VCE Methods Question Thread!

[Talia2144]

Hi, can anyone please help me with this attachment file question 16 b thank you

[lzxnl]

Alright, I'll let you know the next time I see a complex function in a methods paper.

You know I like going on tangents

The equation of the perpendicular bisector of AB is simply the line that passes through the midpoint and has slope which is the negative reciprocal of that of AB.
AB has slope (5-7)/(7-1) = -1/3 so the perpendicular bisector has slope 3
Then it passes through the midpoint, which is (4,6), so you have y - 6 = 3(x-4) => y = 3x-6

Find the equation of BC. y-5 = (-2-5)/(0-7) * (x-7) = x-7
y = x-2

Set y=x-2 and y=3x-6 at the same time => x-2 = 3x-6
3(x-2) = x-2 => x-2 = 0, x = 2 and y = 0

[Talia2144]

Thank you appreciate it

[SE_JM]

With any extension of the natural log to complex numbers, I think that also works for any complex f not equal to zero

What do you mean?

[keltingmeith]

What do you mean?

If we consider the complex number and , then we say that is the polar form of z, with and .

From this, it is possible to define the complex logarithm .

lzxnl is saying that using this, we can actually make the log/exponential rules that normally won't work for negative numbers, work. Which is so far beyond methods it's not funny, so no need to stress yourself over it. (and for those doing spec - no, it's not in there either)

[SE_JM]

If we consider the complex number and , then we say that is the polar form of z, with and .

From this, it is possible to define the complex logarithm .

lzxnl is saying that using this, we can actually make the log/exponential rules that normally won't work for negative numbers, work. Which is so far beyond methods it's not funny, so no need to stress yourself over it. (and for those doing spec - no, it's not in there either)

thank god

[lzxnl]

If we consider the complex number and , then we say that is the polar form of z, with and .

From this, it is possible to define the complex logarithm .

lzxnl is saying that using this, we can actually make the log/exponential rules that normally won't work for negative numbers, work. Which is so far beyond methods it's not funny, so no need to stress yourself over it. (and for those doing spec - no, it's not in there either)

You gotta be careful here. e^(ln a) = a works, but ln(e^a) = a doesn't work

[SE_JM]

Hello
I'm currently studying exponential growth and decay

In my textbook it says:
A= A₀e^kt, where A is quantity at time t, and A₀ is a constant...

but in an example about compound interest problem which states:
"A bank pays 10% interest compound annually. You invest $1000. How does this $1000 grow as a result of the interest added?"

They say the answer is:
A=P(1+(r/100))ⁿ

This equation looks really different from this equation A= A₀e^kt
like where is the k and t, and where did the n come from?

I kinda confused about this whole thing.

Any kind of help would be appreciated

[SE_JM]

You gotta be careful here. e^(ln a) = a works, but ln(e^a) = a doesn't work

I get that  e^(ln a) = a works.
But why wouldn't ln(e^a) this not work?
It's a logarithm, and by using log laws couldn't this be:
log_ee^a
= a log_ee
=a

could you explain this to me? I don't get it

[Zues]

why is this not reflected in the x axis? i have (2-x)^2 in the denominator so (-x+2)^2 therefore -(x-2) ^2 therefore a = -1

or

is it?
(2-x)^2 in the denominator so [(-x+2)]^2 therefore [-(x-2)] ^2 .-> (x-2)^2 ?

[lzxnl]

I get that  e^(ln a) = a works.
But why wouldn't ln(e^a) this not work?
It's a logarithm, and by using log laws couldn't this be:
log_ee^a
= a log_ee
=a

could you explain this to me? I don't get it

In the complex plane, the natural logarithm is multi-valued. ln -1 = i*pi. However, e^(3i*pi) is also -1. Therefore, ln(e^(3i*pi)) = ln(-1) = i*pi which isn't 3i*pi.

You don't need to worry about this until at least first year uni so don't stress. I was joking with EulerFan

[SE_JM]

haha
I freaked out. I  used that all along and I thought I was solving the problems all wrong.
thanks

[cosine]

why is this not reflected in the x axis? i have (2-x)^2 in the denominator so (-x+2)^2 therefore -(x-2) ^2 therefore a = -1

or

is it?
(2-x)^2 in the denominator so [(-x+2)]^2 therefore [-(x-2)] ^2 .-> (x-2)^2 ?

Im pretty sure you mean reflection in the y-axis, as the negative sign is infront of the 'x' and not the whole function. Anyways, the reason its not reflected is because ultimately if you expand the brackets:

  you should obtain



=

=

So, from now, you see that the 'x' is positive, and theres a translation of 2 units in the positive direction parallel to the x-axis.

[Zues]

if i have the expression square root (6+3x) + 2 = 0 , and i am solving for x , why is it that i have no solutions

square root (6+3x) = -2
6+3x = 4
3x = 2/3
x = 2/9

is it because in the step square root (6+3x) = -2

we have to think foward a little bit and if we had 6+3x = 4 then this would become square root (6+3x) = 2, and not -2? hope that makes sense

[cosine]

if i have the expression square root (6+3x) + 2 = 0 , and i am solving for x , why is it that i have no solutions

square root (6+3x) = -2
6+3x = 4
3x = 2/3
x = 2/9

is it because in the step square root (6+3x) = -2

we have to think foward a little bit and if we had 6+3x = 4 then this would become square root (6+3x) = 2, and not -2? hope that makes sense

Well consider the transformations from the normal
to 

There is a translation by 2 units in the positive directin parallel to the y-axis (up), and there is also a translation of 2 units in the negative direction parallel to the x-axis (left), so the end point is [-2, 2]. With no reflection in the x-axis, the graph normally shoots out to the right, intersecting the y-axis at (0, 2+)

So, to answer your question. How can there be a x-value if the graph actually starts at (-2,2)?

Hope it helped