VCE Methods Question Thread!

[Maz]

Thankyou- it seems so simple looking at how u did it haha
Can u help me on another one?
(2x-1)(x^2-x+3)^4

[keltingmeith]

Thankyou- it seems so simple looking at how u did it haha
Can u help me on another one?
(2x-1)(x^2-x+3)^4

This is so spec that it hurts... In full honesty, unless you're doing specialist, I would just ignore this entirely. If you ARE doing specialist, use u-substitution.

[Maz]

This is in my methods textbook and I'm not doing spec...?

[keltingmeith]

This is in my methods textbook and I'm not doing spec...?

Yeah, but the textbook DOES get things wrong - particularly around here, where the lines between methods and spec start to blur. It's the study design that you need to trust in terms of what you can or can't be asked, not your textbook.

[Maz]

Maybe the course is different from WA ( where i live) and Vic...since there is 6 out of 50 questions along this line- thats a pretty big percentage to skip- can u please tell me how to do it...please

[lzxnl]

Notice that you have the derivative of x^2 - x + 3 present there as well. That's looking suspiciously like that du/dx term in the chain rule. If so, that would mean u = x^2 - x + 3. In total, this means that if we associate dy/dx = dy/du * du/dx and du/dx = 2x - 1, then dy/du = u^4 if we let u = x^2 - x + 3

y = u^5 / 5 + c= (x^2 - x + 3)^5 + c

[Maz]

Thankyou

[bills]

A biathlon event involves running and cycling. Kim can cycle 30 km/h faster than she
can run. If Kim spends 48 minutes running and a third as much time again cycling in an
event that covers a total distance of 60 km, how fast can she run?

I got 48.75 km/h (some how?) but the answer is 15 km/h. Can someone show me the correct steps to solving this question?

[toodles]

That's just a really badly written question. Firstly, since it is a biathlon, you would assume that both legs of the race are equal length, and therefore solve these two simultaneous equations together:
4/5*x = 30
and                              (convert time from minutes to hours)
4/15(x+30) = 30             
which gives    4/5*x = 4/15(x+30)     , therefore  x = 15 km/h.
However, if you just solve    4/5*x = 30   , you get    x = 37.5 km/h. Therefore, you cannot travel an equal distance for both legs of the race and making that assumption would be wrong. (you can test that when x = 15 km/h, the distance is not 60 km)
Therefore, you have to solve for different lengths:
4/5*x + 4/15(x+30) = 60           , therefore x = 48.75 km/h
The correct answer to the question is that Kim runs at 48.75 km/h, but the question is super dodgy

[Peanut Butter]

Can someone please help me with this question:

If g(x)=sin(x) and h(x)=loge(x), find the derivative of g(h(1))

I attempted the question but got an answer of 0 (this sounds wrong?)

Could someone please show me how to answer the question as well as the correct solution?

Thank you!!

[brightsky]

Can someone please help me with this question:

If g(x)=sin(x) and h(x)=loge(x), find the derivative of g(h(1))

I attempted the question but got an answer of 0 (this sounds wrong?)

Could someone please show me how to answer the question as well as the correct solution?

Thank you!!

If this is how the question is worded, then you are entirely correct.

g(h(x)) = sin(ln(x))
g(h(1)) = sin(ln(1)) = sin(0) = 0
d/dx(g(h(1)) = 0

However, I'm assuming that the question is asking for d/dx(g(h(x)) | x = 1, which is something slightly different.

d/dx(g(h(x)) = d/dx(sin(ln(x)) = cos(ln(x))*(1/x)
d/dx(g(h(x)) | x = 1 = cos(ln(1))*(1/1) = cos(0) = 1

[Peanut Butter]

If this is how the question is worded, then you are entirely correct.

g(h(x)) = sin(ln(x))
g(h(1)) = sin(ln(1)) = sin(0) = 0
d/dx(g(h(1)) = 0

However, I'm assuming that the question is asking for d/dx(g(h(x)) | x = 1, which is something slightly different.

d/dx(g(h(x)) = d/dx(sin(ln(x)) = cos(ln(x))*(1/x)
d/dx(g(h(x)) | x = 1 = cos(ln(1))*(1/1) = cos(0) = 1

The question on the worksheet is worded exactly as I wrote above.

Could you please tell me where I went wrong (because now the answer I'm getting is undefined...)

Firstly, g(1) = 0
g(h(x)) = sin(loge(x))

u = loge(x)      y= sin(u)
u' = 1/x          y'= cos(u)

Therefore, g'(h(x)) = cos(loge(x))/x

Then I subbed 0 into the equation above....
g'(0) = cos(loge(0))/0 which is undefined

Could someone please tell me what I've done wrong? TIA

[Swagadaktal]

The question on the worksheet is worded exactly as I wrote above.

Could you please tell me where I went wrong (because now the answer I'm getting is undefined...)

Firstly, g(1) = 0
g(h(x)) = sin(loge(x))

u = loge(x)      y= sin(u)
u' = 1/x          y'= cos(u)

Therefore, g'(h(x)) = cos(loge(x))/x

Then I subbed 0 into the equation above....
g'(0) = cos(loge(0))/0 which is undefined

Could someone please tell me what I've done wrong? TIA

Wasn't it x =1 not x = 0?

[Peanut Butter]

Wasn't it x =1 not x = 0?

but the 'x' is actually h(x) (where the x of h(x) is 1) so that it is 0

does that make sense?

[brightsky]

The question on the worksheet is worded exactly as I wrote above.

Could you please tell me where I went wrong (because now the answer I'm getting is undefined...)

Firstly, g(1) = 0
g(h(x)) = sin(loge(x))

u = loge(x)      y= sin(u)
u' = 1/x          y'= cos(u)

Therefore, g'(h(x)) = cos(loge(x))/x

Then I subbed 0 into the equation above....
g'(0) = cos(loge(0))/0 which is undefined

Could someone please tell me what I've done wrong? TIA

but the 'x' is actually h(x) (where the x of h(x) is 1) so that it is 0

does that make sense?

In that case, the question is really poorly worded. As mentioned above, I think what the question is meant to ask for is the rate of change of the composite function y = g(h(x)) at the point x = 1. Imagine the graph of y = g(h(x)) in your head. In order to find the rate of change at x = 1, we need to find the derivative of y = g(h(x)) first, and then substitute x = 1. As Swagadaktal pointed out, it seems that you've found the derivative, but substituted x = 0 instead of x = 1. It is true that the value of the function y = h(x) at x = 1 is 0, but y = h(x) is not the function that we are trying to find the rate of change of, nor is x = 0 the x-coordinate that we are interested in.

Hope this makes sense! I have a feeling you might be thinking too deeply hahaha!