+97 By our standards, Xenia-Lepton aliens are huge, strange creatures and their reproductive method is fascinating: while under the influence of a massive gravitational field, a mature Xenia-Lepton sheds several billion sub-atomic particles. These particles combine to form a seed which, if viable, grows rapidly by direct conversion of dark energy to matter. The rate of growth is directly proportional to how much the juvenile has yet to grow before reaching the height of a Xenia-Lepton adult. All adult Xenia-Leptons have exactly the same height, which is 38m. The coefficient of proportionality for the growth rate has been determined to be 0.049 .
1) Determine the height of a Xenia-Lepton alien that is 94 years old.
2) When a Xenia-Lepton reaches exactly 99.0% of the adult height, they change colour from bright, shining gold to pitch black. At what age does this occur?
| Height at 94 years (m) = | [correct answer is 37.62] |
| Age at colour-change (years) = | [correct answer is 93.98] |
Thanks Guys
+33616 Growth rate of juvenile is given by dh/dt = 0.049*(38 - h) where h is juvenile height and t is time.
This ordinary differential equation may be solved to find height, h, as a function of time, t, as follows:
Rewrite as dh/dt + 0.049h = 0.049*38
Multiply both sides by e0.049t
e0.049tdh/dt + e0.049t*0.049h = 0.049*38*e0.049t
Rewrite this as
d(e0.049th)/dt = 0.049*38*e0.049t
Integrate with respect to t
e0.049th = 0.049*38*e0.049t/0.049 + c where c is a constant
Simplify:
e0.049th = 38*e0.049t + c
When t = 0 then h = 0 (i.e. zero height at birth) so
0 = 38 + c or c = -38
Therefore
e0.049th = 38*e0.049t - 38
Multiply both sides by e-0.049t
h = 38(1 - e-0.049t)
1. Plug in 94 for t on the right-hand side to find height at 94years.
2. Plug in 0.99*38 for h on the left-hand side and solve the resulting equation for t to find the age at colour-change.
+33616 Growth rate of juvenile is given by dh/dt = 0.049*(38 - h) where h is juvenile height and t is time.
This ordinary differential equation may be solved to find height, h, as a function of time, t, as follows:
Rewrite as dh/dt + 0.049h = 0.049*38
Multiply both sides by e0.049t
e0.049tdh/dt + e0.049t*0.049h = 0.049*38*e0.049t
Rewrite this as
d(e0.049th)/dt = 0.049*38*e0.049t
Integrate with respect to t
e0.049th = 0.049*38*e0.049t/0.049 + c where c is a constant
Simplify:
e0.049th = 38*e0.049t + c
When t = 0 then h = 0 (i.e. zero height at birth) so
0 = 38 + c or c = -38
Therefore
e0.049th = 38*e0.049t - 38
Multiply both sides by e-0.049t
h = 38(1 - e-0.049t)
1. Plug in 94 for t on the right-hand side to find height at 94years.
2. Plug in 0.99*38 for h on the left-hand side and solve the resulting equation for t to find the age at colour-change.
