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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: justsoundslikeaworn-outcliche on May 20, 2013, 06:52:51 pm

Title: help me [ Application task] urgent + struggling atm
Post by: justsoundslikeaworn-outcliche on May 20, 2013, 06:52:51 pm

So basically I’ve started doing some revision for my SAC in a next week (it’s an application task) and I’ve been really stuck on a sheet I got given before my teacher who inconveniently left on leave for some family reason So who do I turn to you say???? ATARNOTES OFC anyways here it is (note ill put my working out in bold) It’s kind of long, and for my answers can you please tell me if I’m right or not??? I'm really struggling with methods righht now at the point were I cry and want to give up and join the circus :S

Title: Re: help me [ Application task] urgent + struggling atm
Post by: justsoundslikeaworn-outcliche on May 20, 2013, 07:59:21 pm

PLEASE SOMEONE

Title: Re: help me [ Application task] urgent + struggling atm
Post by: Zealous on May 20, 2013, 09:43:42 pm

Hey, is it possible for you to take a photo or scan the worksheet that you've got this from?
Me personally, I think it might be a little better to help you with.

For now, I can help you with the first question. Because the parabola is negative (upside down to touch the ground) we know A has to be negative. Now you know that B is 5, you can sub in one of the x intercepts in order to find a.
Using the point (4,0)
$0=16A+5$

$A=\frac{-5}{16}$

For sketching, the domain will probably be from $x\in [-4,4]$ as I am assuming we don't look at the parts of the building that is "below" the ground.
[EDITS]clarification

Title: Re: help me [ Application task] urgent + struggling atm
Post by: Zealous on May 20, 2013, 09:56:16 pm

Quote from: justsoundslikeaworn-outcliche on May 20, 2013, 09:54:07 pm

Hey sure! But note that the actual diagram doesn't have any intercepts you have to use the info given to find them :S
P.S THANK YOU SO MUCH

kk thanks, was incorrect about the intercept thing, I misread it before, I shall edit this post as I continue onto question 2, I will do as much as I can now, someone else can probably help you onwards.
'
See laserblued's post below, a lot more detailed.

Spoiler

I don't have much time now, but I shall give you some hints as to how to solve most of the questions.

I had to solve the first part of the load bearing arch on the calculator.

So I used the point (4,0) and put 4 in the x values and let it equal 0, solving for C and I got a value of 2cosh(2). Which I have no idea what it is, although approximately it is 7.5.

To find the y intercept just sub in x=0 for the new equation.

Use calculus to derive the equation, then let the derivative = 0 and see where x is when it occurs, this point will be a point of 0 gradient and therefore min/max/inflection (in this case it is a maximum).

Sorry not much time to give you some detailed answers, hopefully someone else can help you later tonight, hopefully what I have said makes some sense to you =p

Title: Re: help me [ Application task] urgent + struggling atm
Post by: justsoundslikeaworn-outcliche on May 20, 2013, 10:00:53 pm

Quote from: sushi. on May 20, 2013, 09:56:16 pm

kk thanks, was incorrect about the intercept thing, I misread it before, I shall edit this post as I continue onto question 2, I will do as much as I can now, someone else can probably help you onwards.

An honest mistake :) Thanks your a legend!! and yeah hopefully!

Title: Re: help me [ Application task] urgent + struggling atm
Post by: Lasercookie on May 20, 2013, 10:05:58 pm

Quote

This shape is the basis of the building you are trying to build which is an archway and a spire – rough drawing attached
[Spire-lower component-the parabola in blue]
Its vertex is 5m above ground level and touches the ground 4m to either side of it’s axis of symmetry. The equation of the parabola if of the form f(x)=Ax2+B.
   Write down coordinates of the vertex. (0,5)
   Write down the coordinates of both the x-intercepts. (-4,0) and (4,0)
   Determine the values of A and B in the eq of parabola.
When x=0, y=5, sub into eq: 5=0+B, B=5
Can’t find A :S

You've found all the information you need to find A. You've figured out that $f(x) = Ax^2 + 5$ , so just sub a point into the function e.g. $f(4) = 16A + 5 = 0$ and solve for A.

Quote

[Load bearing arch]
The upper component of the arch is to be designed to bear the load of the wall above and around. the best shape therefore is a catenary- which is a name given to a curve formed by 2 exponentials added together. It’s formula being g(x) =-eX/2 – e-X/2+C, the x intercepts of the catenary being (-4, 0) and (4,0) – NEED HELP WITH THIS WHOLE PART DON’T KNOW HOW TO SOLVE I know you let y=e^x but still doesn’t work for me :
   Use information to find the exact values of C =
   Write coordinates of y-intercept of g(x) correct to 2 decimal places
   Show that the mx value occurs at this y-intercept (use calc to find derivative of g’(x))
   Sketch over appropriate domain and state it in set notation

So $g(x) = -e^\frac{x}{2} - e^\frac{-x}{2} + c$ Note that this is equal to $g(x) = c -2\cosh(x/2)$ which is what the calculator might spit out. That's hyperbolic cosine (instead of trig functions defined on the unit circle, these are defined on the 'unit hyperbola', http://en.wikipedia.org/wiki/Hyperbolic_cosine if you're interested).

We can use g(4) = 0 for example to find what c is. We only need to find the exact value too.

$g(4) = -e^2 - e^{-2} + c = 0 \implies c = e^2 - e^{-2}$

y-intercept, sub in x = 0: $g(x) = -e^\frac{x}{2} - e^\frac{-x}{2} + e^2 - e^{-2}$ Remember that $e^0 = 1$
The question asks for a decimal answer, but calculate an exact value by hand first.

So we want to show that the maximum value of the function occurs at the y-intercept.
It says that to use your calculator to do this, but we'll do as much of it as we can by hand.

$g(x) = -e^\frac{x}{2} - e^\frac{-x}{2} + e^2 - e^{-2}$ Note that the last two terms (the value of 'c') are constants.

My answer here:

Spoiler

$g'(x) = -\frac{1}{2}e^\frac{x}{2} +\frac{1}{2}e^\frac{-x}{2}$

So sub in the y-intercept you calculated earlier in there. If g'(x) is a maximum value, then it should be the turning point - as evidenced by the diagram. So g'(x) should equal 0. From there we need to somehow show that this is indeed the maximum value. You can do this by testing the derivative to the left and right of the point (this is probably where you'd want to use your calculator) and see if this is a local maximum, or by the second derivative test (if you're familiar with it and assuming your teacher is okay with you using it in methods, it's not really on the methods study design so don't worry about it if you don't) or by whatever other method.

Might get around to the rest later.

Edit: More maffs
I guess just talking about application tasks in general and the kind of questions they tend to give you. Application sacs tend to have different marking criteria than usual. You might be given marks for explaining each step, writing a sentence about what you're going to do next and why. You might get given marks for drawing a diagram etc. And those are all good things, because the questions you get are the type where it really helps to write out all the information you've be given, and what you can figure out from it. The questions do tend to build on top of each other too. The first few parts of a question might be something relatively simple, calculate x coordinate etc. but the next few parts will probably try to mix up the wording a bit, so you have to pick apart what the question is trying to get at.

A big part of these kind of sacs is technology. It's helpful to be good at using your calculator, but it's also helpful to know when and where to use your calculator. You can nearly always use for it checking, aiding you in sketching graphs, having the function there so you know what it looks like etc. But then there's things like where the calculator isn't the greatest at simplifying equations, or maybe just not in the form you want it (as would be the case with those 'show that' questions).

Quote

Consider the right hand side of the construction. The form of the flaring is a √ fn of the form L(x)=J√(K-x). If the terminating point is ground level. Determine the values of J and K

There's been plenty of questions like this on this practice sac. Figure out what information you know about the function, and then use your knowledge about functions in general (in this case the square root function) combined with some algebra work to figure out what L(x) is.

Quote

Sketch over suitable domain

There's questions like this where you try and consider what values are realistic. Like sushi said above, there was that question where negative ground doesn't make sense, so you ignore all values where f(x) < 0. In this case you have to consider again that f(x) > 0 so what does this imply about x? You might also have to remember that we can't have a negative square root in the real numbers.

The task of sketching itself is hopefully straightforward. Sketch the intercepts etc. first, look at how the function behaves etc.

Quote

Use symmetry to determine the hybrid fn that fully defines x

This is the kind of question where you really have to have a good picture of what the graphs look like and what each term means. You'd want to think about why do we need to use a hybrid function to 'fully define' x and why can we use symmetry to figure it out? There's been some talk about right hand and left hand branches, the "lone segment" etc. in this sac too. For some of those quetsions it's just talking about a hyperbola, say y = 1/x, where the right hand branch is when x > 0 and the left hand branch is when x < 0. We can use symmetry because well, they're exactly the same around some axis (in the case of y = 1/x the axis of symmetry is y=-x, and I guess y = x too).

Another example of when this kind of stuff might come up is with the square root function, say the equation of a semicircle, $y = \sqrt{r^2 - (x-h)^2}$ . The positive square root is the top half of the circle. The negative square root is the bottom half. Take $\pm\sqrt{r^2 - (x-h)^2}$ and you have the full circle. You can define this using a hybrid function, where one part is the positive bit, and the other is the bottom bit. The same idea applies to hyperbolas, truncus etc. too

Title: Re: help me [ Application task] urgent + struggling atm
Post by: Jeggz on May 20, 2013, 11:10:39 pm

laserblued = legend.

Title: Re: help me [ Application task] urgent + struggling atm
Post by: justsoundslikeaworn-outcliche on May 20, 2013, 11:15:56 pm

Thanks so much sushi ! And laser thanks heaps and heaps for doing the first part- tomorrow morning first thing I'll do is play around with those numbers and find the answers and get back to you.
And yeah I get you that's what I struggle with integrating all my math knowledge into one application tasks its just easier when they're divided. That's why I wanted some direction on how to approach these questions individually because then id know what to look out for in the SAC like when they ask for tangents and log graphs with gradients I just get confused asf. :/

Title: Re: help me [ Application task] urgent + struggling atm
Post by: Lasercookie on May 21, 2013, 05:57:12 pm

Quote

Use algebra to show that the exact value of x at the point of g(x) =-eX/2 – e-X/2+C where the gradient=-1 is x=2loge(1+√(2))

So this is a question where it's building off a previous questions, using our value of c and g'
$g(x) = -e^\frac{x}{2} - e^\frac{-x}{2} + e^2 - e^{-2}$
$g'(x) = -\frac{1}{2}e^\frac{x}{2} +\frac{1}{2}e^\frac{-x}{2} = -\frac{e^\frac{x}{2}-e^\frac{-x}{2}}{2}$
So we want to solve $g'(x) = -1$

The line of attack I'll take here is to manipulate it into looking like a quadratic. This is a trick that you end up using fair bit in methods.
$ \begin{align} g'(x) &= -\frac{1}{2}e^\frac{x}{2} + \frac{1}{2}e^\frac{-x}{2} = -1 \ \ \ \ \ \ & \\ &= -\frac{1}{2}e^\frac{x}{2} + \frac{1}{2e^\frac{x}{2}} &\text{Negative exponent can go to the bottom of the fraction} \\ &= -\frac{e^\frac{x}{2}e^\frac{x}{2}}{2e^\frac{x}{2}} + \frac{1}{2e^\frac{x}{2}} &\text{Multiply by so that we can add the fractions together} \\ &= \frac{1-e^x}{2e^\frac{x}{2}} &\text{Index laws} \\ &= \frac{1-e^x}{2e^\frac{x}{2}} = -1 \\ &= 1-e^x = -2e^\frac{x}{2} \\ &= -e^x + 2e^\frac{x}{2} + 1 = 0 &\text{this is looking vaguely like a quadratic now} \\ &\text{Let } u = e^\frac{x}{2} \text{, note that } e^x = (e^\frac{x}{2})^2 = u^2 \\ &= -u^2 + 2u + 1 = 0 \end{align} $
Now it's just too easy to solve (quadratic formula). Solve for u and then sub that back into our equation $u = e^\frac{x}{2}$ so you can solve for x (remember that $log_e(e^x) = x$ . You'll have to make sure that you reject extraneous solutions.

Quote

The equation of the tangent to this point =h(x)= -x+d. Find the value of d correct to 2 places. Tangent I know equals f’(x1)(x-x1) need help with subbing values

Find the equation of the tangent like you normally would, equate it to h(x). Use the coordinates of the point when g'(x_1) = -1 since we already have those figured out.

$y - y_1 = \mathbf{g'(x_1)}(x-x_1) \implies y = x\mathbf{g'(x_1)}-x_1\mathbf{g'(x_1)} + y_1 = -x + x_1 +y_1$ (subbing in g'(x_1) = -1 for that last part)

So equating coefficients with y and h(x), it would seem that $d = x_1 + y_1$ , so just sub in your points x and y.

Title: Re: help me [ Application task] urgent + struggling atm
Post by: justsoundslikeaworn-outcliche on May 21, 2013, 06:29:46 pm

Spoiler

Quote from: laserblued on May 21, 2013, 05:57:12 pm

So this is a question where it's building off a previous questions, using our value of c and g'
$g(x) = -e^\frac{x}{2} - e^\frac{-x}{2} + e^2 - e^{-2}$
$g'(x) = -\frac{1}{2}e^\frac{x}{2} +\frac{1}{2}e^\frac{-x}{2} = -\frac{e^\frac{x}{2}-e^\frac{-x}{2}}{2}$
So we want to solve $g'(x) = -1$

The line of attack I'll take here is to manipulate it into looking like a quadratic. This is a trick that you end up using fair bit in methods.
$ \begin{align} g'(x) &= -\frac{1}{2}e^\frac{x}{2} + \frac{1}{2}e^\frac{-x}{2} = -1 \ \ \ \ \ \ & \\ &= -\frac{1}{2}e^\frac{x}{2} + \frac{1}{2e^\frac{x}{2}} &\text{Negative exponent can go to the bottom of the fraction} \\ &= -\frac{e^\frac{x}{2}e^\frac{x}{2}}{2e^\frac{x}{2}} + \frac{1}{2e^\frac{x}{2}} &\text{Multiply by so that we can add the fractions together} \\ &= \frac{1-e^x}{2e^\frac{x}{2}} &\text{Index laws} \\ &= \frac{1-e^x}{2e^\frac{x}{2}} = -1 \\ &= 1-e^x = -2e^\frac{x}{2} \\ &= -e^x + 2e^\frac{x}{2} + 1 = 0 &\text{this is looking vaguely like a quadratic now} \\ &\text{Let } u = e^\frac{x}{2} \text{, note that } e^x = (e^\frac{x}{2})^2 = u^2 \\ &= -u^2 + 2u + 1 = 0 \end{align} $
Now it's just too easy to solve (quadratic formula). Solve for u and then sub that back into our equation $u = e^\frac{x}{2}$ so you can solve for x (remember that $log_e(e^x) = x$ . You'll have to make sure that you reject extraneous solutions.
Find the equation of the tangent like you normally would, equate it to h(x). Use the coordinates of the point when g'(x_1) = -1 since we already have those figured out.

$y - y_1 = \mathbf{g'(x_1)}(x-x_1) \implies y = x\mathbf{g'(x_1)}-x_1\mathbf{g'(x_1)} + y_1 = -x + x_1 +y_1$ (subbing in g'(x_1) = -1 for that last part)

So equating coefficients with y and h(x), it would seem that $d = x_1 + y_1$ , so just sub in your points x and y.

Thanks sooo much!1 Your the best your such a legend to make time to do this :) :) :) I'll try this bit right now and btw

with this bit

Quote from: laserblued on May 20, 2013, 10:05:58 pm

So sub in the y-intercept you calculated earlier in there. If g'(x) is a maximum value, then it should be the turning point - as evidenced by the diagram. So g'(x) should equal 0.

How exactly do I sub in the y-value I'm not getting 0 as my answer :?

Title: Re: help me [ Application task] urgent + struggling atm
Post by: Lasercookie on May 21, 2013, 07:10:14 pm

Quote from: justsoundslikeaworn-outcliche on May 21, 2013, 06:29:46 pm

How exactly do I sub in the y-value I'm not getting 0 as my answer :?

Oops, I should have said sub in the x-coordinate of the y-intercept. So x = 0 into the derivative.

$g'(0) = -\frac{1}{2}e^\frac{0}{2} +\frac{1}{2}e^\frac{0}{2} = -\frac{1}{2} + \frac{1}{2} = 0$

And then test points to the left and right of that to ensure that it is a maximum (to the right g'(x) should be negative, to the left g'(x) should be positive, and then at g'(0) we know it's equal to zero).

Quote from: justsoundslikeaworn-outcliche on May 21, 2013, 06:59:58 pm

I found that u=-1 yeah and i tried subbing into original but I don't know why I'm doing this to find x=2loge(1+√(2))? :(

$-u^2 +2u + 1 = 0$ There's a negative out the front! It doesn't factorise to $(u+1)^2$ like $u^2 + 2u + 1$ does

$u = \frac{-2 \pm \sqrt{8}}{-2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$

So we look at $e^\frac{x}{2} = 1 \pm \sqrt{2}$

$1 - \sqrt{2}$ is negative (since \sqrt{2} is 1.4something) so it can't be a solution to $e^\frac{x}{2}$ since that's always greater than 0.

$e^\frac{x}{2} = 1 + \sqrt{2}$ then solve for x.

Title: Re: help me [ Application task] urgent + struggling atm
Post by: justsoundslikeaworn-outcliche on May 22, 2013, 05:12:59 pm

Need help please esp with sketching and if your wondering why I'm nagging so much it's because if i fail this SAC next week I can no longer do methods anymore :(

Title: Re: help me [ Application task] urgent + struggling atm
Post by: jojongber on June 24, 2013, 06:42:55 pm

Hey an i ask from which source did you get those questions from? I have my sac coming up as well (we're behind...i know)

Title: Re: help me [ Application task] urgent + struggling atm
Post by: satya on June 26, 2013, 04:43:40 pm

i cant see the questions?? all i can see is the picture of the graph is that the whole question?? :(

Title: Re: help me [ Application task] urgent + struggling atm
Post by: Lasercookie on June 26, 2013, 05:21:03 pm

Quote from: satya on June 26, 2013, 04:43:40 pm

i cant see the questions?? all i can see is the picture of the graph is that the whole question?? :(

Eh, looks like justsoundslikeaworn-outcliche edited them out. Why, I'm not sure. Some of the questions are quoted in the posts above.

Title: Re: help me [ Application task] urgent + struggling atm
Post by: Phy124 on June 26, 2013, 05:24:29 pm

Additionally, justsoundslikeaworn-outcliche, please chose more appropriate thread titles in the future as to coincide with the follow posting rule;

Quote from: ninwa on December 09, 2011, 03:41:46 pm

Avoid putting "Read this need help urgently" or similar in your thread title. Everyone is entitled to the same amount of attention.

Title: Re: help me [ Application task] urgent + struggling atm
Post by: satya on June 26, 2013, 06:16:47 pm

ohh, but lazyred by any chance do u remember any of these questions, i like to practice them??

Title: Re: help me [ Application task] urgent + struggling atm
Post by: Lasercookie on June 26, 2013, 06:54:47 pm

Quote from: satya on June 26, 2013, 06:16:47 pm

ohh, but lazyred by any chance do u remember any of these questions, i like to practice them??

Nah, other than the questions quoted in the above posts I don't remember what they were. If you want some practice questions, you could try googling:
"application task site:atarnotes.com" and maybe throw in a 'methods', 'calculus', whatever there too. That would presumably come up with some previous threads with application task-esque questions.

If you have the essentials text book, I thought the chapter review questions there were pretty good for revising for sacs in general.

ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: justsoundslikeaworn-outcliche on May 20, 2013, 06:52:51 pm