VCE Specialist 3/4 Question Thread!

[Hok999]

Can anyone answer question 4 pls

Thank you in advanced !

Formula for Euler's Method is: y_n+1 = y_n + h.f'(x_n,y_n)
I can link you videos on youtube if you need to understand how this formula came about.

Answering your question,
x₀=0                                          y₀=1
x₁= 0+0.1=0.1                           y1= 1 + 0.1 x  (1³) = 1.1
x₂= 0.1+0.1=0.2                        y2= 1.1 + 0.1 x  (1.1³) = 1.2331
x₃= 0.2+0.1=0..3                       y3= 1.2331 + 0.1 x (1.2331³) = 1.4206
x₄= 0.3+0.1=0.4                        y4= 1.4206 + 0.1 x  (1.4206³) = 1.7073
x₅= 0.4+0.1=0.5                        y5= 1.7073 + 0.1 x  (1.7073³) = 2.2049

y is approximately 2.2049

You can also use the spreadsheet function in the CAS/Ti-Inspire to make this process a lot faster. Hope it helps!

[Mattjbr2]

Could someone please double check for me whether this is correct? How is it possible for c to equal 4?

We know dx/dt=4 when x=0
but x=0 when t=4
hence dx/dt=4 when t=4

So then aren't we supposed to sub in t=4 and dx/dt=4 to get c=44?

Thanks in advance!

[RuiAce]

Could someone please double check for me whether this is correct? How is it possible for c to equal 4?

We know dx/dt=4 when x=0
but x=0 when t=4
hence dx/dt=4 when t=4

So then aren't we supposed to sub in t=4 and dx/dt=4 to get c=44?

Thanks in advance!

At first glance, I think that \( \frac{dx}{dt} = 4\) when \( \boxed{x=0}\) has been typo'd. That should've read \( t=0\).

Else, yeah I agree the answers are wrong.

[Mattjbr2]

At first glance, I think that \( \frac{dx}{dt} = 4\) when \( \boxed{x=0}\) has been typo'd. That should've read \( t=0\).

Else, yeah I agree the answers are wrong.

Both the old and the new (corrected) versions of the worked solutions - as well as the back of the textbook - all agree on the same supposedly incorrect answer!  Haha. The back of the textbook has never been wrong as far as i remember.  That's so strange

[RuiAce]

Both the old and the new (corrected) versions of the worked solutions - as well as the back of the textbook - all agree on the same supposedly incorrect answer!  Haha. The back of the textbook has never been wrong as far as i remember.  That's so strange

If the answers have never been wrong then the question is what's wrong.

(But tbh, yeah, the question is wrong here imo. We know that \(x=0\) when \(t=4\), but that doesn't necessarily mean \(x=0\) ONLY when \(t=4\). What if \(x=0\) when \(t=4, 8\)?

When you consider that, the question suddenly doesn't become so doable.)

[Mattjbr2]

If the answers have never been wrong then the question is what's wrong.

(But tbh, yeah, the question is wrong here imo. We know that \(x=0\) when \(t=4\), but that doesn't necessarily mean \(x=0\) ONLY when \(t=4\). What if \(x=0\) when \(t=4, 8\)?

When you consider that, the question suddenly doesn't become so doable.)

Thanks so much!

[Yertle the Turtle]

Could someone explain how to reach an equation for this question, I've been struggling with it for a while.
Thanks

[S_R_K]

Could someone explain how to reach an equation for this question, I've been struggling with it for a while.
Thanks

Rate of population increase is proportional to current population size can be expressed as

where k is some real constant.

Hence, where k' = 1/k.

Then, antidifferentiating with respect to N, we have (from context we can assume N > 0).

At this stage, before solving simultaneously for k' and c, it is convenient to write N in terms of t. This gives



We then use the facts that:
and to solve simultaneously for k' and c in terms of initial population. This gives:

and

And substituting back into our equation for N, we have (with a bit of tidying up):



An alternative (much slicker) method is to begin by recognising that any solution to an equation of the form will be a function of the form Then substitute in known values (t = 0, f(0) = A, and t = 5, f(5) = 2A) to solve for b in terms of A. This gives This agrees with the above approach, because b is the reciprocal of k'.

[Mattjbr2]

Hey guys, if we want to give the interval for which f(x)=x^2 is strictly increasing, do we say (0,infinity) or [0,infinity)? In other words, do we include a stationary point? The A+ Methods Exam 1 practice exams book did include the stationary point. I didn't. Do i mark myself wrong, or nah?

[TheBigC]

Hey guys, if we want to give the interval for which f(x)=x^2 is strictly increasing, do we say (0,infinity) or [0,infinity)? In other words, do we include a stationary point? The A+ Methods Exam 1 practice exams book did include the stationary point. I didn't. Do i mark myself wrong, or nah?

Strictly increasing means NOT decreasing (hence, stationary points are included)... so yes - minus 1 mark.

[Mattjbr2]

Strictly increasing means NOT decreasing (hence, stationary points are included)... so yes - minus 1 mark.

On pages 43,44 of the Cambridge Methods 3&4 book the stationary points are never included. However on page 363 they note that both [0,inf) and (0,inf) are valid. But I guess [0,inf) is more valid than (0,inf) because the former is a larger set.. ¯\_(ツ)_/¯..they say that f is strictly decreasing in (-inf,0] and strictly increasing in [0,inf) so that means it's both strictly increasing and decreasing at x=0 

[S_R_K]

they say that f is strictly decreasing in (-inf,0] and strictly increasing in [0,inf) so that means it's both strictly increasing and decreasing at x=0 

The first part of the sentence is correct, but the inference is invalid.

f(x) is strictly increasing over the interval [a, b] just in case if, for all x1, x2 in [a, b], if x1 < x2, then f(x1) < f(x2). (The definition for strictly decreasing is obvious). Hence, if f(x) = x^2, f(x) is strictly increasing for all x ≥ 0, and is strictly decreasing for all x ≤ 0.

Notice that this prohibits a function being strictly increasing (or strictly decreasing) at a single point, since we never have f(x) < f(x).

[Mattjbr2]

The first part of the sentence is correct, but the inference is invalid.

f(x) is strictly increasing over the interval [a, b] just in case if, for all x1, x2 in [a, b], if x1 < x2, then f(x1) < f(x2). (The definition for strictly decreasing is obvious). Hence, if f(x) = x^2, f(x) is strictly increasing for all x ≥ 0, and is strictly decreasing for all x ≤ 0.

Notice that this prohibits a function being strictly increasing (or strictly decreasing) at a single point, since we never have f(x) < f(x).

Thanks guys

[Yertle the Turtle]

Could someone give me a hand with these two related rates questions? I just always struggle with related rates questions for some reason.
Thanks 

[DrDin213]

How does this question imply that the system has a net force of 0?