FPM Rates of Change [8] | Scoins.net | DJS

FPM Rates of Change [8]

Rates of change

1. A raindrop is considered to grow at a rate dependent by direct proportion upon its surface area. Its mass is dependent upon its volume. Find an expression for the rate of change of radius for a spherical raindrop.

2. A spherical crystal, radius r, grows at a constant change of volume, a. Find an expression for the rate of change of surface area in terms of r and a.

3.  A different crystal has projections which can be modelled as cones, excluding the base. If the cones are assumed to be sufficiently polygonal to be treated as having circular bases then, assuming that growth is directly proportional to surface area, dA/dt = µ say, explore how the volume changes. Start with a situation where the slant height, l, of a cone is expressed in terms of the radius, l=kr

4.  An iceberg melts so that it loses volume in direct proportion to the submerged surface area. Assuming that  a berg is initally a cube of side r metres and that the top face is parallel to and clear of any sea surface and that 90% of such a berg is submerged, find an expression for the short-term rate of change of volume, showing that dr/dt is a constant.

In practice, a berg will rotate when sufficient erosion has taken place below. Assume that the initial position has side R.  It is suggested that the melting berg could be considered to be a constant cuboid on the surface plus a (lower, smaller) melting cube of side s= r-R/10.  Rotation will occur at around the time when the residual subsurface volume is similar to the super-surface volume, so occurs roughly when s³ = R³/10. Show that this suggests an approximate value of r=0.564R for the point of rotation. At this moment, what is the rate of loss of volume?

Assuming that the melting part of the iceberg is only the small cube side s and the underside of the remainder, find an expression for ds/dt in terms of R,s and k. At the assumed moment of rotation, what is the value of dr/dt (=ds/dt).
if the berg rotates at this point so that the cube side s is uppoer most, find where the water line comes to.  
Note that the modelling rapidly becomes difficult; applied maths is difficult, as we so rarely use it to attempt real-world problems.

5.   Variable x is changing at a rate in direct proportion to x, so that dx/dt=kx. Show that x = µeᵏᵗ is a solution to this differential equation.

If x is the side of a cube that is expanding in this way, show that the rate of change of area, A, and of volume, V, are also directly proportional to A and V respectively, and find the expansion rates in terms of k.

6.  A population of size n varying in a cyclical manner across t months  can be modelled as n=A+Bsin(wt+c). If the initial population is N and briefly unchanging and if the period of the cycles is annual, then rewrite the model, trying to eliminate A, B, w and c. If also the minimum population is half of the maximum population, express the population size, n, entirely in terms of N and t. Using a sketch, explain how there are two possible solutions to this problem.

7.  Write an equation to model the height of water in a tidal basin. Think of water height against a pier or jetty. Use units of metres and hours. As help, the time between tides is about 6.5 hours. Let y be the height, L the lowest tide level and H the high tide level.

If the tidal range changes in a repeating monthly cycle, how would you modify your model?

DJS 20130417

1 dV/dt = kA => dr/dt=k    2 dV/dt=a = 4πr² dr/dt ….. dA/dt=2a/r
3  let Area be A=Nπrl = Nπkr². then dr/dt = µ/2rNπk dV/dt = 4πr² dr/dt = 2µr/Nk.
4 dV/dt = d/dt (r³) = 3r² dr/dt;  dV/dt = kr²(1+4x09) = 4.6kr² => dr/dt = 46k/30
s=r-R/10=R ³√0.1   => r=0.564R  and dV/dt = 1.564 kR²
V=0.1R³ + s³. dV/dt = k (R²+4s²) => ds/dt = 4k+ R²/s²  (k-R/10).  At the moment of rotation, ds/dt=dr/dt = ³√100  (k-0.1R).  depth under water is 9(R³+s³)/10R² but s³=0.1R³, so this is 0.99R, which means only some of the s³ is above the surface.
5 dA/dt=2kA,  dV/dt=3kV
6 either n=3N/4 +N/4  cos2πt   or n = 3N/2 -N/2  cos2πt7  something like y=H+ (H-L)/2  cos (6.5t+c)   Tidal range is (H-L)/2 also of form A cos (wt+d), where w is of size say 28 days or 672 hours (700 hours to 1 s.f). this goves a form of the general sort y = P + Q cos S cos T, which might be expressed in a form y = P + Z cos (S+T)

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