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Various P3 Q's.
Could someone show me how to do these three please.
1.)The number of people, n, in a queue at a post office t minutes after it opens is modelled by the differential equation:
dn/dt = e^(0.5t) - 5, t>0
a.) Find, to the nearest second, the time when the model predicts that there will be the least number of people in the queue.
b.) Given that there are 20 people in the queue when the post office opens, solve the differential equation.
c.) Explain why this model would not be appropriate for large values of t.
(See Attachment for other two Q's).
Many thanks.
1.)The number of people, n, in a queue at a post office t minutes after it opens is modelled by the differential equation:
dn/dt = e^(0.5t) - 5, t>0
a.) Find, to the nearest second, the time when the model predicts that there will be the least number of people in the queue.
b.) Given that there are 20 people in the queue when the post office opens, solve the differential equation.
c.) Explain why this model would not be appropriate for large values of t.
(See Attachment for other two Q's).
Many thanks.
Slice'N'Dice
1.)The number of people, n, in a queue at a post office t minutes after it opens is modelled by the differential equation:
dn/dt = e^(0.5t) - 5, t>0
a.) Find, to the nearest second, the time when the model predicts that there will be the least number of people in the queue.
b.) Given that there are 20 people in the queue when the post office opens, solve the differential equation.
c.) Explain why this model would not be appropriate for large values of t.
dn/dt = e^(0.5t) - 5, t>0
a.) Find, to the nearest second, the time when the model predicts that there will be the least number of people in the queue.
b.) Given that there are 20 people in the queue when the post office opens, solve the differential equation.
c.) Explain why this model would not be appropriate for large values of t.
1.) dn/dt = e^(0.5t) - 5
When n is at minimum:
-> e^(0.5t) - 5 = 0
-> e^(0.5t) = 5
-> 0.5t = ln5
-> t = 2ln5
-> t = ln(25)
-> t = 3.22 mins
-> t = 3 mins 13 s
b.) dn/dt = e^(0.5t) - 5
-> n = 2e^(0.5t) - 5t + k
When t = 0, n = 20:
-> 20 = 2 + k
-> k = 18
-> Solution: n = 2e^(0.5t) - 5t + 18
c.) For large values of t the model would imply an extremely large value for n, which is not pheasible given the size of the post office. Hence the model is not suitable for large t.
2.) a.) x = t + sint, y = sint.
dx/dt = 1 + cost, dy/dt = cost.
dy/dx = dy/dt * dt/dx = cost * [1/(1 + cost)] = cost(1 + cost).
b.) At this point dy/dx = 0:
-> cost = 0 -> t = Pi/2
Hence: x = Pi/2 + 1, y = 1
-> Co-ordinates: (Pi/2 + 1, 1).
c.) When curve crosses x-axis y = 0:
-> 0 = sint -> t = Pi.
Hence: Area = Int (Pi to 0) sint dt. = [-cost] (Pi to 0) = [-(-1)] - [-1] = 1 + 1 = 2
1.
a) dn/dt = 0
b) dn = e^(0.5t) - 5 dt, now integrate both sides and add a constant. To find the constant, use the fact that n=20 when t=0.
c) Because for large values of t, n will be very large. Besides, the post office will eventually close down after a few hours.
2.
a) Find dx/dt and dy/dt, then use the chain rule: dy/dx = (dy/dt) . (dt/dx) = (dy/dt)/(dx/dt).
b) That would be at the turning point. So set dy/dx=0 and find the corresponding value of t, and then plug this value to find x and y.
c) A = ∫ y dx = ∫ y (dx/dt) dt. The limits are 0 and pi.
3.
a) Find B-A. An equation for l1 would then be: r = A + t(B-A).
b) Write both lines in the form: r = (a+xt)i + (b+yt)j + (c+zt)k, e.g. l2 becomes: r = (-2+7u)i + (10-4u)j + (6+6u)k. Now equate the coefficients of i, j and k of both lines, then solve two of those equations to find t and u. Once you find values for both, sub them in the third equation to and see if it works.
c) You should have values for u and t at the point of intersection, so either plug in the value of u in the equation of l2, or the value of t in the equation of l1.
d) Let the position vector of C be (-2+7u, 10-4u, 6+6u) since it lies on l2. Now:
AC = C-A
AB = B-A
AB . AC = 0 (dot product), since they're perpendicular. This should give you a value for u which you can now plug in (-2+7u, 10-4u, 6+6u) to get the position vector of C.
I hope that helps.
a) dn/dt = 0
b) dn = e^(0.5t) - 5 dt, now integrate both sides and add a constant. To find the constant, use the fact that n=20 when t=0.
c) Because for large values of t, n will be very large. Besides, the post office will eventually close down after a few hours.
2.
a) Find dx/dt and dy/dt, then use the chain rule: dy/dx = (dy/dt) . (dt/dx) = (dy/dt)/(dx/dt).
b) That would be at the turning point. So set dy/dx=0 and find the corresponding value of t, and then plug this value to find x and y.
c) A = ∫ y dx = ∫ y (dx/dt) dt. The limits are 0 and pi.
3.
a) Find B-A. An equation for l1 would then be: r = A + t(B-A).
b) Write both lines in the form: r = (a+xt)i + (b+yt)j + (c+zt)k, e.g. l2 becomes: r = (-2+7u)i + (10-4u)j + (6+6u)k. Now equate the coefficients of i, j and k of both lines, then solve two of those equations to find t and u. Once you find values for both, sub them in the third equation to and see if it works.
c) You should have values for u and t at the point of intersection, so either plug in the value of u in the equation of l2, or the value of t in the equation of l1.
d) Let the position vector of C be (-2+7u, 10-4u, 6+6u) since it lies on l2. Now:
AC = C-A
AB = B-A
AB . AC = 0 (dot product), since they're perpendicular. This should give you a value for u which you can now plug in (-2+7u, 10-4u, 6+6u) to get the position vector of C.
I hope that helps.
Thankyou very much guys.
Just another quick Q:
a.) Use the substitution u=2-x² to find:
(Int)x/(2-x²) dx
So, du/dx = -2x
Do you then re-arrange it so you get du/-2 = xdx Then what do you do with this?
Or am I just confusing this?
Just another quick Q:
a.) Use the substitution u=2-x² to find:
(Int)x/(2-x²) dx
So, du/dx = -2x
Do you then re-arrange it so you get du/-2 = xdx Then what do you do with this?
Or am I just confusing this?
Slice'N'Dice
Thankyou very much guys.
Just another quick Q:
a.) Use the substitution u=2-x² to find:
(Int)x/(2-x²) dx
So, du/dx = -2x
Do you then re-arrange it so you get du/-2 = xdx Then what do you do with this?
Or am I just confusing this?
Just another quick Q:
a.) Use the substitution u=2-x² to find:
(Int)x/(2-x²) dx
So, du/dx = -2x
Do you then re-arrange it so you get du/-2 = xdx Then what do you do with this?
Or am I just confusing this?
That's correct so far, but you want dx = du/(-2x)
then substitute that into the integral:
so..
∫ x/(u).(du/-2x) = ∫ (xdu)/(-u2x)
= -1/2 ∫ 1/u du
= -1/2 lnu + c
= -1/2 ln|2-x²| + c
endeavour
That's correct so far, but you want dx = du/(-2x)
then substitute that into the integral:
so..
∫ x/(u).(du/-2x) = ∫ (xdu)/(-u2x)
= -1/2 ∫ 1/u du
= -1/2 lnu + c
= -1/2 ln|2-x²| + c
then substitute that into the integral:
so..
∫ x/(u).(du/-2x) = ∫ (xdu)/(-u2x)
= -1/2 ∫ 1/u du
= -1/2 lnu + c
= -1/2 ln|2-x²| + c
Oh!
I'm with it now. Thanks endeavour.
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