemy · « **Reply #3096 on:** April 26, 2014, 08:48:34 pm »

Quote from: Cort on April 26, 2014, 08:37:54 pm

Stupid question time! Yay!

What is the purpose of double differentiation any way? In what context, in relation to the specialist maths course, would it ever be used?

Definitively not a stupid question as there are plenty of uses in spesh and in real life too!

Spoiler

^I know, how often do you hear that eh?

For starters, finding the double derivative is necessary for when you do problems that ask you to find stationary points and points of inflection on a graph. Rates problems in spesh essentially depend on calculating the double derivative. For example, finding the rate of change of volume of different 3-dimensional shapes and objects. So far that's all I've come across, but I would bet for there to be more uses revealed later nearing the end of the course, particularly in kinematics questions.

lzxnl · « **Reply #3097 on:** April 26, 2014, 09:48:31 pm »

Acceleration is the second derivative of position with respect to time.
Second derivatives also do simplify determining types of stationary points.

Cort · « **Reply #3098 on:** April 26, 2014, 09:49:33 pm »

Thank you Alchemy, that definitely cleared my worries up. Now my other worries! It's space magic!

I am confuzzled when you differentiate sec^2 (x). I know it goes to 1/cos^2(x), but what happens from there on wards? What do you do with the ^2? Do I handle this like a normal 1(cos^2(x))^-1?

kinslayer · « **Reply #3099 on:** April 26, 2014, 10:10:16 pm »

Quote from: Cort on April 26, 2014, 09:49:33 pm

Thank you Alchemy, that definitely cleared my worries up. Now my other worries! It's space magic!

I am confuzzled when you differentiate sec^2 (x). I know it goes to 1/cos^2(x), but what happens from there on wards? What do you do with the ^2? Do I handle this like a normal 1(cos^2(x))^-1?

$\sec^2 x = \frac{1}{\cos^2{x}}$

Start with the chain rule: $y = 1/u, u = \cos^2 x$ .

So $\frac{\text{d}y}{\text{d}x} = \frac{\text{d}y}{\text{d}u} \cdot \frac{\text{d}u}{\text{d}x}$ . But what is $\frac{\text{d}u}{\text{d}x}$ ? Well, let $u = w^2$ where $w = \cos(x)$ . Then $\frac{\text{d}u}{\text{d}w} = 2w, \frac{\text{d}w}{\text{d}x} = -\sin(x)$ .

So $\frac{\text{d}y}{\text{d}x} = \frac{\text{d}y}{\text{d}u} \cdot \frac{\text{d}u}{\text{d}w} \cdot \frac{\text{d}w}{\text{d}x}$

You can actually write this down immediately if we examine what each of those terms is telling us.

We multiply the derivative of the reciprocal by the derivative of the polynomial (square) by the derivative of the cosine. That is, one term for each function composition. $\sec^2(x) = f(g(h(x)))$ where $f(x) = \frac{1}{x}$ , $g(x) = x^2$ and $h(x) = \cos x$ .

So the solution to the original problem is $\frac{\text{d}y}{\text{d}x} = -\frac{1}{\cos^4x}\cdot 2\cos x \cdot -\sin x = \frac{\sin(2x)}{\cos^4x}$ .

lzxnl · « **Reply #3100 on:** April 26, 2014, 11:29:38 pm »

Alternatively you can, like me, just know that the derivative of sec x is sec x tan x
So the derivative of sec^2 x would just be 2 sec x (first part of chain rule) times derivative of sec x (second part of chain rule) which would be 2 sec^2 x tan x

kinslayer · « **Reply #3101 on:** April 26, 2014, 11:37:06 pm »

Quote from: lzxnl on April 26, 2014, 11:29:38 pm

Alternatively you can, like me, just know that the derivative of sec x is sec x tan x
So the derivative of sec^2 x would just be 2 sec x (first part of chain rule) times derivative of sec x (second part of chain rule) which would be 2 sec^2 x tan x

Yeah you can skip a step if you want. But I thought it might be helpful for Cort to see how you could apply the chain rule multiple times, because he was clearly trying to do that and not getting far.

*edited

lzxnl · « **Reply #3102 on:** April 27, 2014, 10:53:19 am »

It's just that not many people doing VCE maths can be bothered knowing that the derivative of sec x is sec x tan x. I'm just a weirdo in that regard xP

Cort · « **Reply #3103 on:** April 27, 2014, 10:55:13 am »

Thank you all for the help!

tiff_tiff · « **Reply #3104 on:** April 30, 2014, 03:52:20 pm »

Can somebody please help me with this question... pretty please

OABC is a parallelogram. D is a point on CB such that CD:DB = p:q (D is closer to C than B)
OD intersects AC at R
OA = a
OC = c
Given that OR = k1OD and CR = k2CA and using OC + RC = 0R find k1 and k2 in terms of p and q

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