09:58 Nov 23, 2017 |
German to English translations [PRO] Science - Mathematics & Statistics / Harmonic oscillations and derivatives algebra | |||||||
---|---|---|---|---|---|---|---|
|
| ||||||
| Selected response from: Annika Hogekamp Germany Local time: 10:50 | ||||||
Grading comment
|
Summary of answers provided | ||||
---|---|---|---|---|
5 +2 | to differentiate |
| ||
4 +1 | derive |
|
Summary of reference entries provided | |||
---|---|---|---|
Differentialquotient |
| ||
Ableitung einer Funktion |
|
Discussion entries: 8 | |
---|---|
to differentiate Explanation: In your context, it will be "differentiate" or "calculate / find the derivative", because in physics, let's say, you get the acceleration function a(t) by differentiating the velocity function v(t) with respect to t . In Maths it's the same as long as you are dealing with functions (and their derivatives). So, in your Maths context, it's also "differentiate" because the cosine function is the derivative of the sine function. (I am a former Maths and Physics teacher, so I should know.) |
| ||||||||||||||||||||||||||||||||
Grading comment
| |||||||||||||||||||||||||||||||||
20 mins confidence: peer agreement (net): +1
|
3 hrs peer agreement (net): +1 |
Reference: Differentialquotient Reference information: Wie die Kollegen, so vermute ich auch, dass es um Differentialrechnungen geht. Grundbegriff der Differentialrechnung ist die Ableitung einer Funktion (auch Differentialquotient genannt), deren geometrische Entsprechung die Tangentensteigung ist. https://de.wikipedia.org/wiki/Differentialrechnung The primary objects of study in differential calculus are the derivative of a function, related notions such as the differential, and their applications. The derivative of a function at a chosen input value describes the rate of change of the function near that input value. The process of finding a derivative is called differentiation. https://en.wikipedia.org/wiki/Differential_calculus |
| |
Login to enter a peer comment (or grade) |
17 hrs |
Reference: Ableitung einer Funktion Reference information: There isn't an easy plug-in for all the forms of ableiten. The following are some examples and my suggested translation of each. https://application.wiley-vch.de/halliday/pdf/352740645X_c16... Durch Ableiten erhalten wir: By differentiating (or by taking the derivative), we obtain: Wenn wir Gl. 16-3 einmal nach der Zeit ableiten, erhalten wir einen Ausdruck für die Geschwindigkeit des Teilchens bei einer harmonischen Schwingung If we differentiate eq. 16-3 once with respect to time, we obtain an expression for the velocity of a particle oscillating in simple harmonic motion Aus der Geschwindigkeit v(t) der harmonischen Bewegung erhalten wir durch erneute Ableitung nach der Zeit einen Ausdruck für die Beschleunigung des oszillierenden Teilchens: From the velocity v(t) of the harmonic motion, we obtain through another differentiation (in other words by differentiating again) with respect to time an expression for the acceleration of the oscillating particle: Die erste und zweite Ableitung von Gl. 16-3 führen auf die Geschwindigkeit und die Beschleunigung des Teilchens bei einer harmonischen Schwingung als Funktion der Zeit: The first and second derivatives of eq. 16-3 give the velocity and acceleration, respectively, of a particle oscillating in simple harmonic motion as functions of time: http://www.peter-junglas.de/fh/vorlesungen/skripte/physik1.p... Welche Funktion s(t) ergibt, zweimal abgeleitet, genau -g? Which function s(t), when twice differentiated (or after taking the second derivative), yields exactly -g? s(t) = -1/2 gt^2 => v(t) = ds/dt = -2*1/2 gt = -gt => a(t) = d^2 s/dt^2 = -g For expressions in English: https://en.wikibooks.org/wiki/Calculus/Implicit_differentiat... -------------------------------------------------- Note added at 17 hrs (2017-11-24 03:22:43 GMT) -------------------------------------------------- Math is Fun: https://www.mathsisfun.com/calculus/derivatives-rules.html |
| |
Login to enter a peer comment (or grade) |
Login or register (free and only takes a few minutes) to participate in this question.
You will also have access to many other tools and opportunities designed for those who have language-related jobs (or are passionate about them). Participation is free and the site has a strict confidentiality policy.