VCE Methods Question Thread!

[Lear]

Is there a more concrete method of creating a transformation matrix to transform g(x) into h(x) for the question attached OTHER than doing it by recognition (visually seeing how it needs to be dilated/translated etc)

[fruitbowl34]

Hi I'm a bit puzzled by this question "find the equations of the normals to the curve y=x2−5x+6 at the points where it cuts the x-axis." I understand how to do it but what does it mean when they say 'at the points where it cuts the x-axis'? what are the points exactly?

[darkz]

Hi I'm a bit puzzled by this question "find the equations of the normals to the curve y=x2−5x+6 at the points where it cuts the x-axis." I understand how to do it but what does it mean when they say 'at the points where it cuts the x-axis'? what are the points exactly?

Pretty sure it just means to find the equations of the normal at the x intercepts of the curve

[michaeljacksonftw]

differentiate
y = x + (1-x^2)^(1/2)
dy/dx for x = 1

dy/dx for (1-x^2)^(1/2) using the chain rule is ((-x)/(1-x^2)^(1/2))
so dy/dx for x + (1-x^2)^(1/2) is ((-x)/(1-x^2)^(1/2))+1
but the answer said it was (-(x-(-x^2+1)^(1/2))/(-x^2+1)^(1/2))
Can someone please explain what's wrong in my working out?
Thanks

[darkz]

differentiate
y = x + (1-x^2)^(1/2)
dy/dx for x = 1

dy/dx for (1-x^2)^(1/2) using the chain rule is ((-x)/(1-x^2)^(1/2))
so dy/dx for x + (1-x^2)^(1/2) is ((-x)/(1-x^2)^(1/2))+1
but the answer said it was (-(x-(-x^2+1)^(1/2))/(-x^2+1)^(1/2))
Can someone please explain what's wrong in my working out?
Thanks

You forgot to -1 from the power (1/2 => -1/2)

[michaeljacksonftw]

if y = f(x+2), but they don't day what f(x+2) is equal to, then how do we find the derivative?

[S200]

if y = f(x+2), but they don't day what f(x+2) is equal to, then how do we find the derivative?

Wait. You aren't given f(x)?

Could you attach a picture?

[pha0015]

if y = f(x+2), but they don't day what f(x+2) is equal to, then how do we find the derivative?

I might be interpreting this wrongly but (from prior experience)...

Is this one of those questions where it gives you the tangent to original  f(x) at a point (as well as a few other points), and you supposedly need to find the derivative of the f(x+2) in order to find the new tangent of f(x+2)? If so, then you just apply the transformation of (x+2) to the original tangent - if I remember correctly.

Sorry in advance as this may be of no use to you at all.

[michaeljacksonftw]

if a question asks to differentiate cos(x/2) with respect to x, does it matter if we use the chain rule, product rule, or standard differentiation?

[darkz]

if a question asks to differentiate cos(x/2) with respect to x, does it matter if we use the chain rule, product rule, or standard differentiation?

The question would most likely just be 1mark for the answer - so doesn't really matter about your working. And there is no need to use the product rule.

[michaeljacksonftw]

The question would most likely just be 1mark for the answer - so doesn't really matter about your working. And there is no need to use the product rule.

Thanks
Also, for differentiating a log function, if y = loge(kx), dy/dx = 1/x
so if y = loge(3-x), y = loge(-x+3), y = loge(-(x-3))
so in this case "k" = -1, so dy/dx = 1/(x-3)
Am i correct?
note: the "e" is meant to be base, don't know how to type that


Also, how do you expand cos(2pix) *(2pi)?
Thanks

[darkz]

Thanks
Also, for differentiating a log function, if y = loge(kx), dy/dx = 1/x
so if y = loge(3-x), y = loge(-x+3), y = loge(-(x-3))
so in this case "k" = -1, so dy/dx = 1/(x-3)
Am i correct?
note: the "e" is meant to be base, don't know how to type that

Well a better way to think of it is to just put whatever is in the loge i.e. (3-x) in the denominator, then differentiate that so (3-x)' = -1, and put that in the numerator. So it becomes -1/(3-x) = 1/(x-3). But I suppose your method works as well

[michaeljacksonftw]

Well a better way to think of it is to just put whatever is in the loge i.e. (3-x) in the denominator, then differentiate that so (3-x)' = -1, and put that in the numerator. So it becomes -1/(3-x) = 1/(x-3). But I suppose your method works as well

Thanks
find the derivate of the following function (quotient rule)
(2loge(2x))/(x)

let u = 2loge(2x)
let v = x

du/dx = 2(1/x) = 2/x
dv/dx = 1
v(du/dx) = (x)(2/x) = (2x)/(x) = 2
u(dv/dx) = (2loge(2x))(1) = 2loge(2x)
v^2 = x^2

so ((2) - (2loge(2x))/(x^2) = (2-2loge(2x)/(x^2)

Is my working out and answer correct?

[darkz]

Thanks
find the derivate of the following function (quotient rule)
(2loge(2x))/(x)

let u = 2loge(2x)
let v = x

du/dx = 2(1/x) = 2/x
dv/dx = 1
v(du/dx) = (x)(2/x) = (2x)/(x) = 2
u(dv/dx) = (2loge(2x))(1) = 2loge(2x)
v^2 = x^2

so ((2) - (2loge(2x))/(x^2) = (2-2loge(2x)/(x^2)

Is my working out and answer correct?

Well in my opinion, the question is really only like a 2 line solution, one for applying the quotient rule and one for the solution - you could add in more lines b/w the quotient rule and the answer for extra working, but isn't really needed. And since you got the same answer, I suppose that your method works as well. I've attached a picture of my working

[michaeljacksonftw]

is related rates still in the methods study design?