﻿ Y Meaning In Math – Anna Hein

## Y Meaning In Math

Cost per purchased scoop of ice cream We want to create a function that represents the relationship between the number of scoops of ice cream purchased and the price. Let’s assume that one scoop of ice cream costs $0.80. Now, based on this information, we can calculate the cost of two, three, four, etc. scoops.  Number of scoops Cost (in €) 0 0 1 0,80 2 1,60 3 2,40 4 3,20 5 4,00 6 4,80 7 5,60 We have entered the costs in relation to the number of scoops in a table. Thus we get the corresponding table of values. We can now represent this relationship between cost and number in a function. Since the costs are increase proportionally we get a linear function. Draw the corresponding function yourself first! You can simply read the values from the table and draw them into a suitable coordinate system. ### Method ### Method Click here to expand In the table of values the x-values on the left and the corresponding y-values on the right. In our example, we assign the number of balls to the cost. Thus, the number of balls must correspond to the$x-value$to which a$y$, the price, is assigned. In a coordinate system, the$x-axis$always runs from left to right and the$y-axis$from bottom to top. We take now e.g. the point$(2/1,6)$and look first for the$2$on the x-axis and draw mentally a line upwards and then the$1,6$on the y-axis and draw again a mentally line to the right. Where the two “thought lines” meet we set the point.$P (x-value / y-value)$Now we do this with several points, connect them and get a function. This is how the function should look like: The figure shows a linear function representing the relationship between the number of balls and the price. On the$x$-axis is the number of balls and on the$y$-axis the cost. The function maps the relationship between them. We can see that the function connects the points on the table of values, creating a straight line. ## Linear functions: Special features of the variables$n$: The y-axis intercept – the point of intersection with the y-axis – is zero, since no ball of ice cream also costs nothing. In general, the y-intercept shows the relationship between no$x$and$y$.$m$: The slope is positive – the larger the$x$values become, the larger the$y$values become. Of course, because the more balls are bought, the more expensive it becomes. The slope can also be negative. Then$m$is a negative value.$x$and$y$: The two variables here are the number of balls and the price. Both variables are related to each other. Here$x$is the independent variable, also called function argument, and$y$is the dependent variable. ## Determine the linear function equation We can calculate the Function Equationwhich represents the relationship between balls of ice cream and price. This has the advantage that you can calculate the price for any number of balls as well as the number of balls for any price. For this we take any two points, for example$P(0/0)$and$Q(1/0,8)$. We now put the points one after the other into the “empty” linear equation$f(x) = m\cdot x +n$. 1.$P(0/0)$This point says that the y-axis intercept,$n$, is zero. As mentioned above, the price of no ball is also$0 €$. Mathematically, we can simply insert the point. Then we get the equation:$0 = m \cdot 0 + n0 = n$So the$n$is dropped from the equation. 2.$Q(1/0,8)$Now to the second point$Q(1/0,8)$. Objectively, this point has the meaning that one ball costs$0.80 $. Therefore, the slope must be$0.8$. Let’s look at this mathematically by substituting the point into the equation$y = m \cdot x0.8 = m \cdot 10.8 = m$Thus, we have now also shown mathematically that the slope is$0.8$. Now we still have to insert the two calculated variables into our equation. It follows that our linear function equation is$f(x) = 0.8 \cdot x$. Now you know the definition and application of linear functions and function equations. You can check if you have understood these explanations with the exercises. Have fun with it! ### Periodic Function A time-varying function is called periodic with the period T, if the function is transformed into itself when shifted by T, i.e. if it is congruent. A oscillation comprises a positive and a negative half-wave and lasts one period T long. The time T is called the period or the oscillation period of the system $$x\left( {t + T} \right) = x\left( t \right)$$ Function p: p(x) = f(x) + g(x) + h(x) Vector u: Vector(H, I) Vector u: Vector(H, I) Vector v: Vector(I, H) Vector v: Vector(I, H) Text9 = “T” ### Frequency Frequency is a measure of the “frequency” of repetitions of an oscillation per unit time. Its unit is therefore the number of oscillations per second and is expressed in Hertz (Hz). $$f = \dfrac{1}{T}$$ ### Periodic time A function is called periodic with period T if the function is transformed into itself when shifted by T, i.e. congruent. $$f\left( {x + T} \right) = f\left( x \right)$$ f(x) = 2sin(2x – 10) Distance g: Distance D, E Point A: Intersection of f, xaxis with start value (-1.28, 0) Point A: Intersection of f, xaxis with start value (-1. 28, 0) Point F: Intersection of xaxis, yaxis Point F: Intersection of xaxis, yaxis text1 = “T =\frac{λ}{c}” text1 = “T =\frac{λ}{c}” text1 = “T =\frac{λ}{c}” text1 = “T =\frac{λ}{c}” text2 = “A_0” text2 = “A_0” text3 = “-A_0” text3 = “- A_0” text4 = “y=A_0 sin (\omega t + φ)” text4 = “y=A_0 sin (\omega t + φ)” text4 = “y=A_0 sin (\omega t + φ)” text4 = “y=A_0 sin (\omega t + φ)” text4 = “y=A_0 sin (\omega t + φ)” text4 = “y=A_0 sin (\omega t + φ)” text4 = “y=A_0 sin (\omega t + φ)” text4 = “y=A_0 sin (\omega t + φ)” text4 = “y=A_0 sin (\omega t + φ)” text4 = “y=A_0 sin (\omega t + φ)” text4 = “y=A_0 sin (\omega t + φ)” text4 = “y=A_0 sin (\omega t + φ)” text4 = “y=A_0 sin (\omega t + φ)” text5 = “${\varphi }$” text1 = “$\lambda =. \frac{c}{f}$” text1 = “$\lambda = \frac{c}{f}$” text1 = “$\lambda = \frac{c}{f}$” text1 = “$\lambda = \frac{c}{f}$” text1 = “$\lambda = \frac{c}{f}$” text2 = “$\begin{array}{l} \omega t\\\ s \end{array}$” Text2 = “$\begin{array}{l} \omega t\\\ t\\\ s \end{array}$” Text2 = “$\begin{array}{l} \omega t\\\ t\\\ s \end{array}$” Text2 = “$\begin{array}{l} \omega t\\\ t\\\ s \end{array}” Text3 = “A” An illustrative example from electrical engineering: in electrical engineering, the period duration at a frequency commonly used in Europe is 50 Hz alternating or three-phase current 20 msec (1sec divided by 50 Hz). A half period, that is the time from a zero crossing (= sign change) to the next zero crossing is therefore 10 msec (20msec : 2 half waves). I.e. one must wait maximally 10 msec, until the considered electrical quantity becomes zero for a short time. ### Wavelength The wavelength is the smallest distance between two points of the same phase in a wave propagation. The wavelength is calculated by dividing the propagation speed c in the respective medium divided by the frequency. In the case of two-dimensional propagation, one speaks of oscillations and their periodic times. With three-dimensional propagation one speaks of waves (e.g..: Sound, div. fields) and of their wavelength. $$\lambda = \dfrac{c}{f}$$ Examples for propagation velocities: • For sound waves: c = 343 m/s • For electromagnetic waves: c = 299 792 458 m/s ### Relationship between period, frequency and wavelength The period T corresponds to the reciprocal of the frequency, or the quotient of the wavelength and the propagation velocity in the respective medium. $$T = \dfrac{1}{f} = \dfrac{1}{{\dfrac{c}{\lambda }} = \dfrac{\lambda }{c}$$ ### Vibration A oscillation comprises a positive and a negative half-wave and lasts for a period T long. With two-dimensional propagation one speaks of oscillations and their period duration. With three-dimensional propagation one speaks of waves (e.g.: sound, div. fields) and of their wavelength. $$T = \dfrac{1}{f}$$ ### Harmonic oscillation Harmonic oscillations are a special case of the periodic oscillation, because they can be completely described by sine or cosine functions. One calls the temporal change of the horizontal and/or the vertical distance of a point P on a circular path a harmonic oscillation. The representation of the point by its position vector is called vector or pointer diagram. • The time change of the horizontal distance from the rotating point P from the y-axis produces a pure cosine oscillation. • The change in time of the vertical distance from the rotating point P from the x-axis produces a pure sine oscillation Circle c: Circle through B with center A Angle ωt: Angle between D’, A, D Angle ωt: Angle between D’, A, D Angle ωt: Angle between D’, A, D f(x) = If(x > 0, sin(x)) g(x) = If(x > 0, cos(x)) Distance k: distance C, E Distance j: distance D, F Distance m: distance F, G Distance n: Distance A, H Vector u: Vector(A, D) Vector u: Vector(A, D) A = (-3, 0) A = (-3, 0) Point D: Intersection of c, i Point D: Intersection of c, i Point H: Intersection of k, xaxis Point H: Intersection of k, xaxis Point D’: D rotated by angle -(45°) Point D’: D rotated by angle -(45°) Text1 = “x(t) = sin(ωt)” Text2 = “x(t) = cos(ωt)” The function u(t) describes an oscillation process, as it is the case for mechanical or electrical oscillating circuits occurs. \eqalign{ & u\left( t \right) = U \cdot \cos \left( {wt + \varphi } \right) \cr & u\left( t \right) = a \cdot \cos \left( {\omega t} \right) + b \cdot \sin \left( {\omega t} \right) \cr & u\left( t \right) = U \cdot {e^{\left( {\omega t + \varphi } \right)}} \cr}  U the amplitude of the vibration (its maximum displacement) $$\omega$$ the angular frequency Here applies: \eqalign{ & \omega = 2\pi f = \dfrac{{2\pi }}{T} \cr & f = \dfrac{1}{T} \cr}  T the period of oscillation $$\varphi$$ the zero phase angle, i.e. the angle at time t=0 ### Changing parameters of a harmonic oscillation Parameters can be used to change the shape of the oscillation. $$f\left( x \right) = a \cdot \sin \left( {bx + c} \right) + d$$ • The factor a causes a stretching or compression of the “height” – the so-called amplitude – of the oscillation. • The factor b causes a change of the period duration – the reciprocal of the frequency – thus a stretching or compression in the direction of the x-axis The factor b corresponds to the number of periods in the interval $$\left[ {0;\,\,2\pi } \right]$$. If you double the factor, there are twice as many periods in this interval. $$b = \dfrac{{2 \cdot \pi }}{T}$$ • The summand c in the argument causes a phase shift (time of the “zero-crossing) in direction of the x-axis (=parallel shift in direction of the x-axis). • If c is positive, the function under consideration is shifted to the left. • If c is negative, the function under consideration is shifted to the right. • The summand d causes a parallel shift of the oscillation in the direction of the y-axis. The oscillation is then no longer symmetrical to the x-axis, but symmetrical to the straight line y=d. ##### ​Illustration • In red the sine function. • In green the sine function phase-shifted by +90° and thus to the left, which thus becomes a pure cosine function (blue) in phase. • In blue the cosine function. We have reduced its amplitude to 75% so that the green and the blue graph are not congruent. f(x) = sin(x) g(x) = 0.75cos(x) h(x) = sin(x + 1.57) Vector u: Vector(F, E) Vector u: Vector(F, E) Text1 = “sin(x)” Text2 = “0.75 \cdot cos(x)” Text2 = “0.75 \cdot cos(x)” Text2 = “0.75 \cdot cos(x)” Text2 = “0. 75 \cdot cos(x)” Text2 = “0.75 \cdot cos(x)” Text2 = “0.75 \cdot cos(x)” Text2 = “0.75 \cdot cos(x)” Text2 = “0.75 \cdot cos(x)” Text2 = “0.75 \cdot cos(x)” Text2 = “0.75 \cdot cos(x)” Text2 = “0. 75 \cdot cos(x)” Text3 = “ \sin \left( {x + \frac{\pi }{2}} \right)$” Text3 = “$ \sin \left( {x + \frac{\pi }{2}} \right)$” Text3 = “$ \sin \left( {x + \frac{\pi }{2}} \right)$” Text3 = “$ \sin \left( {x + \frac{\pi }{2}} \right)$” Text3 = “$ \sin \left( {x + \frac{\pi }{2}} \right)$” Text3 = “$ \sin \left( {x + \frac{\pi }{2}} \right)$” Text3 = “$ \sin \left( {x + \frac{\pi }{2}} \right)$” Text3 = “$ \sin \left( {x + \frac{\pi }{2}} \right)$” Text3 = “$ \sin \left( {x + \frac{\pi }{2}} \right)$” Text3 = “$ \sin \left( {x + \frac{\pi }{2}} \right)\$”
• $$\sin \left( x \right) = \cos \left( {x + \dfrac{{3\pi }}{2}} \right) = \cos \left( {x – \dfrac{\pi }{2}} \right)$$
• $$\cos \left( x \right) = \sin \left( {x + \dfrac{\pi }{2}} \right) = \sin \left( {x – \dfrac{{3\pi }}{2}} \right)$$