PHAS0012 - Numbers in Square Brackets Show The Provisional Allocation - Mathematics Assignment Help

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Assignment Task

 

Section A

( Answer ALL SIX questions from this section)

1. Using Around, or otherwise, plot a graph of the points (0.0, 0.19 ± .01), (1.0, 0.76 ± 0.1), (2.0, 5.4 ± 0.5), (3.0, 9.0 ± 1), (4.0, 25. ± 2) and (5.0, 76 ± 3), and superpose a plot of sinh(x) for 0 < x < 5. Ensure that the whole graph, including all the error bars, is visible.

 

2. Find the values of a, b and c which provide the best fit of a + bx + cx 2 to cos(x) over the range 0 < x < π/2 by integrating a + bx + cx 2 − cos(x) 2 over the required range and using Find Minimum to find the optimal values of a, b and c. Also give the result of expanding cos(x) about x = 0 up to the term in x 2 using Series.

 

3. Use ElementData[n,"AbsoluteMeltingPoint"] to construct a list of ordered pairs {element number, melting point} for all the elements from 1 to 92. Plot the associated points, labelling the axes appropriately. You should see no consistent pattern, as the bonding between the atoms varies in character from column to column in the Periodic Table. By using ElementData[n,"Series"] generate another table of the form {series, element number, melting point}, select the elements from the series "AlkaliMetal", and generate a line plot of their melting points as a function of atomic number. You may find Select, Map and Rest useful.

 

4. Nonlinear differential equations are usually not analytically soluble, but there are exceptions such as the separable equation dx/dt = sin(x). Solve this equation analytically for the initial conditions x(0) = 0, x(0) = 1/2 and x(0) = 1, and plot all three solutions on the same graph, with appropriate legends.

 

5. Create three vectors −→a 1 = (a11, a12, a13) and similarly for−→a 2 and −→a 3. From these create −→b 1 =
−→a 2 ×
−→a 3/(
−→a 1.
−→a 2 ×
−→a 3). Also form −→b 2 and

−→b 3 by cyclically permuting the indices. Confirm that−→a 1.

−→b 1 = 1 and

−→a 1.
−→b 2 = 0, −→a 1.

−→b 3 = 0 : you may need to use Simplify. Confirm that if this is generalised to four dimensions by appending 0 to all the −→a i and −→b i , and defining −→b 4 = (0, 0, 0, 1), the scalar product of −→a 1 with all
four −→b i has the same behaviour.

 

6. Write a function which will accept one argument and which has the following properties:
(a) it will return “Error” if the argument is not an integer between 1000 and 9999 inclusive
(b) it will return “Correct” if the last digit is equal to the sum of the first three digits modulo 10;
(c) it will return “CheckSum Failed” otherwise. Check that your function works correctly (this will require at least 3 checks). You may find the IntegerDigits, Most and Mod functions useful.

 

Section B

7. Gerald North set up a model for average sea-level temperature. With some simplifications the evolution over time of the edge of an ice cap, xs, may be modelled by dxs dt = (1 − xs) Q − 335 + 3  xs − 0.89 0.09 2 !

(7.1)where Q is the incoming solar energy in W m−2 and the time t is measured in centuries. Note that xs = 1 corresponds to the icecap vanishing and xs < 1 corresponding to a growing icecap.

(a) Solve the ordinary differential equation (7.1) numerically for Q = 333 and 0 < t < 1 with the initial condition xs(0) = 0, and plot a graph of your result. Repeat the solution and the plot for xs(0) = 1. [4]

(b) Consider the steady state equation, obtained by setting the time derivative to zero.

i. Find the numerical solution of the equation for the case Q = 333. [2]
ii. Find the percentage difference between the smallest steady-state result and the first solution obtained in part (a) at t = 1. [2]

(c) The intermediate solution in part (b) giving xs = 0.963485 corresponds to an unstable state of the system, whereas the other two solutions are stable.
i. By numerically solving the differential equation for 0 < t < 10 for initial values of xs just greater and just less than this intermediate value, confirm that the system evolves to the other, stable, states.

ii. One way of capturing the unstable state is to propagate backwards in time: find the numerical solution for xs for t running from 0 to −10 with an initial value of xs = 0.9. Find the percentage difference between the solution at
t = −10 and your intermediate solution in part (b).

(d) We can investigate the effect of climate change by altering Q to represent the effect of emissions on the balance between incoming solar energy and outgoing energy. Modify your differential equation in part (a) to allow Q to be a function of time,

Q(t) =
333 if t < 5
336 if t ≥ 5
and solve for 0 < t < 10 with xs = 0.8 at t = 0. Comment on your result. [4]

(e) If we take action on climate change, we may be able to alleviate some of the damage we are doing. Modify your calculations in part (d) to investigate what happens if the same increase in Q again happens instantaneously at t = 5, but Q then reverts to 333 after a further 0.5, 1 or 2 centuries. Plot all three results on one graph, with appropriate legends, and comment on your result. [4]

 

8. Newton’s method of finding a root of a function works by repeatedly using the gradient to move towards zero.

(a) Write a function which will accept two arguments: the name of a function, f, and an estimate of the zero, xi. Your function should return xo = xi − f(xi)/f 0 (xi), where the prime denotes differentiation. Check your function by defining another function which will return x − 1 and confirm that your first function immediately returns the value 1 for the position of the root of this second function when given 1 or 0 as the starting value. [4]

(b) Use FindRoot to locate the zero of ExpIntegralEi, starting from the point x = 1 and store the value as a variable for later use.

(c) Use Nest and your first function, with a starting value of 0.5, to estimate the position of the zero of ExpIntegralEi by Nesting your function three times. What is the numerical difference between this result and the result from (b)?

 

 

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