VCE Methods Question Thread!

[Professor Polonsky]

If we divide using long division, we will get it in the form

I found this method easier, myself, so in case this is helpful to anyone.

[darklight]

An adventurous man is marooned on an island, with the depth of water around the island modelled by h(t) = 2 cos ((pi*t)/12) +3.

A channel separates the island from the mainland. The depth of water in this channel is the same at all points between the island and the main land, and is given by h(t). The man decides to attract the attention of passing ship. A ship will need a depth of at least 4 metres.

If the mainland is 10 km away from the island, what speed (in km/hour) must the ship average to rescue the man and take him to the mainland?

Is the answer 5 km/hr?

[zhe0001]

Please help with this indicial equation, stuck



Thanks!

[Alwin]

Please help with this indicial equation, stuck



Thanks!

Hint: times both sides by 2^x

Edit: Dont think my hint was enough, so here's some more:

[zhe0001]

hahah thats where i was stuck

Well lets break it up first



That should be enough to help you get started

[zhe0001]

Thank you! This is what i was trying to do, now suddenly your hint makes sense.
Thanks!

Hint: times both sides by 2^x

Edit: Dont think my hint was enough, so heres some more:

[lzxnl]

An adventurous man is marooned on an island, with the depth of water around the island modelled by h(t) = 2 cos ((pi*t)/12) +3.

A channel separates the island from the mainland. The depth of water in this channel is the same at all points between the island and the main land, and is given by h(t). The man decides to attract the attention of passing ship. A ship will need a depth of at least 4 metres.

If the mainland is 10 km away from the island, what speed (in km/hour) must the ship average to rescue the man and take him to the mainland?

Is the answer 5 km/hr?

OK. Let's solve for h(t) = 4.
cos(pi t/12) = 1/2
pi t/12 = pi/3, 5pi/3, 7pi/3, 11pi/3 etc
t = 4, 20, 28, 44 etc

Now h(12) = 2 cos pi + 3 < 4 so we need the time between t = 20 and t = 28, for instance, for the height to be over 4 m.
This implies that we only have eight hours. If the ship has to make a return journey in that time, then its average speed would be 20 km/ 8 hrs = 2.5 km/hr

If it just has to make a one-way trip in 8 hours, then its minimum average speed would be 1.25 km/hr. Again, I'm not sure which interpretation it is.

[Alwin]

An adventurous man is marooned on an island, with the depth of water around the island modelled by h(t) = 2 cos ((pi*t)/12) +3.

A channel separates the island from the mainland. The depth of water in this channel is the same at all points between the island and the main land, and is given by h(t). The man decides to attract the attention of passing ship. A ship will need a depth of at least 4 metres.

If the mainland is 10 km away from the island, what speed (in km/hour) must the ship average to rescue the man and take him to the mainland?

Is the answer 5 km/hr?

OK. Let's solve for h(t) = 4.
cos(pi t/12) = 1/2
pi t/12 = pi/3, 5pi/3, 7pi/3, 11pi/3 etc
t = 4, 20, 28, 44 etc

Now h(12) = 2 cos pi + 3 < 4 so we need the time between t = 20 and t = 28, for instance, for the height to be over 4 m.
This implies that we only have eight hours. If the ship has to make a return journey in that time, then its average speed would be 20 km/ 8 hrs = 2.5 km/hr

If it just has to make a one-way trip in 8 hours, then its minimum average speed would be 1.25 km/hr. Again, I'm not sure which interpretation it is.

Hmm, I agree. it is rather ambiguous.
Another different interpretation that would lead to 5km/h is if you chose the time from t=0 to t=4.
This yields a return trip speed of: 20 / 4 = 5 km/h
A one way trip is 10 / 4 = 2.5 km/h

Or, if the adventurous man does not mind being marooned on the island for longer waiting for t=20, then we get nliu's answers of t=20 and t=20... speeds of 0.25 km/h (return trip) and 0.125 km/h (one way)

[lzxnl]

If this was an exam question...I swear...

[Alwin]

If this was an exam question...I swear...

This question was not accessible to the majority of students. As such a variety of answer were accepted. . .

[BubbleWrapMan]

This question was not accessible to the majority of students. As such a variety of answer were accepted. . .

I doubt VCAA would ever admit to that

[tote.moore]

A long trough whose cross-section is parabolic is 1 1/2 metres wide at the top and 2 metres deep. find the depth of water when it is half -full....

I tried finding a quadratic equation using 3 points, then antidiffing area.. then halving it.. not sure guys abit stuck.

[lzxnl]

So...let's place the origin at the vertex of the parabola. It passes through (1.5,2) and its equation is of the form y=ax^2
This means that 2=a*2.25
a=2/2.25=8/9
y=8x^2/9

It's a parabola; antidifferentiating this yields 8x^3/27 + c
Now if evaluated from 0 to 1.5 to find the area under the curve (it's an even function, so we only have to consider the right half of the trough), we end up with 8*1.5^3/27 = 8*(1.5/3)^3 = 1 square units
So half full => area = 0.5 square units = 8x^3/27
2x/3 = (0.5)^1/3
x = 1.5*(0.5)^1/3
That's your height.

[tote.moore]

Thanks for your reply, the answer at the back of the book says 1.26m which is slightly higher value than yours

[tote.moore]

ahh with your help i got it.. So the equation was 32/9x^2... antidiff between -.75 to .75 for 2 - f(x) dx. giving area of 2.. half the area for 1.

I then let H = F(x) and solve for x for my new height.. and put into the calc solve(between my new height in terms of h for h-f(x) dx,h) h=1.26