Fluid dynamics of animal locomotion
PROBLEM SET 3
Due October 20, 2004
1. Show that, if ui, i = 1, 2, 3 denotes the velocity eld given by ui = [ 2 xixj
2
ij ]aj
with a = i1 , i2 , i3 respectively and the same function in each case, then x1u1 + x2 u2 + x3u
Fluid dynamics of animal locomotion
PROBLEM SET 5
Due November 24, 2004
1. In the following two examples the object is to compute the average x-directed thrust given that the side forces on the sh are somehow in equilibrium.vWe will compare the thrust res
Fluid dynamics of animal locomotion
PROBLEM SET 2
Due October 6, 2004
1. Dene the complex derivatives 1 1 d d = +i = i , . dz 2 x y d z 2 x y Applied to a complex-valued function w = F (x, y) + iG(x, y), where f, g are very smooth, verify that (a) dw = 0
Fluid dynamics of animal locomotion
PROBLEM SET 1
Due September 29, 2004
1. In one dimension, the Eulerian velocity is given to be u(x, t) = 2x/(1 + t). (a) Find the Lagrangian coordinate x(a, t). (b) Find the Lagrangian velocity as a function of a, t. (c
Fluid dynamics of animal locomotion
PROBLEM SET 4
Due November 3, 2004
1. Dene an eciency of the ciliated sphere by
2 6r0 Usc . W Where W is computed by integrating Ws = 2 k(a2 + b2 ) over the surface of the sphere. Here Usc = 1 k(b2 + 10 a0 b0 cos a2 ) i
Hypotheses
The null hypothesis will be that all population means are equal, the alternative
hypothesis is that at least one mean is different.
The null hypothesis can be written as
, but the alternative can not be
written as
, all it takes is for one of t
Hypotheses
There are three sets of hypothesis with the two-way ANOVA.
The null hypotheses for each of the sets are given below.
1. The population means of the first factor are equal. This is like the one-way
ANOVA for the row factor.
2. The population mea
Tukey Tests
When the decision from the One-Way Analysis of Variance is to reject the null
hypothesis, it means that at least one of the means isn't the same as the other means.
What we need is a way to figure out where the differences lie, not just that t
Scatter Plots
Enter the x values into L1 and the y variables into L2.
Go to Stat Plot (2nd y=)
Turn Plot 1 on
Choose the type to be scatter plot (1st type)
Set Xlist to L1
Set Ylist to L2
Set the Mark to any of the three choices
Zoom to the Stat setting (
Grand Mean
The grand mean doesn't care which sample the data originally came from, it dumps all
the data into one pot and then finds the mean of those values.
The grand mean of a set of samples is the total of all the data values
divided by the total samp
The idea behind the chi-square goodness-of-fit test is to see if the sample comes from
the population with the claimed distribution. Another way of looking at that is to ask if
the frequency distribution fits a specific pattern.
Two values are involved, a