VCE Specialist 3/4 Question Thread!

[keltingmeith]

When they are using triangles and shapes to describe angles.

What does it mean when they say and and

what angles are they referring to in a triangle.

Say they let one point be a, and the other two b and c.

Can someone show me an example with a triangle and what exactly these angles are referring to?

Thanks 

This is



is the upper angle, is the lower angle. refers to the angle formed by adding those two angles together.

So, the letter in the middle tells you at which point the angle forms. The first and last letter tell you how the angle extends out.

Make sense?

[Eiffel]

That's my first line, not my second line, see common parts in bold!

Maximum gradient is what is written in the second line, if your "gradient function" is f'(x) (ie. derivative of f(x)), then you need to derive one more to find the maximum of that GRADIENT function.

i kind of get it. So if we let f''(x) = 0, this will tell us the x value in which the f'(x) graph is either maximum or minimum? can you expand on this, i think im nearly there with its understanding.

Also, from your summary book, why do need to convert to radians before undergoing diff?

[keltingmeith]

Also, from your summary book, why do need to convert to radians before undergoing diff?

Yes. Radians are numbers, degrees are some shitty thing that high school maths kids learn about and I still don't know whynot numbers, and so aren't suitable for arithmetic with numbers.

[Eiffel]

if i was to do diff with degrees will i be penalised?

[pi]

if i was to do diff with degrees will i be penalised?

Yes, because it doesn't work as it isn't mathematically correct.

This link explains it quite well via first principles -> http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func.htm

Worth reading, it's not a concept taught well in school

i kind of get it. So if we let f''(x) = 0, this will tell us the x value in which the f'(x) graph is either maximum or minimum?

Yep.

[lzxnl]

Yes, because it doesn't work as it isn't mathematically correct.

This link explains it quite well via first principles -> http://www.phengkimving.com/calc_of_one_real_var/06_the_trig_func_and_their_inv/06_01_the_trig_func/06_01_04_diffn_of_trig_func.htm

Worth reading, it's not a concept taught well in school

Yep.

Well...sometimes when f''(x) = 0, f'(x) has a point of inflection

[Eiffel]

Yeah exactly that is what's annoying. Graphically can someone show me how it all links? e.g. the maximum and minimum gradient. I would've though that if f''(x) = 0 we are finding points in which f'(x) has a gradient of 0 (how is this a maximum or minimum gradient?) - isnt a negative or positive number more "minimum" and more "larger" respectively?

So back to -  if f''(x) = 0 we are finding points in which f'(x) has a gradient of 0. If we sub these points into f'(x) doesnt this provide a zero gradient?

Well...sometimes when f''(x) = 0, f'(x) has a point of inflection

and do you mean f(x) has a POI?

[pi]

Graphically can someone show me how it all links?

You know, you could start off with a function, derive it, derive it again, graph the three and see for yourself? Super easy on CAS I'd highly encourage trying things out for yourself (or trying, taking a break, and then trying some more!) to get such concepts!

Anyway, quick google found this (ignore the random picture bottom-left):

To explain it in case you're still confused:
> f(x) has a non-stationary point of inflection, that is where the gradient of that function is the greatest, hence f'(x) has a maximum at that x-value
> f'(x) has a maximum (turning point) as aforementioned, hence the gradient at that point is 0, hence its gradient function which is f''(x) will have an x-int at that point
> f'(x) also has maximum/minimum gradients either side of its turning point, hence f''(x) will have maximums turning points and minimum turning points on either side of its x-int as they represent the gradient of f'(x)

That's really all there is to it.

[SE_JM]

No worries. First go to the link I posted. Circumcircle is another word for "circumscribed circle" and the radius of the circle is given in the article (sqrt(3)/2 * a, where a = 10).

Now you have the radius of the circle, and you can work out the area of the circle. The area of the bit between the circle and the triangle is the area of the circle take away the area of the triangle, which is half the base times the height. The base you know is equal to 10cm. Now make a right-angled triangle with adjacent side 5cm and angle 60 degrees, and use trigonometry (remember SOHCAHTOAH) to calculate the height.

Now you have both areas and you can subtract one from the other to get the answer.

Thank you Kinslayer!

[SE_JM]

Hello,
Could someone please help me with these questions?
I'm getting really frustrated with this

Two circles of radii 3 cm and 4 cm have their centres 5 cm apart. Calculate the area of the region common to both circles

and

Two wheels (pulleys) have radii of length 15 cm and 25 cm and have their centres 60 cm apart. What is the length of the belt required to pass tighlt around the pulleys without crossing?

Thank you

[pinklemonade]

Hey,
Could someone help me find how to work out the turning point for

I've done this in my book before and got x intercepts of 4 and -2 but I'm confused as to how to do the turning point

Thank you

[IndefatigableLover]

Hey,
Could someone help me find how to work out the turning point for

I've done this in my book before and got x intercepts of 4 and -2 but I'm confused as to how to do the turning point

Thank you

There's two a few ways you could do it (one being the easier and the other takes a bit more time).

1. Just use the formula which you should know as the TP formula. Using this and subbing in the values from your function, you'd get x=1. From here you just sub it back in and you'll get your TP (1, -2/9).

2. You can differentiate the whole function and then let the derivative equal zero and solve for x (which will be x=1). Then you would just sub it back in and get the same answer as above

EDIT: point gives another example of how you could find the turning point as well (I guess I shouldn't restrict the number of ways you can solve a particular question )

[point]

There's two ways you could do it (one being the easier and the other takes a bit more time).

1. Just use the formula which you should know as the TP formula. Using this and subbing in the values from your function, you'd get x=1. From here you just sub it back in and you'll get your TP (1, -2/9).

2. You can differentiate the whole function and then let the derivative equal zero and solve for x (which will be x=1). Then you would just sub it back in and get the same answer as above

You can also use the original equation , Complete the square to get , Which will give you a turning point of , then take the reciprocal of that which will give you a final turning point of

[IndefatigableLover]

So we're doing Implicit Differentiation at school and one question in the chapter I'm doing has stumped me a bit:

Find    for:



So I have no issue with differentiating on the right hand side but differentiating on the left side is what's troubling me because I don't know how the answers got from:

to

[abcdqd]

So we're doing Implicit Differentiation at school and one question in the chapter I'm doing has stumped me a bit:

Find    for:



So I have no issue with differentiating on the right hand side but differentiating on the left side is what's troubling me because I don't know how the answers got from:

to

Chain rule. Let u=x-y