VCE Methods Question Thread!

[Escobar]

This is a really dumb question but how do you evaluate cos (23pi)? And how do you know which quadrant 7pi\2 is?

remember that every 2pi is one revolution

[scarletmoon]

remember that every 2pi is one revolution

How do solv it though?

[Eiffel]

i know that the basic shape of a y = x^(even/odd) function is like bird wings (like a square root and also its reflection in y axis) such as y = x^(2/3) however when graphing it on my casio it has a sort of dilation y = x^(1/3) function. why is this so? i tried clearing it and also all variables still no hope.

[Escobar]

How do solv it though?

cos (23pi)=cos(22pi + pi)=cos(2pi*11+pi)
because each revolution (2pi) brings it back to the same place,
cos(2pi*11+pi)=cos(pi)=-1 (check unit circle)

7pi\2=7pi/2-2pi=3pi/2 which is in the 3rd quadrant (check unit circle)

[scarletmoon]

cos (23pi)=cos(22pi + pi)=cos(2pi*11+pi)
because each revolution (2pi) brings it back to the same place,
cos(2pi*11+pi)=cos(pi)=-1 (check unit circle)

7pi\2=7pi/2-2pi=3pi/2 which is in the 3rd quadrant (check unit circle)

Thanks that makes sense now

[knightrider]

when asked to find related rates equtions.

say we had to find and we knew

we would write

does it matter if we write

Or do we have to write it in the first order i stated?

[Escobar]

when asked to find related rates equtions.

say we had to find and we knew

we would write

does it matter if we write

Or do we have to write it in the first order i stated?

order doesn't matter

[Eiffel]

can someone please explain this?

[Escobar]

can someone please explain this?

even functions are symmetrical about the y axis so there cannot be any horizontal translation (h=0)
vertical translation doesnt matter (k is an element of R)

[Eiffel]

even functions are symmetrical about the y axis so there cannot be any horizontal translation (h=0)
vertical translation doesnt matter (k is an element of R)

what do you mean symmetrical about the y axis?

y = (x-1)^2 + 4

1,4 TP it is not symmetric about the  y axis, but rather symmetric at the vertex?

[dankfrank420]

Even functions mean that f(x) = f (-x). Therefore, they are concerned with reflection in the y axis.

Think about it, what is the only value of x will always be true for x = -x... 0!

1 =/= -1
2 =/= -2
etc...

But 0 = -0
0 = 0

Therefore, if your function is to be even, then x can have no horizontal translation.

X - h can only be equal to -(x-h) if the value of h is 0.

Hope that made sense.

Edit: Oh and K can be any real number because symmetry around the y-axis is not affected by vertical translations.

[Escobar]

what do you mean symmetrical about the y axis?

y = (x-1)^2 + 4

1,4 TP it is not symmetric about the  y axis, but rather symmetric at the vertex?

this is a property of even functions: f(-x)=f(x)
it means that: the y value of an x value is equal to the y value of -x
for y=(x-h)^2+k to be symmetrical, h can only be 0 because if it is shifted left or right, f(-x)=f(x) is not true

[Eiffel]

thanks!

just a self doubt with dialations

y= 3/(-5x+2) + 1

can we say 1/5 from y and 3 from x OR 3/5 from x, are both correct? (i will only mention one, not the or part)

[cosine]

thanks!

just a self doubt with dialations

y= 3/(-5x+2) + 1

can we say 1/5 from y and 3 from x OR 3/5 from x, are both correct? (i will only mention one, not the or part)

They are both correct man, but the second one is more accurate and more preferred. But definitely both are right.

[cosine]

Can someone help with:



Thank you in advance!