Post by: vintagea on June 28, 2014, 10:30:06 pm
i need help in explaining how the below two images make sense???
I don't get how they have split the absolute values?
can someone please explain what they have done?
thank you
Post by: vintagea on June 28, 2014, 10:32:31 pm
Post by: kinslayer on June 29, 2014, 01:25:04 am
hello !
i need help in explaining how the below two images make sense???
I don't get how they have split the absolute values?
can someone please explain what they have done?
thank you
The first image is just the definition of the absolute value function.
The second image is using the above definition also but it could be a little trickier to see, because you have to use some logic to determine when (y-1)/(y+1) is negative/positive.
Post by: kinslayer on June 29, 2014, 01:29:39 am
its basically this question e and f thank you
For the first one write cot(y) = cos(y) / sin(y). Then you make a substitution of u = sin(y) then continue as normal
The second one is a separable differential equation, you should get
Integrate RHS using partial fractions then solve for y
Post by: vintagea on June 29, 2014, 01:58:06 pm
The gradient of the normal to a curve at any point (x, y) is three times the gradient of
the line joining the same point to the origin.
isn't it just (3y)/x
becasue the gradient of the line joining is y/x and the gradient of normal is three times this and there fore the ans is (3y)/x?
help :(
Post by: Conic on June 29, 2014, 04:11:19 pm
The gradient of the line joining (x,y) to the origin is
These are equal, so we have
Post by: vintagea on July 04, 2014, 09:51:59 pm
can someone please explain this question?
the population (N) of a town increases at a rate proportional to its current size. If the population initially is No and its doubled in 5 years find:
a) population of N at any time, t years, in terms of No
so what ive tried to do is: N = kN?
b) time required for the population to increase to 6 times the initial population, No
please help with part a in particular i'll try to do part b later once i get part a) thank you
Post by: keltingmeith on July 04, 2014, 10:09:01 pm
The key here is the term "rate" - this ALWAYS means d(something)/dt. So, in this instance, we have
For part a
Then, for b, you've just gotta use the equation above to find what N is when t=6.
Post by: vintagea on July 04, 2014, 10:18:35 pm
i have another question:
A bowl of soup initially at temp. 40 degrees cel. Is placed in freezer. The temp of freezer is a constant – 10 degrees cel. If the temp. of the soup drops to 20 degrees in 5 mins, find the temp. of the soup after another 3 mins. Assume Newton’s law of cooling applies.
So I know newton’s law of cooling is dT/dt = -k(T-20)
t = -1/k loge(T-20) + c
At t = 0 T = 40
0 = -1/k loge(40-20) + c [1]
At t = 5 T = 20
5 = -1/k loge(20-20) + c
But loge(0) is undefined, im stuck :(
Post by: vintagea on July 04, 2014, 10:31:00 pm
isn't e^(1/5 ln(2) t)
2^1/5 t?
can you please show step by step how come its 2^t/5?
thank you
Post by: keltingmeith on July 04, 2014, 10:33:43 pm
Post by: vintagea on July 04, 2014, 10:36:48 pm
thank you so much eulerfan101!!!!!
A bowl of soup initially at temp. 40 degrees cel. Is placed in freezer. The temp of freezer is a constant – 10 degrees cel. If the temp. of the soup drops to 20 degrees in 5 mins, find the temp. of the soup after another 3 mins. Assume Newton’s law of cooling applies.
So I know newton’s law of cooling is dT/dt = -k(T-20)
t = -1/k loge(T-20) + c
At t = 0 T = 40
0 = -1/k loge(40-20) + c [1]
At t = 5 T = 20
5 = -1/k loge(20-20) + c
But loge(0) is undefined, im stuck :(
Post by: keltingmeith on July 04, 2014, 11:04:19 pm
Post by: vintagea on July 04, 2014, 11:08:08 pm
For a particular solution of
F’(x) = sqroot(1-y^2) if f(0) = ½
I solved it to find x = sin^-1(y) +c
With respect to 0 and ½ which is the x value and which is the y value?
Post by: keltingmeith on July 04, 2014, 11:12:00 pm
The convention is y = f(x), so if you have f(a) = b, a is the x value, and b is the y value. Similarly, if they say y(a) = b, a is the x value (again) and b is the y value (again).
Post by: vintagea on July 04, 2014, 11:19:50 pm
a saline sol. Of conc. 0.05kg/L is added to 0.5L of water in a drip bag at constant rate 0.005L/min. The solution is also being drained at a rate of 0.006L/min and the mixture is kept uniform. Find an equation for the rate of change of saline conc. With time.
So from what I have done so far:
The conc of saline added is: 0.00025kg/min to 0.5L water
0.05kg - - - 1 L
X kg - - - - - 0.005L
X = 0.00025kg
So the conc. of saline in the drip bag at time t mins is:
(0.00025 t)/0.5 in kg/L
Overall water change is loss of 0.001L/min
I have no idea what im doing….
dx/dt = inflow - outflow (let x be the amount of saline in sol)
0.005 - (x/0.5)0.006
Post by: vintagea on July 05, 2014, 05:44:49 pm
Post by: nhmn0301 on July 05, 2014, 06:32:43 pm
someone please help?Notice that you only need to find the equation for this question because these are the types that VCAA regards as "unsolvable" for Specialist Math, anw,let me know if there is any errors!
Post by: vintagea on July 05, 2014, 10:17:39 pm
im just wondering why is it 0.006x on the top line? isn't the 0.006 already taken into consideration in the 0.001 in the bottom line?
Post by: nhmn0301 on July 05, 2014, 10:22:03 pm
thank you for helpingThe "0.5 - 0.001t" does take 0.006 into consideration, but the whole thing is just the "V" term in dx/dV at"t" time. So, we still need dV/dt as usual because without it, you cannot get dx/dt(out). I hope that makes sense.
im just wondering why is it 0.006x on the top line? isn't the 0.006 already taken into consideration in the 0.001 in the bottom line?