VCE Methods Question Thread!

[swagsxcboi]

I am getting no where with this question... someone pls help

[Phy124]

Distance through bush is equal to and distance through clear is equal to

[b^3]

Firstly start by trying to make a function that gives you the time taken to travel the distance given, in terms of . You can do this by finding the time taken for each to travel each section using , adding those together to get the total time taken. Then finally minimising this function, by checking the endpoints of the function for the domain it's defined on and checking where you have minimum turning points.

Spoiler

So the distance that she walks through the bushland we'll call and the distance through clear land . Then we have

Then the taken taken to go through bushland will be

And through clear land

So the total time is then given by

Checking the graph we can see that the minimum will be at a turning point and not the endpoints (https://www.desmos.com/calculator/yzfwjgrvt6), so we need to first differentiate it,

So the value of for which she gets there in the least time is

EDIT: Beaten.

[MNM101]

For exponential functions what would c*c^-2 be?

[kinslayer]

For exponential functions what would c*c^-2 be?

You need to use the exponent laws: and

In particular:



Alternatively,

[MNM101]

How would u deal with (1/8)^1/3 ?

[Phy124]

[paper-back]

Can someone help me with the second post on page 288?
Thanks

[swagsxcboi]

h^3.. what?

[EspoirTron]

h^3.. what?

So you know that Volume=1/3pir^2h
Okay well tan(a)=3/4. Tan is just opposite/adjacent, which in this case is just the radius/height. Hence r/h=3/4. Rearranging we find that r=3h/4.
Let us sub r=3h/4 into our volume formula.
If follows that V=3/16pih^3 as required.

[soNasty]

For what values of k, where k is a real constant, does the equation , have two distinct solutions?

i get its a quadratic and stuff and to solve using determinant, however the answer is and i dont get how it'd be that because i initially had

[Frozone]

Could someone please help with with question 1.3.9?
I get lost when I am up to the point where cos(2x-pi/2)=1
Because cos 1=0.

[Thorium]

Could someone please help with with question 1.3.9?
I get lost when I am up to the point where cos(2x-pi/2)=1
Because cos 1=0.

If cos(2x-pi/2)=1
2x-pi/2=0
x=Pi/4

Also pi/4-(2pi/2)=-3pi/4, (same point occurs at 2pi intervals for cos. Since we have dilation of factor half from y-axis, same point occurs at pi intervals for our graph)

So x-intercepts are at x=pi/4,-3pi/4

[Zealous]

For what values of k, where k is a real constant, does the equation , have two distinct solutions?

i get its a quadratic and stuff and to solve using determinant, however the answer is and i dont get how it'd be that because i initially had

So many of us have had trouble with this question in this thread =p. Here's b^3's explanation from ages ago (Go vote up his post!):

The biggest trick is making sure your exponential function (y in his working out) is greater than 0 at all times to avoid anything undefined.

Now as we said before, has to be greater than zero, since we cannot power something and get zero or a negative number. So we need to check for this.

Now for both solutions to be positive, we need the part before the to be greater than the part after the , as if this occurs, when we take the second part away we will still get a positive solution (as long as our denominator is positive).


So the possible values of are

This is represented by the following graph, where the intercepts have to be greater than zero.
https://www.desmos.com/calculator/gxxmhbo2v6

EDIT: Added in the dropped explanation.

[soNasty]

thanks so much Zealous and B^3! That Q was annoying lol