VCE Methods Question Thread!

[StupidProdigy]

How did you get equation of tangent to equal q-4??

Also, I graphed it with x^2+4x+4, unfortunately, that's not the correct answer

Bloody neap, so hard 

I didn't think of graphing it, but then I tried just then and it looks good surprisingly. Your graph had positive 4 not neg 4!

edit: equation of tangent I got by using y-y1=dy/dx(x-x2)-->y-(q-4)=0(x-2)-->y-(q-4)=0-->y=q-4   (I got y1 by subbing x=2 into f(x) ie (2)^2-4(2)+q which gives q-4). Look better?

[qwerty101]

Hey,

could someone help me out by explaining what is needed for a graph to be continuous and for its gradient to exist at a point.

At this time, for it to be continuous we essentially will have no breaks in the break - what if there is an open point at say x=3 for y=x?

Next, for gradients i properly know that end points and kinks have undefined gradients, what else can be said? say we have a break in a hybrid graph, how can i deduce that it is not defined at this point?

Lastly, could someone also explain the incorrect options for this question and what the right option is. (E)

Thanks

[cosine]

I didn't think of graphing it, but then I tried just then and it looks good surprisingly. Your graph had positive 4 not neg 4!

edit: equation of tangent I got by using y-y1=dy/dx(x-x2)-->y-(q-4)=0(x-2)-->y-(q-4)=0-->y=4-q   (I got y1 by subbing x=2 into f(x) ie (2)^2-4(2)+q which gives q-4

Ah, yes, sorry!

But still, the graph that it is supposed to look like has no x-intercepts, so basically the graph is like yours but it does not touch the x-axis, there is a vertical translation?!

Thanks for the help, by the way

[qwerty101]

Could someone also help me out on these two please 

[kinslayer]

Hey,

could someone help me out by explaining what is needed for a graph to be continuous and for its gradient to exist at a point.

At this time, for it to be continuous we essentially will have no breaks in the break - what if there is an open point at say x=3 for y=x?

Next, for gradients i properly know that end points and kinks have undefined gradients, what else can be said? say we have a break in a hybrid graph, how can i deduce that it is not defined at this point?

Lastly, could someone also explain the incorrect options for this question and what the right option is. (E)

Thanks

A good way to think about it is that if you can draw the graph without taking your pen off the page, then the graph is continuous.

If a function is differentiable at a point, then it must be continuous there, but even if the function is continuous at a point, it need not be differentiable.

Again, I like to think of a differentiable function as one that "looks like a straight line" as you zoom in. All polynomial functions have this property at any point since they smooth out. Likewise trig, exponential and logarithms. The classical example of a non-differentiable point is the point x = 0 in the graph of y = |x|. No matter how closely you look at that point, it's always going to look like a corner, so no dice.

To answer your first question, a function can't be continuous at a point unless it is defined there.

[StupidProdigy]

Ah, yes, sorry!

But still, the graph that it is supposed to look like has no x-intercepts, so basically the graph is like yours but it does not touch the x-axis, there is a vertical translation?!

Thanks for the help, by the way

Sorry I don't really understand what you mean :/
I've attached a picture of the graph I got, is this what you had? Anyway I'll leave it to someone else to confirm. No worries by the way cosine, just returning the favour

[cosine]

Hey,

could someone help me out by explaining what is needed for a graph to be continuous and for its gradient to exist at a point.

At this time, for it to be continuous we essentially will have no breaks in the break - what if there is an open point at say x=3 for y=x?

Next, for gradients i properly know that end points and kinks have undefined gradients, what else can be said? say we have a break in a hybrid graph, how can i deduce that it is not defined at this point?

Lastly, could someone also explain the incorrect options for this question and what the right option is. (E)

Thanks

For a graph to be continuous at a point, x=a:

- f(a) must exist, that is, the graph must be defined at x=a
- must exist, that is, the limit from the left hand side must equal the same limit from the right hand side
- that is, the limit must also equal to f(a).

A graph can only have a gradient at points where the tangent can be drawn, so no gradient at:

- Sharp points
- End points
- Discontinuous points

Option A is incorrect, you tell me why, but here is also why:

Spoiler

The point x=a is not continuous. Let's test the three rules for continuity to make sure. Is x=a defined? Yes, it is. One part isn't, but the top part is defined for x=a, so overall this box is checked. Now let's test the next rule. Is the right hand limit as the left hand limit? No. As x approaches a from the right hand side, it is approaching y=0. As the limit approaches x=a from the left hand side, it is approaching a negative y value. Clearly these are not the same limits. So x=a is not continuous.

Option B is incorrect. This is because the derivative is not defined at end points. And at x=a, that is an end point and is not defined, no tangent can be drawn. So it is not differentiable at

Option C is incorrect. This cannot be true because it says f(a)>f(b). But b is greater than x=a, so x=b is any x value after x=a, and the graph on the right side of x=a is above f(a)

Option D is obviously not correct. This function requires at least three funcitons

Option E is correct. This is because assuming a-2, we are on the left side of the graph, see that line at about y=-2? That's the part im talking about. Now does this have a zero gradient? Yes it does, hence option E is correct!

[cosine]

Sorry I don't really understand what you mean :/
I've attached a picture of the graph I got, is this what you had? Anyway I'll leave it to someone else to confirm. No worries by the way cosine, just returning the favour

Sorry, I was unclear. The graph is given, it is exactly as you drew it, but it's a bit above the x-axis. It says at x=2, the tangent is parallel to the x-axis, so the gradient at x=2 is zero. The parabola equation was x^2+px+q and asked to find q.

[StupidProdigy]

Sorry, I was unclear. The graph is given, it is exactly as you drew it, but it's a bit above the x-axis. It says at x=2, the tangent is parallel to the x-axis, so the gradient at x=2 is zero. The parabola equation was x^2+px+q and asked to find q.

Ah I think I get what you mean..so qR\{4} then I think. Nice question, I like it

[garytheasian]

Hey im struggling with this question i looked at the solutions and it still makes little sense to me.

[Floatzel98]

I've only seem to run into this problem now, but when using quotient rule does it actually matter if we derive the numerator or the denominator first?

[garytheasian]

I've only seem to run into this problem now, but when using quotient rule does it actually matter if we derive the numerator or the denominator first?

Yes, you must derive the numerator first. So vu' if u=numerator. However it doesn't matter for product rule, the order in which its derived won't change the end result as you are adding.

[Floatzel98]

I don't really know how to do this question:

The area of a circular disc in increasing from 100pi to 101pi. Find the corresponding increases in the radius. It is a linear approximation question if that helps

[garytheasian]

I don't really know how to do this question:

The area of a circular disc in increasing from 100pi to 101pi. Find the corresponding increases in the radius. It is a linear approximation question if that helps

I think this is right if i remember right.

[Floatzel98]

I think this is right if i remember right.

Thanks for that. I'm also stuck on another question. I'm trying to find dT/dl for the equation of a period of a pendulum swing. I keep getting dT/dl as . Where am i going wrong?

Thanks

EDIT: I worked out my mistake dw.

EDIT 2: I worked out dT/dl properly but i'm having trouble simplifying this the same as the answer in my book. Find the approximate increase of T when l is increased from 1.6 to 1.7. Give the answer in terms of . It carries over from the question above btw.

I have and then , but i can't get the same 'simplified answer as the book which is for the first one and then for the final answer