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
 = Ax^2 + 5)
, so just sub a point into the function e.g.
 = 16A + 5 = 0)
and solve for A.
[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
 = -e^\frac{x}{2} - e^\frac{-x}{2} + c)
Note that this is equal to
 = 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.
 = -e^2 - e^{-2} + c = 0 \implies c = e^2 - e^{-2})
y-intercept, sub in x = 0:
 = -e^\frac{x}{2} - e^\frac{-x}{2} + e^2 - e^{-2})
Remember that
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.
Note that the last two terms (the value of 'c') are constants.
My answer here:
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 maffsI 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).
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.
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.
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,
^2})
. The positive square root is the top half of the circle. The negative square root is the bottom half. Take
^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
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