VCE Specialist 3/4 Question Thread!

[lzxnl]

I actually mean differentiate x*cos(x/5). I was clearly not thinking straight when I typed that up

[Limista]

Nah it's okay I worked that part out myself 

Yeah, because if I differentiate the sine function as you said originally, I'd get a cosine function which can't be antidifferentiated and well...we'd end up going in circles.

[saba.ay]

Could someone please explain how to do the following two? I have no clue as to where I should begin. :/

[lzxnl]

14:
Let the person accelerate for time t. If he runs 30m in time t, as he starts from rest, 30m = 1/2*at^2
a = 60/t^2

Which means that his velocity at the end of the 30 metres is v*t = 60/t

Now we also know that he runs at this velocity for (10.4-t) seconds as that's the amount of time left.

So he's already run 30 metres, has 70 metres to go, and runs this 70 metres in 60/t * (10.4-t) seconds
70t = 624 - 60t
130 t = 624
t = 4.77s = time taken to reach top speed

Top speed = 60/t = 12.58 m/s


15:

60 metres in 10 seconds and the next 60 metres in 15 seconds? OK.

Let its initial speed be u and his acceleration be a. Then 60 = ut + 1/2*at^2 = u*10 + 1/2*a*100 = 10u + 50a
But it travels 120 metres in 25 seconds in total
120 = ut + 1/2*at^2 = 25u + 1/2*625a
So 240 = 50u + 625a
Multiplying the first equation by five, we get 300 = 50u + 250a
Subtracting, -60 = 375a
a=-60/375 = -4/25 ms^-2
60 = 10u + 50a = 10u - 8
u = 6.8 m/s
Decelerating at a rate of -4/25 m/s^2 => 6.8 = 4/25 t, t = 6.8*25/4 = 1.7*25 = 42.5 seconds
But 25 seconds have elapsed
So 17.5 seconds left before it stops
As for how much further to travel, v^2 = u^2 + 2ax. v=0
6.8^2 - 2*4/25*x=0
46.24 = 8/25 * x
x = 46.24*25/8 = 144.5m
But 120 metres have already been travelled, so 24.5 metres left.

Someone please check my working.

[Stick]

Could someone please explain Q7b? My teacher didn't really discuss that particular type of question in class today (only perpendicular scenarios). I'm thinking I need to use the dot product, although my attempts have so far been unsuccessful. Any help would be great.

[SocialRhubarb]

r''(t) only has a vertical component. Thus, the angle between r'(t) and r''(t) will be 45 degrees when the 'i' component of r'(t) is equal to the 'j' component of r'(t). If you're having trouble seeing this, draw a diagram.

[bucklr]

Could someone please explain Q7b? My teacher didn't really discuss that particular type of question in class today (only perpendicular scenarios). I'm thinking I need to use the dot product, although my attempts have so far been unsuccessful. Any help would be great.

(Image removed from quote.)

or solving accel.velocity=|accel||velocity|cos45 for t in terms of a gives the answer.
something like this should be your first line:

(a⁸t²)/16=(a⁴/8)(a²+(a⁴t²)/4)

[Stick]

r''(t) only has a vertical component. Thus, the angle between r'(t) and r''(t) will be 45 degrees when the 'i' component of r'(t) is equal to the 'j' component of r'(t). If you're having trouble seeing this, draw a diagram.

That's a great method if the angle is 45 degrees. I'll make sure I remember that.

or solving accel.velocity=|accel||velocity|cos45 for t in terms of a gives the answer.
something like this should be your first line:

(a⁸t²)/16=(a⁴/8)(a²+(a⁴t²)/4)

I'll give that a go for these questions.

[zvezda]

Hey,
Just having some trouble with q15 c in the extended responses of the ch 8 review on essential.
Help is appreciated
Thanks

[Homer]

Euler's method with a step size of 0.1 is used to solve the differential equation dy/dx =
 with initial condition y=1 at x=2. When x=2.2, the value x=2.2, the value obtained for y, correct to four decimal places, is?

I keep getting 1.0258 the ans is 1.0122

[09Ti08]

Euler's method with a step size of 0.1 is used to solve the differential equation dy/dx =
 with initial condition y=1 at x=2. When x=2.2, the value x=2.2, the value obtained for y, correct to four decimal places, is?

I keep getting 1.0258 the ans is 1.0122

I got the same answer as you (either by using a program on cas calculator or by hand).

Hey,
Just having some trouble with q15 c in the extended responses of the ch 8 review on essential.
Help is appreciated
Thanks

I think this question has the same procedure as part b, just changing the upper limit.
This is how I solved it:
Find equation of line AB which has m=H/(b-a)
So the equation is: y=Hx/(b-a) - Ha/(b-a)
Rearranging the equation gives x=(y+Ha/(b-a))*(b-a)/H (1)
Now we need the upper limit: you draw the side view of the bowl, it should look like a trapezium with the larger base on top, now draw two vertical lines starting from (b,H) going down and another one from (a,0) going up. Now you pick an arbitrary value of y, and use similar triangles: y/H=(r-a)/(b-a) => y=H*(r-a)/(b-a) (2)
Now just put this on your calculator: pi*integration of x^2 (given by (1)) dy from 0 to y=H*(r-a)/(b-a) (given by (2))
Sorry if my explanation is hard to understand, I don't have access to a computer atm.

[b^3]

Euler's method with a step size of 0.1 is used to solve the differential equation dy/dx =
 with initial condition y=1 at x=2. When x=2.2, the value x=2.2, the value obtained for y, correct to four decimal places, is?

I keep getting 1.0258 the ans is 1.0122

I got the same answer as you (either by using a program on cas calculator or by hand).

Yeah I get the same too, probably a mistake in the book.

[bucklr]

Would someone mind giving me their answer to this question:
why is it when you have two vectors r1(t) and r2(t) and your trying to find when their paths cross, you have to redefine one of the vectors to differentiate between the two time variables (i.e r1(t), r2(s))?

i'm struggling to visually and perceptively understand the logic.

[lzxnl]

Let's put it this way. You have vector functions given by r1(t) and r2(t) that describe the position of objects at any given time. There are two parts to the function r1(t). Firstly, the input is a time. Secondly, the output is a position. When you have two paths meeting, the outputs are the same, but we don't care about the inputs. We only care about if the two functions ever have the same value regardless of the time. If we use the same time variable, then we are implying that both objects move in the same time, but this requirement is not necessary. That's why we can change the time variable.

If that didn't make sense, try this. r1 and r2 are vector functions that can be rewritten as Cartesian relations relating y and x. If we are able to eliminate the time variable altogether, it makes no difference whether we defined the time variable as t or as s (cos t i + sin t j and cos m i + sin m j give the same Cartesian relation but the variable is different), and this works because we only care about the intersection of two Cartesian relations in space, not in space and time.

[b^3]

This might help to illustrate it better. Move the slider for .
https://www.desmos.com/calculator/9chmh7ui7x
At the point , the two particles collide, since they are in the same place at the same time, but at the points and , the paths cross, but the particles are not necessarily in that same position at the same time. As nliu1995 has said above, if you do redefine the equations, then you're only solving for when they are in the same places at the same time, where as redefining one of them allows you to only need to consider the position, that is when the paths cross.