bestie's question thread

[Bestie]

help!
i don't get these flow in flow out type questions???
please refer to the attachment

[nhmn0301]

help!
i don't get these flow in flow out type questions???
please refer to the attachment

Inflow/outflow question might be confusing a bit in the first go so don't worry !
Summarise what we have:
V = 100 L  x_o=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.

[Bestie]

did you end up getting a really weird number like 0.081873 kg?

thank you so much btw very much appreciated

[nhmn0301]

did you end up getting a really weird number like 0.081873 kg?

thank you so much btw very much appreciated

The exact number I get is 10e^(-1/5), which is around 8.187 kg
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

[Bestie]

yup i got it now thank you:)

can you please help me here as well?

[nhmn0301]

yup i got it now thank you:)

can you please help me here as well?

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 . 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 !

[Bestie]

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?

[keltingmeith]

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.

[Bestie]

oh thank you!

[Bestie]

a solid sphere of radius 16cm has a cylindrical hole of radius 1cm bored through the centre. What is the volume of the remainder of the cylinder?
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

[Conic]

Rather than finding the volume of the caps and a cylinder, you can find the volume of the remaining solid with one integral. If you revolve the area bound by and the circle around the y-axis, you will form a sphere with a radius of 16 with a hole with radius 1 removed. First you find the points of intersection of the circle and the line,



so the circle and the line intersect at and . Now, the volume bound by this area is given by



     

     

If you need to find the volume of a cap, you can revolve the area bound by the line and the previous circle around the y-axis. Then double this to find the total volume of the caps.

[Bestie]

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?

[keltingmeith]

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.

[Bestie]

thank you!

how would I do the dw/dt method?

this was on a revision sheet for a applications SAC

[keltingmeith]

It's literally just like every other time you would've done it.