solve-the-given-problems-by-finding-the-appropriate-derivatives-the-radio-waveguide-wavelength-lam-3

Question

Solve the given problems by finding the appropriate derivatives.
The radio waveguide wavelength $\lambda_{r}$, is related to its free-space wavelength $\lambda$ by
$\lambda_{r}=\frac{2 a \lambda}{\sqrt{4 a^{2}-\lambda^{2}}}$
where $a$ is a constant. Find $d \lambda_{r} / d \lambda$

Alex Moore · Accepted Answer

all right here. We&#x27;ve been told the frequencies of different wavelengths and of different electromagnetic waves, and we want to find their difference in wavelengths. So for all electromagnetic waves, this speed of the wave, it&#x27;s going to be equal to the waves wavelength times its frequency. So if we want to find the wavelength of a wave, you know that the wavelength is equal to with speed divided by the frequency. Okay, if you want to know the difference in wavelengths w one wavelength one minus wavelength to we also can call this Lambda one mind slammed the two. It&#x27;s gonna be equal to see over F one Linus. See over f too. Speed of electromagnetic radiation is always the same, but the frequencies in the wavelengths could be different. Okay, this is the same thing. A C. Times one over. F one, minus one over two. Okay, so this is three. This is three times 10 to the ace meters per second times for part. A frequency one is 50 kilohertz. So this is one over 50 times 10 to the three hurts. So this isn&#x27;t seconds. One over seconds is seconds. Minus one over 52 times 10 to the three hurts. Hurts is one over seconds. One over one over seconds. Two seconds. I know units get kind of crazy sometimes. So the answer to part A is one over 52 the 3 50 times, 10 to the three. So one over. Five times. 10 to the fourth, minus one over 5.2 times. 10 to the fourth times, three times 10 to be eight. So this is gonna be 200 and sir T point eight meters. Okay, that&#x27;s your answer to part a week. Also around this to being about two kilometers. About 0.2 Cloutier&#x27;s. All right, that&#x27;s your answer to part A is 0.2 kilometers. So for part B, we&#x27;re gonna use the same trick. It&#x27;s gonna be C Times one over f when minus one over F two is equal to wavelength one minus. We have links to All right. See, It&#x27;s still the speed of light, which is we haven&#x27;t been told any different we&#x27;re gonna assume is traveling through air. So its feet is three times 10 to the ace meters per second times won over 500 kilohertz. So that&#x27;s 500 groups. 500 times 10 to the three. One over one over second. So this is all gonna be in units of seconds. Minus won over. 502 killers tend to the three. This is all in units of seconds. Seconds. You&#x27;re gonna cancel. We&#x27;re gonna be left with meters, right? This is three times 10 to me is times one over 500 times 10 to the three minus. Won over 502 times. 10 to the three. It&#x27;s gonna be times point 7.968 times 10 to the minus nine. All right, so three times 10 to the eighth time, 7.9. 68% of the minus nine. It&#x27;s gonna be 2.3 nine meters, which you know we can around to being around two meters. We don&#x27;t feel like being very precise.

Ren Jie Tuieng · Answer

no discussion. We have two different wavelengths passing through the same slit. They want a fight. What is one of the, uh, we think given did it? Uh uh. They&#x27;re free. Just actually overlap at a specific auto, uh, stable king it. So that&#x27;s all I got right down the equation for the defection creating decided to I&#x27;m gonna consider the first sweet thing I know. About what? So for them, the one we have the fourth order, which actually overlaps 55th order off them. Not to so far. The forefather off number one, the ankle iss Scient eater. Close to four times number one over deep. And this is actually D. C. The tango for number two. But for the fifth other sister actually over. Let me if you shudder. So So it goes to five times two over from here. We can infer yet for them that one must be close to find up to. And so to find that too take four or five times one. And we know that I loved on what is 500 40 centimeters stays given So substituting that gets funded. That too millimeters for two

Chris Trentman · Answer

were given in equation in a variable in this equation and were asked to sell this equation for this variable. The equation is Lambda equals two l over end and the variable miss equation is l first multiply both sides by the least common denominator, which in this case is in. So we get that in Lambda is equal to two l and then by the multiplication of property of the quality divide both sides by two to get and lambda over two equals l and this is our in.

Mayukh Banik · Answer

the optical path and difference is given by de scientists are one miners de scientific tool. It was D scientist, a one minus sign that I do now. Constructive interference means Gajah Optical bathroom difference. Easy with integer much of the present. So the condition that we have is G scientist a one minus sign that I do. It was

Gregory Higby · Answer

um So what we have is this Greek letter. Hey, you plus me over. See you, plus de all to the six Power Now on Lee, The Greek letter And you are variable. So a, B, c and D you can treat us Constance s. Oh, this is just a quotient rule because there&#x27;s a question right here inside of a change in The big thing is taken to the six power. So I asked for the derivative. You want to bring that six in front first and now it&#x27;s to the fifth Power. Could you subtract one from the exponents? Leave the inter function alone. But now you need a multiplied by the derivative. And that&#x27;s where your chain rule comes in. So the derivative of the top is a sorry I said chain him and quotient the drift of a be easier. Remember, it&#x27;s a constant. You leave the denominator alone, minus the Dhruv Dev of the Bottom, Miss See America Dribs of of a constant is zero. Uh, you leave the top alone all over the denominator. See you plus D Square. Now you can actually clean this up pretty nicely as you distribute a to this you have a C U plus a D minus. As you distribute this fun a c U minus BC It&#x27;s like going in alphabetical order. So anyway, you can see the A C use. Cancel. Um, so the only thing left to do is if you want to clean this up a bit six and the red work here, green or chi, have a D minus B. C um, this au plus B, I can leave us to the fifth power. And then right here, by the way and then the sea, you plus D. I can write to the seventh power cause you would have to take this one to the fifth, this one to the second you add the exponents to get that answer. Um, now, your teacher might explain a little different, but this is sufficient.

Problem

Solve the given problems by finding the appropria…

Problem 55 Easy Difficulty

Solve the given problems by finding the appropriate derivatives.
The radio waveguide wavelength $\lambda_{r}$, is related to its free-space wavelength $\lambda$ by
$\lambda_{r}=\frac{2 a \lambda}{\sqrt{4 a^{2}-\lambda^{2}}}$
where $a$ is a constant. Find $d \lambda_{r} / d \lambda$

Answer

$$\frac{8 a^{3}}{\left(4 a^{2}-x^{2}\right)^{3 / 2}}$$

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$$\frac{8 a^{3}}{\left(4 a^{2}-x^{2}\right)^{3 / 2}}$$

Basic Technical Mathematics with Calculus

Anna Marie V.

Heather Z.

Caleb E.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Anna Marie V.

Heather Z.

Caleb E.

Michael J.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus