Post by: Bestie on July 04, 2014, 10:01:45 pm
i don't get these flow in flow out type questions???
please refer to the attachment
Post by: nhmn0301 on July 04, 2014, 10:49:23 pm
help!Inflow/outflow question might be confusing a bit in the first go so don't worry :D!
i don't get these flow in flow out type questions???
please refer to the attachment
Summarise what we have:
V = 100 L xo=10
In : dV/dt = 2
Out: dV/dt = 2
So to the question is asking for dx/dt
We say that dx/dt = dx/dt (in) - dx/dt(out)
dx/dV x dV/dt (in/just using chain rule here) - dx/dV x dV/dt (out)
0 - (x/100) x 2 (the reason we have 0 inflow here is because we are not pumping any salt into the tank)
Hence, dx/dt = -x/50 (negative rate => the salt amount is decreasing, make sense with the question)
Now, take the reciprocal because we can't integrate -x/50 with respect to "t". After that, you will get a ln|.....| + c etc, using the initial condition, which is t=0, x= 10, you can find the value of c. After that, check with the initial condition again to find the right sign for the value within the modulus.
Post by: Bestie on July 04, 2014, 11:28:00 pm
thank you so much btw very much appreciated :)
Post by: nhmn0301 on July 05, 2014, 07:27:36 am
did you end up getting a really weird number like 0.081873 kg?The exact number I get is 10e^(-1/5), which is around 8.187 kg
thank you so much btw very much appreciated :)
I get dx/dt = -x/50 (from above)
dt/dx = -50/x
t = -50 ln |x| + c
t= -50 ln (x) + c (I choose the positive sign here because x>0)
to find the value of c, we have when t=0, x=10
0 = -50ln(10) + c
c = 50ln(10)
Hence, t=-50ln(x) + 50 ln (10)
t= 50ln (10/x)
(Normally if the question specifically ask you for the equation of the salt, you should rearrange this equation to make "x=...." but I'm just lazy here :( )
So I sub t=10 in, 10=50ln(10/x)
Plug in CAS x= 10e^(-1/5) (kg) (left exact value here since the question does not ask you round off)
Hence, remain salt after 10 mins = 10e^(-1/5) kg
Post by: Bestie on July 05, 2014, 11:29:02 am
can you please help me here as well?
Post by: nhmn0301 on July 05, 2014, 06:22:48 pm
yup i got it now thank you:)I'm assuming you are asking question a)? To me I think there is a typo in this question or I must have misunderstood something cause the question says "it will not survive if its sodium concentration rises above 38mg/L" but we have already known from above that "when t=5, S=38", then obviously, the answer for question a) would be t=5? So I think there must be something wrong there, may be it cannot survive if sodium concentration rises above 40mg/L (I'm just making up a number larger than 38). Anw, I'm assuming there's nothing wrong with the question and show you the method :D. I stop at the very last step because you can just simply substitute in any S value from there and find t. Correct me if you see any errors in my working out !
can you please help me here as well?
Post by: Bestie on August 01, 2014, 04:48:27 pm
be modelled by two curves which join at x = e:
y = 2 loge (x), 0 < x ≤ e
y = x2 − 2ex + e2 + c, e ≤ x ≤ 5
Find the volume of wine in the glass when the depth
is 2 cm.
I did pi integral sign with limits e and o (e^(y/2))^2 dy which equates to 44.4669 but the answers says 20.07? how did they get that?
Post by: keltingmeith on August 01, 2014, 04:57:26 pm
The side view of the right side of a wine glass vessel can
be modelled by two curves which join at x = e:
y = 2 loge (x), 0 < x ≤ e
y = x2 − 2ex + e2 + c, e ≤ x ≤ 5
Find the volume of wine in the glass when the depth
is 2 cm.
I did pi integral sign with limits e and o (e^(y/2))^2 dy which equates to 44.4669 but the answers says 20.07? how did they get that?
We want to know the volume at a depth of 2 cm - you've made the calculation for for e cm. Since e > 2, the volume with depth e > volume with depth 2 (which is what you've gotten - higher than the answer). So, redo with terminals from 0 to 2.
Post by: Bestie on August 01, 2014, 05:19:14 pm
Post by: Bestie on August 01, 2014, 05:20:46 pm
I'm trying to find the volume of the two caps at the end of the sphere, how would I do that?
v sphere is 16384pi/3
Post by: Conic on August 01, 2014, 06:23:27 pm
so the circle and the line intersect at
If you need to find the volume of a cap, you can revolve the area bound by the line
Post by: Bestie on August 23, 2014, 02:22:44 pm
week 0 weight 100
week 1 weight 88.3
week 2 weight 75.9
week 3 weight 69.4
week 4 weight 59.1
week 5 weight 51.8
week 6 weight 45.5
let t equals the week and w equals the weight
After finding the exponential model, find the half life of the material?
Post by: keltingmeith on August 23, 2014, 02:36:35 pm
how would i find exponential model using the following info:
week 0 weight 100
week 1 weight 88.3
week 2 weight 75.9
week 3 weight 69.4
week 4 weight 59.1
week 5 weight 51.8
week 6 weight 45.5
let t equals the week and w equals the weight
After finding the exponential model, find the half life of the material?
Personally, I'd use a regression technique with assistance of a computer package such as excel. You won't get questions like this in specialist.
Otherwise, you can let dw/dt = kw and solve as per usual, which I'm pretty sure is what they're going for.
Post by: Bestie on August 23, 2014, 02:44:26 pm
how would I do the dw/dt method?
this was on a revision sheet for a applications SAC
Post by: keltingmeith on August 23, 2014, 02:53:33 pm
Post by: Bestie on August 23, 2014, 02:59:19 pm
but the ans have something completely different from us:
w = 100.3(0.877)^t
Post by: keltingmeith on August 23, 2014, 03:12:57 pm
Post by: Bestie on August 30, 2014, 09:32:43 pm
thanks in advance
Post by: Phenomenol on August 31, 2014, 09:49:59 am
how is sin(2x +pi/2) equal to cos(2t)?
thanks in advance
sin(2x + pi/2) = sin(2(x + pi/4))
sin(2x + pi/2) is sin(2x) translated pi/4 units to the left, which graphically can be seen as equivalent to cos(2x). Try plotting both by hand and seeing for yourself.
Alternatively, use compound angles:
sin(2x + pi/2) = sin(2x)cos(pi/2) + cos(2x)sin(pi/2) = sin(2x)*0 + cos(2x)*1 = cos(2x)
Post by: Bestie on October 05, 2014, 05:36:58 pm
thanks in advance