wildareal's questions thread

[wildareal]

Given vector OA=i-2j-2k and vector OB=4i+3j-4k. A lies on vector OB, such that it is the closest point to A. Find the vector OP.

[enpassant]

P lies on vector OB?

[enpassant]

Vector OP is the vector component of vector OA in the direction of vector OB.

[wildareal]

P lies on vector OB?

Yes P lies on B. is it something like OP vector=(a.bhat)bhat? why is this the case?

Thanks.

[kamil9876]

Why? this is because AP is perpendicular to OB. There are two says to see this:

(1) draw a circle of radius |AP| around A, now you see that OB is tangent to this circle meeting at P hence OB is perpendicular to the radius.

(2) let C be the point on OB such that OB is perpendicular to CA, now by Pythagoras hence and only when (which is exactly when )

[wildareal]

The Points X, Y and Z correspond to the numbers 4root(3), +2i, 5root(3)+i and 6root(3)+4i. Find the Vector XY and vector XZ.

Let z1 and z2 be the complex numbers corresponding to the vectors XY and XZ. Find z3 such that z2=z3z1.

By writing z3 in modulus argument form, show that XYZ is half an equilateral triangle XWZ and give the complex number to which W corresponds.

The triangle XYZ is rotated through an angle of of TT/3 about Y. Find the new position of X (this is the part I'm not sure about).

Cheers.

[kamil9876]

to do what you are unsure about:

Let X' be the new position of X. Hence we know that the vector YX' is the vector YX rotated by an angle of pi/3. Hence as a complex number we have and then just use to find the position.

[wildareal]

Prove that y=xe^(2x) is a solution to the differential equation:

d^(2)y/dx^(2) -dy/dx-3e^(2x)=2xe^(2)

[Mao]

"prove"? do you mean "show"?

if so, just sub it in.

[wildareal]

14.
A hot air balloon is ascending at a speed of 2m/s. When the hot air balloon is at a height above the ground of 300 metres, a passenger accidentally drops their wallet over the side of the balloon basket.
(a)   Assuming no air resistance, how long does it take the wallet to strike the ground? Give your answer correct to 2 decimal places.

In an effort to save the wallet the burners on the balloon are turned off. The balloon stops rising and remains steady at an altitude of 350 metres. The wallet’s owner dons an emergency parachute and goes overboard. Because of the low altitude he releases the parachute immediately.
(b)   (i) Assuming that the retardation due to air resistance is equal to 1.2v2, where v m/s is the   velocity at time t seconds, express the acceleration of the parachutist in terms of his velocity.
(ii) Hence express the velocity in terms of the position x metres below the parachutist’s starting point and find the speed of the parachutist as he hits the ground.

Thanks

[Mao]

a. The wallet has an initial velocity of 2m/s upwards. It accelerates downwards at 10 m/s^2, covering a total distance of 300m. u=2, a=-10, d=300

b. i. dv/dt = -10 - 1.2v^2*sign(v) <-- the sign of v is important. if v is positive (going up), retardation is down. if v is negative (going down), retardation increases velocity (decrease in magnitude). In this application, initial v is 0, and the parachutist will be descending, i.e. v<0 for all cases. Thus, dv/dt = -10 + 1.2v^2

ii. trivial.

[wildareal]

1. The relation between x and y is defined by the differential equation dy/dx=loge(x), with y=3, x=2. Show how Euler's method, with a step size of 0.1 can be used to approximate the value of y when x=2.2. Give your answer correct to 3 decimal places.

2. Using three strips and the midpoint rule, find an approximation to integral(sec(x/2),0,1.5)

[luken93]

1) f(2.0 + 0.1) = f(2.0) + h x f'(2.0)
 = 3 + 0.1 x loge(2.0)

f(2.1 + 0.1) = f(2.1) + h x f'(2.1)
 = 3 + 0.1loge(2) + 0.1 x loge(2.1)
cbf doing it on the calc haha

[wildareal]

A large tank holding 300 litres of water has 50 kg of salt dissolved in it. A brine solution is pumped into the tank at a rate of 3 litres per minute. The mixture is continuously stirred and pumped out at a constant rate of k litres per minute. The concentration of the solution entering the tank is 2 kg per litre.
   Let A kg be the amount of salt in the tank at any time t minutes after the process has started.
   (a)   Find the differential equation for A in terms of t. Also state the initial condition.
   (b)   (i)   If k = 3, show that   = 600 – 550  , t   0.
      (ii)   How much salt will there be in the tank eventually?
   (c)   (i)   If k = 2, show that   = 6 - 
(ii)   Show that  A = 2(300 + t) +  , t   0, (where c is a constant) is a solution to this differential equation.
(iii)   Hence find the amount of salt in the tank 10 minutes after the process started.

[tony3272]

A large tank holding 300 litres of water has 50 kg of salt dissolved in it. A brine solution is pumped into the tank at a rate of 3 litres per minute. The mixture is continuously stirred and pumped out at a constant rate of k litres per minute. The concentration of the solution entering the tank is 2 kg per litre.
   Let A kg be the amount of salt in the tank at any time t minutes after the process has started.
   (a)   Find the differential equation for A in terms of t. Also state the initial condition.
   (b)   (i)   If k = 3, show that   = 600 – 550  , t   0.
      (ii)   How much salt will there be in the tank eventually?
   (c)   (i)   If k = 2, show that   = 6 - 
(ii)   Show that  A = 2(300 + t) +  , t   0, (where c is a constant) is a solution to this differential equation.
(iii)   Hence find the amount of salt in the tank 10 minutes after the process started.

Lol i'm surprised that im stopping my dramas to do maths at this hour 
But for part a:

First you want to find , which is just equal to the inflow of substance (3) minus the outflow of substance (k)




and you know at


now that you've got that sorted you want to find , which is the inflow of salt minus outflow of salt

Inflow of salt =

Outflow of salt =