VCE Methods Question Thread!

[Holmes]

For the composite function to be defined we need 'the range of the second to be equal to or a subset of the domain of the first', way to remember it : R2D1.
So here what are our ranges and domains for and ?

So we need

Sorry, I'm a bit confused still. How do we get the range of gof? After gof has been aptly defined, as you demonstrated, is the range of the whole function just the range of g(x)?

[b^3]

...from there you should be able to do a little sketch and work out the range of the resulting function with that domain.

So you take the domain of the second function, here that is , that will be your domain for . From that you work out the equation for , and given the domain we have, work out the range from that graph.

The range of the composite function won't be the range of (well in some cases it might). There's no rule for the range of the composite function from the range or domain of the others like there is for the domain, you need to work out the range of from the domain of .

[DetteAmelie]

Hey guys!
So, I am stuck on the following question:
If y = , show that

I've tried to identify where my mistake lies within my working out, but I honestly can't. So, I was hoping someone could help me with that. Here is my working out so far:
1) http://oi41.tinypic.com/att2fa.jpg
2) http://oi40.tinypic.com/9t0k5l.jpg

[b^3]

Note: and are not the same, on a second glance you probably just left the power off the
Then you have the derivative of with respect to incorrect.

... actually well if your then the derivative is right but then the term isn't doing too much...

..ok now I'm confused as to what you've defined as what...

This is how you could set it out, although I've added steps in there. Meanwhile let me see if I can work out what you've done


EDIT: found it, in the when you've multiplied by , it should only be the second term multiplied by , not the , as the isn't included as part of that you were differentiating. The second problem is when you've done , you've said that , which is really the derivative of with respect to not .

[DetteAmelie]

Ohhhhhhhhhhhhhhhhhhhhhhhhh! Yep, I understand now. Argh, I feel so silly. Anyways, thanks b^3! I really appreciate it

[Holmes]

This was a question on today's methods sac. 9 marks, and I couldn't do it. From memory, this is the question. Could a pro please demonstrate a way to do this?

A soft drink company manufactures cylindrical cans. The volume of a can has to be
The cost of making the top of the can is $0.50, the cost of making the bottom of the can is $1.50 and the cost of making the side of the can is $1.00.
Using calculus, find the value for height and radius of the can so that the cost is a minimum.

If I've forgotten something or somethings unclear, sorry about that, ask me and I'll try my best to remember it.

[shadows]

Essentials Extended Response Ch 11
Question 26.

Extended Response Ch 13 Q22

bii) Isn't it just  error (change in alpha) multiplied by  ds/d(alpha)?

how to do iv, and v.


Help please.

Omg I am struggling. Are these questions supposed to be really hard or am I just not getting it D:?

[e^1]

Essentials Extended Response Ch 11
Question 26.

Extended Response Ch 13 Q22

bii) Isn't it just  error (change in alpha) multiplied by  ds/d(alpha)?

how to do iv, and v.


Help please.

Omg I am struggling. Are these questions supposed to be really hard or am I just not getting it D:?

This is for bi and bii. The picture below might help you with how you can get the answer.

[Sanguinne]

Related Rates Question

The surface area of a cube is decreasing at a constant rate of 9cm^2/s. Find the rate at which the sides of the cube are decreasing when the sides are the 1.5cm long.

I  worked out that the rate was 0.5. But the answer says -0.5. Is there a reason for this and if so why?

[Daenerys Targaryen]

Simply because it's decreasing?

[Sanguinne]

I thought it was only necessary to put a negative if it was decreasing and it asked for the rate of change

In this case however, it asks what rate  the sides are decreasing and  hence its implied that  the rate should be positive because the word decreasing is included in what it is asking.

[abcdqd]

This was a question on today's methods sac. 9 marks, and I couldn't do it. From memory, this is the question. Could a pro please demonstrate a way to do this?

A soft drink company manufactures cylindrical cans. The volume of a can has to be
The cost of making the top of the can is $0.50, the cost of making the bottom of the can is $1.50 and the cost of making the side of the can is $1.00.
Using calculus, find the value for height and radius of the can so that the cost is a minimum.

If I've forgotten something or somethings unclear, sorry about that, ask me and I'll try my best to remember it.

Volume of cylinder is pi*r^2*h, so 12pi = pi*r^2*h, therefore h= 12/r^2
The area of the top of the can (same as the bottom), is pi*r^2, so the cost for making the top and bottom is 0.5*pi*r^2+1.5*pi*r^2 = 2pi*r^2
Area of the side is 2*pi*r*h, cost is $1 per cm^2, so the cost to make is just 2*pi*r*h. Sub in 12/r^2 for h to get it all in terms of r.
Now you have the equation for the Cost, C= 2pi*r^2+ 24*pi/r. For minimum cost, find dc/dr, let it equal 0 and solve for r. Then use r to find h. Hope that helps

[jimmy22]

Volume of cylinder is pi*r^2*h, so 12pi = pi*r^2*h, therefore h= 12/r^2
The area of the top of the can (same as the bottom), is pi*r^2, so the cost for making the top and bottom is 0.5*pi*r^2+1.5*pi*r^2 = 2pi*r^2
Area of the side is 2*pi*r*h, cost is $1 per cm^2, so the cost to make is just 2*pi*r*h. Sub in 12/r^2 for h to get it all in terms of r.
Now you have the equation for the Cost, C= 2pi*r^2+ 24*pi/r. For minimum cost, find dc/dr, let it equal 0 and solve for r. Then use r to find h. Hope that helps

Hey, how come you multiplied the cost of the top of the can by the area of the can?
When you did 0.5*pi*r^2, i don't quite get this.

Thanks

[Holmes]

Hey, how come you multiplied the cost of the top of the can by the area of the can?
When you did 0.5*pi*r^2, i don't quite get this.

OP here, I think it's because it costs ($0.50 (as given) to manufacture the top of the can, so we multiply this with the area of the circle, which ends up giving us 0.5 pi r^2. Doing the same with the bottom of the can, which is $1.50, we get the usual measurements of a cylinder, which is that the cost is . So at the end of the question, the TSA actually equals the cost (obviously done deliberately), so that people like me who had no idea that we had to multiply the area of the different parts of the can by the cost could still get the answer.

Edit: Fixing up the english

[Sanguinne]

q4 exercise  8G MATHS quest
An isosceles triangle has its equal sides of length 10 cm with an included angle Ө as shown in the diagram at right. (I don't know how to upload stuff onto AN so it basically is an isosceles triangle with a theta for angle at the top as well has both its diagonal sides having the length of 10cm)

If Ө changes from 60 to 61, find correct to 2 decimal places:
a) the approximate area of the triangle when Ө=61
b) the approximate increse  in the area, A, of the triangle.

My main problem is setting up this equation....