VCE Methods Question Thread!

[Ancora_Imparo]

You should know that:
1) Displacement of a particle is given by the signed area (area over x-axis minus area under x-axis) under the velocity-time graph
2) Distance of a particle is given by the unsigned area (area over x-axis plus area under x-axis) under the velocity-time graph.

So, in this question, you need to find the unsigned area under the graph of from to .
You can use integration, or since this is just a straight line, use triangles.

The t-intercept is at .
When :
When :

Distance travelled = Unsigned area =

[nspire]

Thanks a lot for your help, I understand better now. 

Could you please also help me with this question:

A body moves in a straight line so that its acceleration (a m/s2) after time t seconds
(t ≥ 0) is given by a = 2t − 3. If the initial position of the body is 2 m to the right of a point O and its velocity is 3 m/s, find the particle’s position and velocity after 10 seconds.

Thank you!!

[Ancora_Imparo]

You also should know that:
1) The relation for velocity versus time can be determined by differentiating the relation for displacement versus time:
Thus, the relation for displacement versus time can be determined by integrating the relation for velocity versus time:
2) The relation for acceleration versus time can be determined by differentiating the relation for velocity versus time:
Thus, the relation for velocity versus time can be determined by integrating the relation for acceleration versus time:

You are given the relation for acceleration versus time. So, you can determine the relation for velocity versus time by integration:

You are told that initially its velocity is 3 m/s, that is, at , :


So,

Now that you've found the relation for velocity versus time, you can similarly determine the relation for displacement versus time by integration:

You are told that initially its position is 2 metres right of O, that is, at , :


So,

To find the particle's position and velocity after 10 seconds, simply substitute into and :

right of O

[Planck's constant]

You should know that:
1) Displacement of a particle is given by the signed area (area over x-axis minus area under x-axis) under the velocity-time graph
2) Distance of a particle is given by the unsigned area (area over x-axis plus area under x-axis) under the velocity-time graph.

Very good and lucid explanation.

May I add that there is a commonly used graphical alternative :

1) Antidifferentiate v(t) to obtain an expression for x(t) - using the initial condition t=0, x=0 to determine the constant of integration.
2) sketch the graph of x(t) from t=0 to t=3
3) Add the distances travelled from beginning to end allowing for turning points.

Not as mathematically sound a solution as Ancora-imparo's but it is practical and intuitive

[joey7]

Hey just wondering if someone could give me a hand with this question
say you have a positive cubic with a local maximum at -3 and a local minimum at 5 and you are asked the domain over which y is increasing as x increases and the domain over which y is decreasing as x increases
would the answer be
-3<x<5 for increasing and x<-3 and x>5 for decreasing
or would the greater than and less than signs be greater than or equal to and less than or equal to
just confused as saw a checkpoints question where when asking for the domain for which the graph is strictly decreasing it included the turning point in the answer.

[SocialRhubarb]

At our school, we get taught that strictly increasing is inclusive. The difference between increasing and strictly increasing is that strictly increasing does not permit straight sections of graph, such as a hybrid function with y = 3 as an element.

[Daenerys Targaryen]

Hey just wondering if someone could give me a hand with this question
say you have a positive cubic with a local maximum at -3 and a local minimum at 5 and you are asked the domain over which y is increasing as x increases and the domain over which y is decreasing as x increases
would the answer be
-3<x<5 for increasing and x<-3 and x>5 for decreasing
or would the greater than and less than signs be greater than or equal to and less than or equal to
just confused as saw a checkpoints question where when asking for the domain for which the graph is strictly decreasing it included the turning point in the answer.

other way around

[fleet street]

So, let me just see if this is right. Increasing does not include endpoints, but strictly increasing does. In the case of straight lines, neither are included. Is this right?

[BubbleWrapMan]

Both include endpoints, regardless of the graph.

[Lasercookie]

So, let me just see if this is right. Increasing does not include endpoints, but strictly increasing does. In the case of straight lines, neither are included. Is this right?

The definition VCAA is using is stated here: http://www.vcaa.vic.edu.au/Documents/bulletin/2011AprilSup2.pdf This is the same definition given here: http://en.wikipedia.org/wiki/Monotonic_function#Monotonicity_in_calculus_and_analysis

Strictly increasing is defined as whenever we have we also have . This just says a smaller number will give a smaller output than some other number that's bigger, so it should make sense why we're defining things like this.

You can see this does not include straight lines. If we take the example of f(x) = 3 mentioned earlier, it is true that we can take two x-coordinates, a and b, and have for which . Clearly this doesn't fit with our definition of strictly increasing.

On the other hand, an increasing/decreasing function will include a straight line.  You can think of this term in this sense

increasing = An increasing function will never decrease.
decreasing = A decreasing function will never increase.

Makes sense, right?  In other words, whenever we have we also have then we say that the function is increasing over that interval. Flip the inequalities for the definition of a decreasing function over an interval. 

You can see those two statements would be true for a straight line, we can have an a < b where f(a) = f(b) and still fit the definition. Even though an increasing function will allow the values to be constant for some section, overall the values don't get smaller. For strictly increasing functions, we can see that it is 'strict' because it'll never stop increasing for one moment, every value will be larger than the previous value.

[clıppy]

I have no idea how to begin this question. Can anyone help me out?

[b^3]

"The radius of the sphere is cm". Since A is the center of the sphere, the hypotenuse of that triangle will be the radius of the sphere. So you have a right angled triangle with Hypotenuse , and the adjacent side of length . That should allow you to get a relationship between the variables for part a and b.

c Is just using plugging in the expressions from parts a and b.

d, find the derivative, let it equal zero to find stationary points, check endpoints (probably 0 in this case), find the corresponding theta.

e, Substitute the theta in to get the volume in terms of a.

[clıppy]

"The radius of the sphere is cm". Since A is the center of the sphere, the hypotenuse of that triangle will be the radius of the sphere. So you have a right angled triangle with Hypotenuse , and the adjacent side of length . That should allow you to get a relationship between the variables for part a and b.

I'm still lost. I can see how to do part b, but I can't see the relationship between that triangle and the height of the cone

[b^3]

Since A is the center of the sphere, the length AV will be equal to the radius of the sphere, that is cm. Then to get you just need to add the expression you found for AB.

EDIT: This might help.

[abcdqd]

Hey guys, just to clarify, if X is normally distributed, and Pr (X>a)= 0.025, it means that a is 2 standard deviations from the mean right? Sorry for the stupid question haha