VCE Methods Question Thread!

[keltingmeith]

HELP PLEASE:

For a methods question we are given a table of values:
x   |   y
0   |   10
7   |   6.35
14   | 2.7
21   |   7.05
28   |   11.4

And the graph which passes through these points is in the form y = ax + b + dcos(nx)
We need to find the values of a,b,d&n

First thing:
I created 5 equations and tried to solve them simultaneously on CAS to find the values, the equations were:
10 = b + d
6.35 = 7a + b +dcos(7n)
2.7 = 14a + b +dcos(14n)
7.05 = 21a + b +dcos(21n)
11.4 = 28a + b +dcos(28n)

CAS did not give me any of the values.

So what am I doing wrong?? I have never seen a question in the form y = ax + b +dcos(nx) before and I am seriously stuck!!!!!!

A little unsure on the issue. Were you getting the wrong answer from CAS or nothing at all? Did it display any error messages?

[EulerFan102]

Could I please have some help with ... c for the second?

Okay, so here's how I'd attack part c
So, working with ϴ = 2, as was calculated previously, first calculate the area for the whole 'slice' OPQO:

Next, we can work out the area for the triangular section. Since it's a triangle, we can use A=1/2 bh. The base and height of the triangle can be found through trigonometry:

Comparing these two areas we can see that the triangle will be roughly 45.5% of the total area.

[EulerFan102]

Were you getting the wrong answer from CAS or nothing at all? Did it display any error messages?

I get "ERROR:Insufficient Memory" from mine when I try and compute it

[geminii]

I get "ERROR:Insufficient Memory" from mine when I try and compute it

That means you probably need to clear your memory...because it says 'insufficient memory'...

My question is, how do you write this function:
g(x) - 2x^5 - 2x^4 - 4x^3 +4x^2 + 2x
In the form:
g(x) = A(x-b)^3(x-C)^2+D ?
(aka what is A, B, C and D?)

Also, how many stationary points does it have? On my calculator I get two turning points and a point of inflection - don't know if that is counted as stationary or not - but in the formula it looks like there's two.

So confused, please help!

[Sirius]

Also, how many stationary points does it have? On my calculator I get two turning points and a point of inflection - don't know if that is counted as stationary or not - but in the formula it looks like there's two.

So confused, please help!

Stationary point basically means at what point is the gradient zero. The gradient is zero at both turning point and at point of inflection. Therefore this graph will have 3 stationary points.

[Abbey McMahon]

A little unsure on the issue. Were you getting the wrong answer from CAS or nothing at all? Did it display any error messages?

I was receiving no values at all from CAS, it would just spit out all the equations I put in without finding the variables.

[captkirk]

Hey Everyone.
The graph in the book is wrong.
I believe the answer is wrong too. Can someone please answer this question?
Answer= 16units^2

http://imgur.com/a/xMMWW

[keltingmeith]

I was receiving no values at all from CAS, it would just spit out all the equations I put in without finding the variables.

Interesting. I would assume this means that it just doesn't know what you're writing. Things to check:

1. All pronumerals have multiplication signs between them and anything else (eg, write 2*a*b instead of 2ab)
2. Clear the memory of any stored variables (it's been a while since I've used a CAS - if you don't know how to do this, type what calculator you have here and someone here can help. On my original TI-nspire, I remember there was a command somewhere called "Clear A-Z" or similar)
3. Open a new document and try performing the calculation in there.

If all of that doesn't work, try doing the same thing on a friend's calculator. If it works on their's, reset your calculator (by the button on the back) and try again. If it still doesn't work after that, I've got no clue, sorry.

Hey Everyone.
The graph in the book is wrong.
I believe the answer is wrong too. Can someone please answer this question?
Answer= 16units^2

http://imgur.com/a/xMMWW

lol that's beautifully wrong.

To find the answer for "find the area under the graph" questions, the easiest way to find the answer is to evaluate the integral of the absolute value of that function. In this case, that would be:



I've evaluated it for you, answer is 8*sqrt(13). It's worth double-checking my calculation if your work says otherwise, though.

[Sirius]

Hey Everyone.
The graph in the book is wrong.
I believe the answer is wrong too. Can someone please answer this question?
Answer= 16units^2

http://imgur.com/a/xMMWW

Hey 
I got 16 units^2 
Basically you find the area under the curve between 0 to pi and then multiply by 4 as all the other sections oscillate by the same amplitude and are a pi apart ( from pi to 2pi is a pi apart)

[captkirk]

Basically you find the area under the curve between 0 to pi and then multiply by 4 as all the other sections oscillate by the same amplitude and are a pi apart ( from pi to 2pi is a pi apart) hope this makes sense

Hmm that's what my solution manual says but i don't know. It's confusing mee 

[keltingmeith]

Hey 
I got 16 units^2 
Basically you find the area under the curve between 0 to pi and then multiply by 4 as all the other sections oscillate by the same amplitude and are a pi apart ( from pi to 2pi is a pi apart) hope this makes sense

This isn't the answer - and if it is, it's a fluke, because the method is faulty. You've based your integrands on the  graph they provided - which isn't the actual graph of the function. Note that your answer is slightly less than mine - this is due to the fact that you've got some negative area in your calculation.

[Sirius]

This isn't the answer - and if it is, it's a fluke, because the method is faulty. You've based your integrands on the  graph they provided - which isn't the actual graph of the function. Note that your answer is slightly less than mine - this is due to the fact that you've got some negative area in your calculation.

Yea true, the graph is incorrect! My mistake, I assumed the graph was correct. But if a question asked to find the area between the curve and the x axis from 0 to 4pi, assuming it had a correct equation(something similar to 3sin(x)) then can I use the method above? ?

Hmm that's what my solution manual says but i don't know. It's confusing mee 

So sorry for causing confusion. I only looked at the graph but as meltingkeith said the graph is incorrect. Sorry again.

[keltingmeith]

Yea true, the graph is incorrect! My mistake, I assumed the graph was correct. But if a question asked to find the area between the curve and the x axis from 0 to 4pi, assuming it had a correct equation(something similar to 3sin(x)) then can I use the method above? ?
So sorry for causing confusion. I only looked at the graph but as meltingkeith said the graph is incorrect. Sorry again.

Yup yup - your logic otherwise is flawless. It's a good way to do it, too.

[TheCommando]

Whats the derivative forumla for cos(x)
Doing -sin(x) doesnt seem to work.

[TheCommando]

Whats the derivative forumla for cos(u)
Doing -sin(u) doesnt seem to work.
For eg. Derivative of cos(-2x)