Physics for informatics Lecture 1 Introduction vector calculus

Physics for informatics Lecture 1 Introduction, vector calculus, functions of more variables, differential equations Ing. Jaroslav Jíra, CSc.

Introduction Lecturers: prof. Ing. Stanislav Pekárek, CSc. , pekarek@fel. cvut. cz , room 49 A Ing. Jaroslav Jíra, CSc. , jira@fel. cvut. cz , room 42 Source of information: http: //aldebaran. feld. cvut. cz/ , section Physics for OI Textbooks: Physics I, Pekárek S. , Murla M. Physics I - seminars, Pekárek S. , Murla M. Scoring system of the Physics for OI The maximum reachable amount of points from semester is 100. Points from semester go with each student to the exam, where they create a part of the final grade according to the exam rules. Conditions for assessment: - to gain at least 40 points, - to measure specified number of laboratory works, - to submit specified number of partial problems, - to submit semester work

Points can be gained by: - written tests, max. 50 points. Two tests by 25 points max. (8 th and 13 th week) - semester work for max. 30 points - activity on exercises, partial problems solving, max. 20 points Examination – first part: Every student must solve certain number of problems according to his/her points from the semester. Number of problems to solve Points from the semester 1 90 and more 2 75 – 89 3 65 – 74 4 55 – 64 5 less than 55

Examination - second part: Student answers questions in written form during the written exam. The answers are marked and the total of 30 points can be gained in this way. Then follows oral part of the exam and each student defends a mark according to the table below. The column resulting in better mark is taken into account. written exam semester + written exam A excellent 1 25 120 B very good 1 - 23 110 C good 2 20 100 D satisfactory 2 - 18 90 E sufficient 3 15 80

Vector calculus - basics A vector – standard notation for three dimensions Unit vectors i, j, k are vectors of magnitude 1 in directions of the x, y, z axes. Magnitude of a vector Position vector is a vector r from the origin to the current position where x, y, z, are projections of r to the coordinate axes.

Adding and subtracting vectors Multiplying a vector by a scalar Example of multiplying of a vector by a scalar in a plane

Multiplication of a vector by a scalar in the Mathematica

Example of addition of three vectors in a plane The vectors are given: Numerical addition gives us Graphical solution:

Addition of three vectors in the Mathematica

Example of subtraction of two vectors a plane The vectors are given: Numerical subtraction gives us Graphical solution:

Subtraction of two vectors in the Mathematica

Time derivation and time integration of a vector function

Example of the time derivation of a vector The motion of a particle is described by the vector equation Determine for any time t: a) b) the tangential and the radial accelerations

Time derivation of a vector in the Mathematica

Time derivation of a vector in the Mathematica -continued What would happen without Assuming and Refine What would happen without Simplify Graphical output of the

Example of the time integration of a vector Evaluate the time dependence of the velocity and the position vector for the projectile motion. Initial velocity v 0=(10, 20) m/s and g=(0, -9. 81) m/s 2.

Time integration of a vector in the Mathematica Study of balistic projectile motion, when components of initial velocity are given Projectile motion - trajectory:

Scalar product (dot product) – is defined as Where Θ is a smaller angle between vectors a and b and S is a resulting scalar. For three component vectors we can write Geometric interpretation – scalar product is equal to the area of rectangle having a and b. cosΘ as sides. Blue and red arrows represent original vectors a and b. Basic properties of the scalar product

Vector product (cross product) – is defined as Where Θ is the smaller angle between vectors a and b and n is unit vector perpendicular to the plane containing a and b. Geometric interpretation - the magnitude of the cross product can be interpreted as the positive area A of the parallelogram having a and b as sides Component notation Basic properties of the vector product

Scalar product and vector product in the Mathematica

Direction of the resulting vector of the vector product can be determined either by the right hand rule or by the screw rule Vector triple product Scalar triple product Geometric interpretation of the scalar triple product is a volume of a paralellepiped V

Scalar field and gradient Scalar field associates a scalar quantity to every point in a space. This association can be described by a scalar function f and can be also time dependent. (for instance temperature, density or pressure distribution). The gradient of a scalar field is a vector field that points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is that rate of increase. Example: the gradient of the function f(x, y) = −(cos 2 x + cos 2 y)2 depicted as a projected vector field on the bottom plane.

Example 2 – finding extremes of the scalar field Find extremes of the function: Extremes can be found by assuming: In this case : Answer: there are two extremes

Extremes of the scalar field in the Mathematica

Vector operators Gradient (Nabla operator) Divergence Curl Laplacian

Differential equations The most simple differential equation: where We are looking for the function Solution of such equation is Example: Where x 2+C is general solution of the differential equation Sometimes an additional condition is given like that means the function y(x) must pass through a point We have obtained a particular solution y(x)=x 2 -1

First order homogenous linear differential equation with constant coefficients The general formula for such equation is To solve this equation we assume the solution in the form of exponential function. If then and the equation will change into after dividing by the eλx we obtain aλ+b is the characteristic equation the solution is Where C is a constant resulting from the initial condition

Example of the first order LDE – RC circuit Find the time dependence of the electric current i(t) in the given circuit. Now we take the first derivative of the right equation with respect to time characteristic equation is Constant K can be calculated from initial conditions. We know that general solution is particular solution is

Solution of the RC circuit in the Mathematica Given values are R=1 kΩ; C=100 μF; u=10 V

Second order homogenous linear differential equation with constant coefficients The general formula for such equation is To solve this equation we assume the solution in the form of exponential function: If then and the equation will change into after dividing by the eλx we obtain We obtained a quadratic characteristic equation. The roots are and

There exist three solutions according to the discriminant D 1) If D>0, the roots λ 1, λ 2 are real and different 2) If D=0, the roots are real and identical λ 12 =λ 3) If D<0, the roots are complex conjugate λ 1, λ 2 where α and ω are real and imaginary parts of the root

If we substitute we obtain This is the solution in some cases, but … Further substitution is sometimes used and then considering formula we finally obtain where amplitude A and phase φ are constants which can be obtained from the initial conditions and ω is angular frequency. This example leads to an oscillatory motion.

Example of the second order LDE – a simple harmonic oscillator Evaluate the displacement x(t) of a body of mass m on a horizontal spring with spring constant k. There are no passive resistances. If the body is displaced from its equilibrium position (x=0), it experiences a restoring force F, proportional to the displacement x: From the second Newtons law of motion we know Characteristic equation is We have two complex conjugate roots with no real part

The general solution for our symbols is No real part of λ means α=0, and omega in our case The final general solution of this example is Answer: the body performs simple harmonic motion with amplitude A and phase φ. We need two initial conditions for determination of these constants. These conditions can be for example From the first condition From the secondition The particular solution is

Example 2 of the second order LDE – a damped harmonic oscillator The basic theory is the same like in case of the simple harmonic oscillator, but this time we take into account also damping. The damping is represented by the frictional force Ff, which is proportional to the velocity v. The total force acting on the body is The following substitutions are commonly used Characteristic equation is

Solution of the characteristic equation where δ is damping constant and ω is angular frequency There are three basic solutions according to the δ and ω. 1) δ>ω. Overdamped oscillator. The roots are real and different 2) δ=ω. Critical damping. The roots are real and identical. 3) δ<ω. Underdamped oscillator. The roots are complex conjugate.

Damped harmonic oscillator in the Mathematica Damping constant δ=1 [s-1], angular frequency ω=10 [s-1]

Damped harmonic oscillator in the Mathematica All three basic solutions together for ω=10 s-1 Overdamped oscillator, δ=20 s-1 Critically damped oscillator, δ=10 s-1 Underdamped oscillator, δ=1 s-1