This booklet has been produced to assist second year physics students with the mathematical content of their course. The content in this booklet has been developed using resources such as lecture notes, lecture slides and past articles provided by the University of Birmingham.
About the Authors
It was designed as an interactive resource to supplement lecture material with a special focus on the application of mathematics in physics.
How to use this Booklet
The domain of a function is the set X of all the values we assign to the function. To find the discontinuities of a function, we find the values ex for which the function is undefined.
Curve Sketching
Asxtenders to−∞ we can see that x−21 will turn towards zero again, soy will turn towards the line y=x.
Trigonometry
We know the length of the hypotenuse and want to find the opposite, so we need to use sin(θ) =O. We know the length of the adjacent and want to find the opposite, so we need to use tan(θ) = O.
Hyperbolics
Hyperbolic function identities have very similar shapes to the trigonometric identities, but there is one key difference described in Osborn's rule. This rule states that all identities for the hyperbolic functions are exactly the same as the trigonometric identities, except that when a product of two sinh functions is present, we put a minus sign in front of them.
Parametric Equations
Polar Coordinates
This is now in polar form, since all terms are inror θ and contain no xory. This is now in Cartesian form, since all terms are in xor y and do not contain rorθ.
Conics
In many cases we can manipulate the given equation to take the form of a conic. Solution: Since the question only asks us to describe the motion of the particle for small displacements, we can say that xi is much less than one, so x6 '0.
Review Questions
Imaginary numbers allow us to find the answer to the question 'what is the square root of a negative number?' We define it to be the square root of minus one.
Complex Numbers
Example: Find the imaginary and real parts of the following complex numbers along with their complex conjugations. This makes the denominator a real number and the numerator the multiplication of two complex numbers.
Applications of Complex Numbers
In fact, if we take a complex root for any polynomial equation with real coefficients, the complex conjugate of this root is always another root of the polynomial. The idea of phasors makes the addition of sinusoidal functions a simple task since the real part of the complex number in the phasor at any value of Aeiωt−θ is exactly equivalent to the value of eAcos(ωt−θ) at the same point.
Review Questions
Introduction to Matrices
Matrix Algebra
Suppose we want to find the product AB of the matrices A and B. The number of columns of A must be equal to the number of rows of B. The product matrix has the same number of rows as A and the same number of columns as B. Below is the rule for finding the product of two 2×2 matrices: The number of columns of the matrix on the left is not equal to the number of rows of the matrix on the right, so this product cannot be found.
The Identity Matrix, Determinant and Inverse of a Matrix
The (i,j) minor of a matrixA is the determinant of the submatrix obtained by removing the ith row and jth column of A. Calculate the determinant for the submatrix of elements that are not covered and multiply our element by this determinant.
Review Questions
Any vector in 3D space can be written as a combination of 3 basis vectors; i, j and k, each being the unit vector in the x, y and z directions. The vector is in the same form as unit vector notation (a, bandcare the same numbers), but it is more compact.
Operations with Vectors
Note: The dot product can be thought of as how much one vector points in the direction of the other. It is only possible to take the cross product of two 3-D vectors (and technically also 7-D vectors). The cross product also depends on the order of the vectors that appear in the product.
Vector Equations
We can find a normal vector for a plane by computing the cross product of two direction vectors in the plane. Using the concept of a normal vector in a plane, we can formulate another equation of the plane. Now we can easily derive another plane equation by calculating scalar products.
Intersections and Distances
This value of the parameter corresponds to the point on the line that intersects the plane. This will give us our equation for the intersection of the two planes. Now we take the cross product of these normal vectors to find the direction vector for the line of intersection.
Review Questions
We often want to find the value that the function f(x) approaches when x tends to a certain value. In other words, if we are infinitely close to the point x= 1, then f(x) is infinitely close to 2, but never actually reaches it. We can also have 'one-sided limits', where a function approaches two different values depending on the direction from which we approach the limit point.
Algebra of Limits
Methods for Finding Limits
This technique works when we can remove the part of the function that makes the function undefined. In some cases, we can use L'Hˆopital's rule to help us find the limit of a function that says that for two functions f(x) and g(x) this is true.
Review Questions
Introduction to Differentiation
Differentiation properties arise from the idea of calculating the gradient between two points in the limit that the distance between them tends to zero. Similarly, when we differentiate, we are effectively drawing an infinitesimal triangle to calculate the rate of change (or gradient) of a small part of the function. This works the same way we've seen forxandy, but in this case we have it in terms of a and b and we'd say we're "differentiating with respect to tob".
Standard Derivatives
Note: The first rule that we discussed is just a special case of the above rule, whena= 1. All the examples so far have only contained the xandymen in this question we have the variables dy. Note: In the above rule, the b disappears, and it provides a good exercise in using the laws of logarithms, so we have included the algebra in the next example.
Differentiation Techniques
Simplify this by taking a factor of e2x from dy. e2x)2 Thee2x on top cancels with one on the bottom dy. When we have a function within a function (called a composite function), we differentiate using the chain rule. Note that if we differentiate a term involving y(x) we should use the chain rule where appropriate.
Stationary points
If we have the point x where there is a stationary point, we replace this value with the second derivative. First find tox values just to the left and right of the stationary point and calculate f0(x) for each. To find the minimum V, we need to differentiate it and then set it equal to zero (find.
Review Questions
We do this by differentiating with respect to one variable and treating the other variable as a constant. The expression ∂x informs us that we are doing partial differentiation with respect to x, while the index variable is treated as a constant in this case (not every textbook uses index notation). This is the Gay-Lussac (or pressure) law; at fixed volume, pressure is proportional to temperature.
Higher Order Partial Derivatives
When a function does not have continuous second-order partial derivatives, then the order of partial differentiation matters.
The Chain Rule with Partial Differentiation
Gradients and ∇f
Note that this is the same as the equation for electric field strength, so we have essentially proved the relationship E~ =∇V. Now that we can use ∇f, we can find the gradient of a function f in any direction. Note: ~u ∇f(~a) is a scalar product (or dot product), so the directional derivative is a scalar, not a vector.
Other Operations with ∇
For first year physics, an understanding of the proofs for divergence and curvature of a vector field is not required, however it is useful to remember and be able to apply the formulas.
Applications of Partial Differentiation
Example: Using the method of Lagrange multipliers, find the maximum and minimum distance from the point (-2,1,2) to the sphere axis2+y2+z2= 1. By substituting our three equations for the variables in Lλ, we can find the values ofλ that correspond to the minima and maxima of the problem. So by substituting these values ofλ into our equations forx,yandz, we find that the point on the sphere that gives the minimum distance is and the point that gives the maximum distance is.
Review Questions
Suppose we want to find the area under the curve between the values of x and in the graph above. It creates an infinite number of infinitely small width rectangles under the curve and adds their area to find the total area under the curve. This tells us that we want to find the integral of f(x) and then use that information to find the area under the curve between a and b.
Standard Integrals
It's important in this integral because we can input a negative value of xintox−1, but we can't take the log of a negativex. In integrating we do the opposite, so we divide by the coefficient of power. If we get the gradient of a curve and a point through which the curve passes, we can find the complete equation of the curve and thus a value of C.
Integrals with Limits
Physical example: the force on a charged particle, A, due to a second charged particle, B, is given by.
Integration Techniques
We can also make a substitution u = u(x), but then we have to use the rule for differentiating an inverse function to find dx. Method 1: We can leave the limits in terms of the original variablex and then evaluate the limits after converting our solution back in terms ofx. If we choose to do this, we do not need to rewrite our solution in terms of the original variable; we can only evaluate the limits directly.
Constructing Equations Using Integrals
Using integration and knowing that the area of the sphere is 4πr2, we can derive this equation for the volume. Using integration, we add up all the elementary shells from the center of the sphere to the edge, i.e. Since the width of the disk is infinitely small, we can approximate the volume of the disk at πr2×dx.
Geometric Applications of Integration
Solution: We know that the formula for the area of revolution of a curve is A= 2π. By looking at the diagram in the previous section and considering an infinitesimal part of the volume, we can deduce this. Therefore, the volume of revolution we are asked to work out is equal to π.
Review Questions
We can also deduce what the limits of integration are if we find a region over which we want to integrate. Using the relation r2=x2+y2, remembering that in polar coordinatesS=r dr dθ and inserting the limits, we can write the integral as Remember that we can integrate in any order since all the limits are constant, but in this example we will integrate with respect first.
Triple Integrals (Volume Integrals)
As with double integrals, we can usually find the limits of the region of integration from a given form. Solution: Remember that we can use the integral Z Z Z. to find the volume of the solid represented by V. To do this, we need to find the limits of the three variables x, y, and z. Starting with θ we can see that the limits are θ= 0 to θ= 2π as it describes a full circle.
Review Questions
A differential equation is an equation involving an unknown function and one or more of its derivatives. Our goal in solving an ordinary differential equation is to find a function y(x) that satisfies that equation. A differential equation is linear if the power of the dependent variable and all its derivatives is equal to 1 and there are no terms containing the product of these.
First Order ODEs
This form is advantageous because we can now integrate this directly to find the solution. We can then normally separate the variables and integrate with respect to this new variable. Obviously we can't just separate the variables, but if we use substitutionz=y1−k we can get around this.
Second Order Differential Equations
Setting Up Differential Equations
Review Questions
Sequences
Summations
Series
Expansions
Review Questions
Review Questions
Kinematics
Newton’s Laws
Conservation Laws
Oscillatory Motion
Circular Motion
Variable Mass
Review Questions
