Applications

This section is currently under construction.

1. Einstein's Approximation for Heat Capacity

This is an application from physical chemistry. Einstein's approximation for the heat capacity C of a solid is given by the equation

C = 3Rx²e^x/(e^x - 1)²,................................................(Equation 1)

where R = 1.987 cal/mol.K is the universal gas constant. If an experimental value for C is given, then one can ask for the values of x which generate the given value of C. For this purpose, we rewrite Equation 1 as

C/R = 3x²e^x/(e^x - 1)²,

and define f(x) to be the right hand side: f(x) = 3x²e^x/(e^x - 1)² .

To find a solution x, we are going to graph y = f(x) and y = C/R on the TI-83 calculator. If there is an intersection point for the two graphs, the x-coordinate will be a solution to Equation 1. To prepare for the graphs, we first analyze f(x).

The following properties of f(x) can be shown to be true:

(a) f(x) is an even function of x. (Exercise: Show that f(-x) = f(x). ) Thus, if we have a solution x, then -x is also a solution. Our interest is in positive solutions, since these are the only ones that are physically possible. Thus, we consider only positive values of x.

(b) f(x) > 0 for x positive. (Exercise: Explain why f(x) > 0.)

(c) f(x) decreases for x positive. (Exercise: Introduce g(t) = ( e^t - e^{( - t )} ) / t . Show that f(x) = 12/(g(x/2))² . Use the Maclaurin series for e^t and e^{( - t )} to show that g(t) > 0 and g(t) increases when t > 0. Explain why f(x) decreases for x > 0 . )

(d) As x approaches 0, the limit of f(x) is 3. (Exercise: Use l'Hospital's Rule to show that the limit of g(t) is 2 as t approaches 0. Now show that f(x) -> 3 as x -> 0 . ) (Note: We are using the symbol -> to represent "approaches".)

(e) f(x) approaches 0 as x approaches infinity. (Exercise: Use l'Hospital's Rule to show that g(t) -> (infinity) as t -> (infinity). Now show that f(x) -> 0 as x -> (infinity) . )

Thus, 0 < f(x) < 3 when x > 0, and f(x) decreases from 3 to zero for x > 0 .

Thus, we can solve uniquely for x if and only if 0 < C/R < 3 . The idea is to plot y = f(x) and y = C/R using the graph window Ymin = 0 , Ymax = 3 and Xmin = 0 . You will need to use trial and error to determine Xmax. Start with Xmax = 5 . If the two graphs show an intersection point, the x-coordinate will be a solution. Use a numerical method to get accuracy. (If you do not see an intersection point, make Xmax larger and adjust Ymin, Ymax and Xmin . ) After a solution for x is determined, the vibrational frequency "V_f" of the solid can be found from the equation x = hV_f / kT, where h is Planck's constant , k is Boltzmann's constant, and T is the Kelvin temperature. (Thank you to JS for this question.)

2. A Descriptive Model of Signal Propagation on an Unmyelinated Neural Axon

This is an application from biology. Neural axons carry signal information back and forth from the brain to other parts of a living organism. Roughly speaking, an axon is 1/10 of a millimeter (10^-4 meters) in diameter and could be as long as 3 meters in a giraffe. Along its length, an axon is divided by nodes, where the distance dx from one node to the next is about 1 micrometer (10^-6 meters).

The mechanism for signal propagation along an axon is as follows: Assume the axon is along the x-axis, for x>0 (and x=0). In its rest state, the voltage at every node is -60 millivolts. At a given node, say at x=0, and at a given time t, a stimulus of some type causes an influx of sodium ions into the node. If the voltage at this node rises to above -40 millivolts, a resulting "avalanche" (or impulse) of sodium ions will quickly flow into the node, raising the voltage to approximately +50 millivolts. Sodium ions then begin to flow along the axon into the next node at x+dx, triggering another "avalanche" of ions (at a later time t+dt) at the new node, and raising the voltage again to +50 millivolts, at x+dx. While this is happening, the voltage at the previous node (at x) "slowly" begins to "degrade" and return to its rest state. This process is repeated sequentially at additional nodes along the axon, located at x+2dx, x+3dx, etc... .The effect is to give the apperance of a signal wave propagating along the axon. The front of the signal is at +50 millivolts. The back of the signal will be at nodes that have fully recovered to their rest state values of -60 millivolts.

The recovery time at a node is measured to be about 2 milliseconds. The propagation speed of the sodium ion impulses is measured to be about 5 meters per second in an "unmyelinated" neural axon. Thus, in 2 milliseconds, the front of the signal has moved about 5(10^6)(2)(10^-3) = 10^4 micrometers, or 1 centimeter. Thus, the signal extends over about 10,000 nodes from the front to the back of the signal. As a result, dx (=10^-6 meters) is small compared to the length of the signal wave itself (=10^-2 meters). This justifies the use of the notation dx for the distance between neighboring nodes. The smallness of dx also suggests the use of a continuous model to describe the signal propagation. (Note also that dt, the time between avalanches at adjacent nodes, is small compared to the recovery time at a given node.)

We assume that the axon is initially in its rest state (at -60 millivolts), but that at time t=0, the node at x=0 receives an impulse of sodium ions, raising its voltage to +50 millivolts. Then, there are three main functions to discuss. We call the first function T* = f(x). For a given position x on the axon, f(x) is the time at which an impulse occurs at x. This time must be x/v (distance divided by velocity). If time is measured in milliseconds, and if distance is measured in centimeters, then T*=2x.

The second function describes the degradation of a "node voltage" as a function of time. We use the notation D* = g(t), where t is measured in milliseconds and D* is measured in millivolts. If the impulse occurs at time t=0, then we construct g(t) as follows: g(t)=-60 millivolts if t<0, and g(t)=-60 millivolts if t>2 milliseconds. If t is between 0 and 2 milliseconds, we use the quartic polynomial g(t)=50+At^3+Bt^4 to give a reasonable fit to experimental observation. This quartic polynomial has the property that g(0)=+50 millivolts and g'(0)=0 (0 derivative at t=0). We choose A, B and an unknown point in time T=C such that the following properties hold for g(t): g(2)=-60 millivolts, g'(C)=0 (0 derivative at t=C), and g(C)=-80 millivolts (minimum value of g(t)). These 3 conditions give 3 equations for the three unknowns A, B and C. We solve these equations to find A=-97.79438394, B=42.02219197, and C=1.745406047 (milliseconds for C). (Exercise: Set up the three equations for A,B,C and solve using the TI-83 calculator.)

The third function is the wave function W(x,t). This function gives the value of the voltage at position x and time t, where t>0. Then, we may write W(x,t) as a piecewise function as follows:

W(x,t) = -60 if t < 2x , and

W(x,t) = g(t-2x) if t=2x or t > 2x .

The first line says that if the time t is less than the arrival time T* of the impulse of ions at position x, then W(x,t) is the rest state voltage. The second line says that if the time t is greater than the arrival time T* of the impulse at position x, then W(x,t) is the degradation voltage over a time period t-T* (where T*=2x).

Based upon the way we set up our function g(t), we can simply write W(x,t)=g(t-2x) for t>0. The function W(x,t) then has discontinuities only along the line t=2x in the x,t-plane.

In the TI-83 calculator, we can enter the functions Y1 = Y3(T-Y2) (this is g(t-2x)) , Y2 = 2X (this is f(x)), and Y3 = ((BX+A)*X*X*X+110)(0<X or 0=X)(X<2)-60 (this is g(t) with t replaced by X in the calculator). Set the window to be 0<X<4 and -90<Y<60. On the home screen, enter the values for A and B as given above. Then enter a value for the time T, such as T=3 (but use the instruction 3 ---> T). Now press [GRAPH]. The calculator will graph W(x,t) when t=3, giving the signal graph shape along the axon as a function of x. (Exercise: Use the TI-83 calculator to graph W(x,t) when t=5. Also, use the TI-83 calculator to graph W(x,t), as a function of t, when x=2.) (Thank you to JR for suggesting this topic.)

3. Electric Fields Generated by Stationary Point Charges

This is an application from physics. We are given that the electric field due to a single point charge falls off as 1/(distance)^2 and is equal to E = kQ/(distance)^2. The electric field (a vector quantity) points away from a positive charge and points toward a negative charge. If two or more charges are present, the electric field at some position in space is the sum of the fields due to each of the separate charges. (This "superposition principle" is observed from experiments.)

In each of the following examples, we answer the question: Is the electric field equal to zero at some point?

Example #1. (Two charges.)

A charge of -3 (micro C) is at the origin on the x-axis. A second charge of +5 (micro C) is located at ( 1 cm. , 0 ) on the x-axis. Is there a point on the x-axis (other than + or - infinity) where the electric field is zero?

The electric field on the x-axis due to the charge at the origin is given by the piecewise function f(x) = -3k/(x^2) if x > 0 and f(x) = +3k/(x^2) if x < 0. The electric field due to the charge at ( 1 cm. , 0 ) is g(x) = +5k/(x-1)^2 if x > 1 and g(x) = -5k/(x-1)^2 if x < 1 . When we add f(x) + g(x) to get the total electric field E(x) at position x, we get a piecewise function with three pieces:

Region 1: E(x) = k[+3/(x^2) - 5/(x-1)^2] if x < 0 .
Region 2: E(x) = k[-3/(x^2) - 5/(x-1)^2] if 0 < x < 1 .
Region 3: E(x) = k[-3/(x^2) + 5/(x-1)^2] if 1 < x .

In Region 2, E(x) is negative and is never zero. In Region 3, the larger magnitude charge (+5) is always closer to any points in this region compared to the smaller magnitude charge (-3). Thus, E(x) is positive in Region 3 and never zero.

In Region 1, we set E(x) = 0 . This leads to the quadratic equation 3(x-1)^2 - 5(x^2) = 0. Although there are two solutions, only one will satisfy the condition x < 0 (for Region 1). The solution is [-3-sqrt(15)]/2 = -3.436491673 .

The TI83 calculator can be used to graph E(x) in Region 1 to verify that there is only one solution. For the purposes of this graph, set k = 1, and enter the expression for Region 1 into the function variable Y1. Set the window to -5 < X < -1 and -.02 < Y < .08 . Then press [GRAPH].

Example #2. (Three charges on a line.)

A charge of -5 (micro C) is at ( -1 cm. , 0 ), a charge of +3 (micro C) is at ( 0 , 0 ), and a charge of +5 (micro C) is at ( 1 cm. , 0 ) . Is there a point on the x-axis (other than + or - infinity) where the electric field is zero?

Now we have E(x) equal to a piecewise function with four pieces:

Region 1: E(x) = k[+5/(x+1)^2 - 3/(x^2) - 5/(x-1)^2] if x < -1 .
Region 2: E(x) = k[-5/(x+1)^2 - 3/(x^2) - 5/(x-1)^2] if -1 < x < 0 .
Region 3: E(x) = k[-5/(x+1)^2 + 3/(x^2) - 5/(x-1)^2] if 0 < x < 1 .
Region 4: E(x) = k[-5/(x+1)^2 + 3/(x^2) + 5/(x-1)^2] if 1 < x .

In Region 2, E(x) is negative and is never zero. In Region 4, the charge (+5) is closer to any points in this region than the charge (-5). This, together with the charge of (+3) at x = 0 , will make E(x) positive in Region 4. Thus, E(x) is never zero in Region 4. The graphs of E(x) can be plotted for Regions 2 and 4 separately to confirm these results. (For the purpose of plotting the graphs, enter the 4 expressions for E(x) into Y1, Y2, Y3, Y4 respectively. For Region 2, turn on the Y2 function and turn off Y1, Y3, and Y4. Set the window appropriately and graph Y2. Follow a similar procedure for Region 4.)

In Regions 1 and 3, the functions are sufficiently complicated to justify using the TI83 calculator to plot the graphs in order to search for zeros. We then see that E(x) will be zero near x = -7 and x = +.4 . If we use Newton's iteration method with these values as starting values, we quickly converge to x = -6.951401408 and x = .4173974566 . (Newton's iteration is x(n+1) = x(n) - E(x(n)) / E'(x(n)), where E'(x) = the derivative of E(x) . For Region 1, enter the expression X - Y1/nDeriv(Y1,X,X,.00001) into Y5. On the homescreen, enter -7 ---> X . (We use ---> to mean "STORE"). Then enter Y5 ---> X . Press [ENTER] three more times to get three more iterates and convergence. Follow a similar procedure for Region 3 . )

Example #3. (Three charges at points which form a triangle.)

A charge of -5 (micro C) is at ( -1 cm. , 0 ) , a charge of +3 (micro C) is at ( 0 , 1 cm. ) , and a charge of +5 (micro C) is at ( 1 cm. , 0 ). Is ther a point (x,y) where the electric field is zero?

Now the equations are quite complicated. The electric field due to the charge of +5 at (1,0) is f(x,y) = [ +5 / ( (x-1)^2 + y^2 ) ] u1, where u1 is a unit vector pointing from (1,0) to (x,y). Thus, u1 = [ (x-1) i + y j ] / sqrt( (x-1)^2 + y^2 ), where i , j are the standard unit vectors. Then f(x,y) can be written as

f(x,y) = [ +5 / [ sqrt( (x-1)^2 + y^2 ) ]^3 ] * [ ( x-1 ) i + y j ] .

The other two electric fields are

g(x,y) = [ +3 / [ sqrt( x^2 + (y-1)^2 ) ]^3 ] * [ x i + ( y - 1 ) j ] .
h(x,y) = [ -5 / [ sqrt( (x+1)^2 + y^2 ) ]^3 ] * [ ( x +1 ) i + y j ] .

We add these fields, and set the i , j components to zero. This gives two equations for the two unknowns x,y .

Now use Newton's iteration method for systems of equations. (Click here for a discussion of this method.) If we start the iteration with x = .5 and y = .6 (these are just guesses), we quickly converge to x = .4795564920 and y = .6321398219 .

If we start the iteration with x = -7.0 and y = -1.7 (these are also just guesses), we quickly converge to x = -7.107477936 and y = -1.706771780. (Thank you to FC for suggesting this topic.)

4. Steady State Groundwater Flow in a Rectangular Region

This is an application from hydrogeology, within the field of geology. We assume the rectangular region R is in the xz-plane. (x is the horizontal axis, and z is the vertical axis.) The distance variables are chosen so that the rectangular region is given by 0 < x < P and 0 < z < (.5) P . ( P=Pi, which is approximately 3.1415926535898 ) Thus, the rectangle is in the first quadrant with the origin as one corner. Also, the width of the rectangle is twice the height.

In a steady state flow, the equation for the "hydraulic head" h(x,z) in a homogeneous region is given by the Laplace equation :

[ ( d/dx )² + ( d/dz )² ] h(x,z) = 0 .........................(Equation#1).

Here, d/dx and d/dz are partial derivative operators. The negative of the hydraulic head h(x,z) plays the role of a "potential function" for the velocity field describing the fluid flow. At any point Q(x,z), the velocity vector (u,w) is perpendicular to the curve of constant head which passes through the point Q.

To complete the statement of the problem for the head h(x,z), we need to assign boundary conditions around the boundary of the rectangle. Along x=0 and x=P, we asume there is no flow in the x-direction (because of impermeable vertical boundaries). Thus,

( d/dx ) h(x,z) = 0 ...... (at x = 0 or x = P) .............(Equation#2)

at these two side boundaries. Also, we assume the bottom boundary (z=0) is an impermeable boundary (due to bedrock). Thus,

( d/dz ) h(x,z) = 0 ...... (at z = 0 ) ............................(Equation#3)

at the bottom boundary.

The water table forms the top boundary z = (.5) P . Along the top boundary, we assume that the head h(x,z) is given by

h(x,z) = cos (x) ...... (at z = (.5) P ) .........................(Equation#4).

This top boundary condition represents a fluid motion along this boundary which has a non-zero (but also a non-uniform) x-component when 0 < x < P .

In practice, equations similar to Equations#1,2,3,4 would be solved by finite difference methods. Such methods can be used to solve for the head h(x,z) when the problem becomes much more complicated. For example, the shape of the boundary of the region may be irregular, the porous ground medium may be non-uniform, the flow field (and the region) may be three-dimensional, and the flow field may involve sources and sinks. (With a small amount of programming, the TI-83 calculator can easily handle a grid with 20 to 30 points. More points would involve a bigger programming effort for the TI-83 calculator.)

Amazingly, our above simplified rectangular problem has a (unique) solution which can be written compactly as

h(x,z) = ( k ) ( cos x ) cosh(z) ...............................(Equation#5)

where cosh(z) = ( e^z + e^-z ) / 2 , and where the constant k = 1 / cosh( .5 P ) .

This solution satisfies Equations#1,2,3,4 and can be used to obtain a picture of the streamlines for the flow field using a TI-83 calculator. The velocity u(x,z) in the x-direction is given by

u = - ( d/dx ) h = ( k ) ( sin x ) ( cosh z ) .

The velocity in the z-direction is given by

w = - ( d/dz ) h = - ( k ) ( cos x ) ( sinh z ) .,

where sinh(z) = ( e^z - e^-z ) / 2 .

(That is, the vector (u,w) is equal to - ( grad ( h(x,z) ) ) . )

Thus, a differential equation for the streamlines is given by

dz/dx = w / u = ( - cos x ) ( sinh z ) / [ ( sin x ) ( cosh z ) ] ....(Equation#6).

(That is, the slope of the streamline is the slope of the velocity vector.) This differential equation is separable. Solutions are given implicitly by

ln ( sinh z ) = - ln ( sin x ) + C₀ .

Thus, ( sin x ) ( sinh z ) = C .................................(Equation#7)

(where C = exp ( C₀ ) ).

Or, z = sinh^-1 [ C / ( sin x ) ] ..................................(Equation#8).

(Note: sinh^-1( ) is the "inverse function" for the function sinh z . ) In our rectangular region R , the values for C must satisfy 0 < C < sinh ( .5 P ) = 2.3012989023073 . (The maximum for C occurs when x = P / 2 and z = P / 2 in Equation#7.) Thus, to graph some representative streamlines, set C = .5 and 1.0 and 1.5 for example .

Here is what you should enter into the TI-83 calculator:

Set Y₁ = sinh^-1 ( .5 / sin(x) ) , Y₂ = sinh^-1 ( 1.0 / sin(x) ) , Y₃ = sinh^-1 ( 1.5 / sin(x) ) .
( sinh^-1 may be found in the CATALOGUE. )

Go to the WINDOW menu. Set X_min = 0 , X_max = P , X_scl = 1 , Y_min = 0 , Y_max = .5 P , Y_scl = 1 , X_res = 3 .

Now press GRAPH .

You should see 3 graphs, each in the shape of a "U" . The streamlines start at the water table (top of the rectangle) , go down into the interior of the region, and then come back up to the top of the rectangle (the water table) , but at a different position from the starting position. (Thank you to SC for suggesting this problem.)

A lot more information about the types of problems similar to the one above can be found in the following courses:
EES 347 - Principles of Hydrology
Geology 745 - Hydrology
Geology 746 - Hydrogeology
EES 393 - Summer Hydrology
Geology 799.3 - Summer Hydrology

Exercise#1. Explain why sinh(z) is a one-to-one function.
Exercise#2. Evaluate sinh^-1(2) to 10 decimal places without using the CATALOG in the TI-83 calculator.




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