VCE Methods Question Thread!

[MightyBeh]

Yes that's right, could you share your working?

Attached

[aliannalevsschool]

Thank you the question is quite hard to post but it VCAA 2013 exam 2, question 3d

[StupidProdigy]

For the question attached I've differentiated the function and found the discriminant. Is this the correct working and would the answer be D?

Has anyone done vcaa 2015 recently and can help me with part C of this question?
I could find the d value, but i can't find the set of values of D that they want, thanks in advance..

Quoting both of you because you should do these questions with cas and are both exam 2 questions (hopefully you have the ti-nspire)-however be sure you understand how to do it by hand also. I hope you are aware of 'sliders', if not you can find them in the menu on a graphing page (menu, 1:actions, b:insert slider). Put in the graphs relevant to each of your questions and name the sliders after the unknown variables that you have, then fiddle around with the slider value until you satisfy what the question wants. If you need a deeper explanation on this I'll make some screenshots.
Also yes MB you're working is correct by finding the discriminant for the derivative

When solving simulataneous equations for a and b, and the calculator outputs
{a = 2, b = 3}, {a = 3, b = 2}, do we have to list out both situations? This is assuming there are no restrictions for the values of a or b.

Yep, especially if there are no restrictions!

Is it just me or is the answer to 2013 Exam 2 4diii (very last question) incorrect? When I solve for the turning point, I do not get k = 16/3, but I get k = 5.5937807, with a min area of 1.9183868??

Nothing wrong with their answer, maybe put your working up if it's still bugging you

[sweetcheeks]

For the attached question, would the derivative exist when x=1. The solutions say that the derivative doesn't exist at this point, as the original function is undefined. However, as the limit exists at this point (I think), wouldn't the derivative technically still exist?

Have I misunderstood the concept, or have iTute made a mistake?

Look at the f(x). Although you could theoretically sub in x=1 into the derivative, what would happen if you sub x=1 into f(x)? the bottom would be equal to zero, which cannot occur.

[MB_]

Quoting both of you because you should do these questions with cas and are both exam 2 questions (hopefully you have the ti-nspire)-however be sure you understand how to do it by hand also. I hope you are aware of 'sliders', if not you can find them in the menu on a graphing page (menu, 1:actions, b:insert slider). Put in the graphs relevant to each of your questions and name the sliders after the unknown variables that you have, then fiddle around with the slider value until you satisfy what the question wants. If you need a deeper explanation on this I'll make some screenshots.
Also yes MB you're working is correct by finding the discriminant for the derivative

I don't have the ti-nspire but the classpad does have a slider feature. Also, when working by hand what do I do after finding the discriminant?

[solution]

Sorry for the question spam
Is (1.) sufficient or do we actually need to sub values in (2.)?

[StupidProdigy]

I don't have the ti-nspire but the classpad does have a slider feature. Also, when working by hand what do I do after finding the discriminant?

Oh that's good, I wasn't sure if they had it.
You don't need to do anything after that for this question, just work through the options until you find what is true which you had already done. You found the equation for when the gradient is zero, and then found the necessary discriminant condition to ensure no solutions to the equation-i.e no stationary points

[nadiaaa]

Hi StupidProdigy,
So i put the graph in my ti and it just doesnt come up and i play around with the window settings but neither works..are you able to put the function in your ti and tell what settings its on?

[StupidProdigy]

Sorry for the question spam
Is (1.) sufficient or do we actually need to sub values in (2.)?

Just copy what's in your textbook-they will have one of these tables somewhere. Also I personally like to sub in points closer to where the gradient is zero, i.e if the gradient is 0 at x=2, I would use x=1.95 and x=2.05 as my sign/value check either side of x=2-it's a bit safer as the graph is less likely to change its gradient significantly in a smaller interval usually

[StupidProdigy]

Hi StupidProdigy,
So i put the graph in my ti and it just doesnt come up and i play around with the window settings but neither works..are you able to put the function in your ti and tell what settings its on?

Okay I'm not 100% sure what you mean doesn't come up, however I've gone through the question and set up some screen shots. Also I should clarify that although the 'slider' approach is incredibly good, you have to get the exact answer in a calculator page rather than the graphing page. The graphing page is there to give you a really intuitive understanding of what the question wants, and furthermore it is a checking method, it will guarantee that you have the right answer-this is a pretty good feeling when you know that you make up only 4% of the state who got the answer to both parts and this is why I want you to understand it.

First of all, the 'setting up' photo is just to show you how to graph the function with what range you need and also entering in the slider 'd'. Once this is done, have a look what happens when you range d from >0 to <10. You'll see that the function has a minimum appear and disappear-this is what the question revolves around.

The 'first part' photo demonstrates how the function changes near t=0. You can see that when d is about 6.3 it is just less than V(0). As we increase 'd' there is a change in the function, it's y value becomes larger and larger until it becomes above 10 (i.e V(0)). So from this we have already confirmed that the minimum value of d will be very very close to 6.3. What also happens is that the zero gradient dissapears, and it is when this dissapears that the function becomes greater than 10. It transitions from a local minimum, then instantaneously to a stationary point of inflection, and then to a curve without zero gradient. To find the exact value see the 'cas' picture, which just solves for this turning point value. What we then do to be 100% positive that our answer for our gradient transition is the value for 'd' is to convert it to the approximate (rather than exact) form, purely for our own analysis and check. Remember to write the value of 'd' in exact form for the final answer of course and remember to use an inequality sign (you can't use d=20/3 as this is when the stationary point of inflection instantaneously occurs.

To the next part of the question, its the exact same basic principles-finding where the function forms it's minimum turning point as d increases from greater than 0 to almost 0.4. The function has no minimum when 'd' is close to 0, we want to find the instantaneous point where the gradient changes becomes 0 and the stationary point is introduced. The picture shows that we can expect with certainty that the upper value of d will between 0.1 and 0.4. Again we solve using the calculator page for the exact answer and check it is between 0.1-0.4. And then it's just writing down the values of 'd' again so remembering to make sure your inequalities are correct which you can double check with the slider and your answers from the calc page.

I really hope this all makes sense and you see value in it (because you're guaranteed full marks basically)

[nadiaaa]

Wow thank you for putting so much time into this question!!!
But just a few things,
i get it but i dont really get it? The slider thing is awesome, like im playing around with the d value but i dont know how to interpret what im doing? I feel like i just dont understand this question. The 't=0 and t=5' really confuses me, like do i input t=5 and then sketch the graph for part ii?? And when im doing it, i dont see any like difference between stat point of inflection and zero gradient when i change the d value.
Are u able to explain in like really baby steps what the question like means?
Btw thanks again, this was super duper helpful, i really appreciate it

[YellowTongue]

Look at the f(x). Although you could theoretically sub in x=1 into the derivative, what would happen if you sub x=1 into f(x)? the bottom would be equal to zero, which cannot occur.

So does that mean that because f(x) is undefined at x=1, the derivative is also undefined at x=1?

[StupidProdigy]

Wow thank you for putting so much time into this question!!!
But just a few things,
i get it but i dont really get it? The slider thing is awesome, like im playing around with the d value but i dont know how to interpret what im doing? I feel like i just dont understand this question. The 't=0 and t=5' really confuses me, like do i input t=5 and then sketch the graph for part ii?? And when im doing it, i dont see any like difference between stat point of inflection and zero gradient when i change the d value.
Are u able to explain in like really baby steps what the question like means?
Btw thanks again, this was super duper helpful, i really appreciate it

We don't actually input t=0 or t=5, it's already there (on the graph I have used x instead of t because it doesn't accept t for graphing). So basically we are looking at the point where x=0 on the graph-so the y intercept. In part i, we want the lowest value over the interval 0<x<5 to be the y intercept (which has a fixed value of 10 no matter what-it is independent of when we change 'd'). We must find the values of d such that they will cause the function to always have a y value (or height on the graph) above the line y=10 (that's why I put the red graph y=10 there, this is like a bar that we have to be above to satisfy what the question wants.

You might not see a change from a turning point to a stationary point of inflection very well because it is very small, you have to zoom in lots and lots and you will be able to then see more closely when it changes as you change 'd' (you won't be able to see it exactly though, but pretty close).

Now for t=5, or x=5 as it is shown on the graph, what you should do is draw the vertical line x=5 (on the ti cas graphing page go menu: 3: 3: 1: 2) or visualise it. So now we want our graph to always be below the height of the graph when x=5, or in different words we want the y values between x=0 and x=5 to always be smaller than the value right at the right end point (x=5 exactly).

If I need to keep saying things just tell me, I'm more than happy to try and reword or more concisely convey what I'm saying

[StupidProdigy]

So does that mean that because f(x) is undefined at x=1, the derivative is also undefined at x=1?

Yep spot on. Though the limit exists this does not mean the point exists. You were correct earlier that the limit for f(x) at x=1 exists, this is because as x approaches 1 from the LHS and RHS the function value approaches one. Though it may approach this value it cannot equal it as our function output will become undefined.

[Jesse_ando]

Hi StupidProdigy, just a quick question, do you know why the VCAA answer includes the square bracket around the 9? As the question states 'strictly decreasing', the point where x=9 the curve is stationary, not increasing nor decreasing. It's question 1 of section 2 of 2009 E2 (i've included the pics)