solve-the-given-problems-by-finding-the-appropriate-derivative-the-energy-e-in-mathrmj-dissipated--2

Question

Solve the given problems by finding the appropriate derivative.
The energy $E$ (in $\mathrm{J}$ ) dissipated by a certain resistor after $t$ seconds is given by $E=\ln (t+1)-0.25 t .$ At what time is the energy dissipated the greatest?

Stanley Enemuo · Accepted Answer

we need to find the time at which the energy E. He dissipated by senate resistor is the greatest. So we&#x27;re given the punch and eve he is equal to is equal to Ellen. She plus one -0.25 T. So to find the greatest valuable function first we need to find the first derivative and quote that you know. And we can find that the first deliberate is using the formally DDT. Uh huh. Oh what so D. D. T. Over Ellen of you is equal to one over years time to be on the Bt. Um for here we could say U. is equal to T. Plus one. And we get the E. O. D. Ti that will be equal to 1, 1/1 1/2 plus one- Air .5- Their .25. So then we can set this to be equal to zero and then to self a team And he will be equal to three. So now to find the second derivative to find the second president, once he is equal to three, The second derivative will be one Over T. Plus one. Great. That is equal to negative one over four square Which is negative 1/16. Now, since the value of our second derivative for T. Is equal to three, we can find the maximum value. Uh huh. Once he is equal to three so um is equal to Ln of T plus one -0.5 -0.2. And this happens when C. is equal to three

Ren Jie Tuieng · Answer

honest question. Restart over 50 current equation. You need discharging capacitor my English Qatif. All right, So it I not istea maximum current in this case are the maximum current. Yes, actually was to the MF off the entire circuit. But by our or, you know that&#x27;s, um f can be derived from the equation. Q equals two CV where ve stdm If so, it would be and Q is Q not and in the show Toto Toto church We close to see times he But, um if therefore this would be just goes to Oh, I see So disappear your not RC negative. You know, over RC times Exponential noted 50 of us and what we actually want to do is to find e power politics Peter Body resistor Given this ice cream are it&#x27;s the instant in years pole at any given point in time t and therefore substitute I equals toe this for expression. All right, this is T We want to simplify this a little bit. Grouped together in common tones nature, diesel find expression gonna show that this is key. You will integrate this very to find what s T chota energy dissipated by the resistor so that be taking into Grew off Peter T t right From zero to infinity We take out the terms there are not dependent on time and there the rest will integrate. So infinity, the exponential terracotta cereal and a serially exponential chemical too One and f o E t zero tomb which is when it&#x27;s one Abu we left and their b r c over to we left with not square over to see skc This expression were here It&#x27;s the expression for the total energy It will stop, You know Sebesta. Thank you. Square over to see.

Ben Nicholson · Answer

31.26. We have A nor L C circuit, and we&#x27;d like to find how long it takes for the maximum energy stored in the capacitor during a cycle to decrease toe 1/2 of its initial value. And we&#x27;re assuming that the, uh, initial condition is that the charge on the capacitor is at the tightest at people&#x27;s heroes of the current zero. And so the phases zero declares in 31 25. So for a given cycle, where the maximum charges few bags, it&#x27;s energy is this. We wanted to be equal 1/2 the initial maximum energy, which means that we want maximum charge during that cycle, be one over the square root of two times the initial charge. So the maximum few in a cycle is going to be. The initial few times are decaying exponential over to L. And so then taking the dividing by capital Q and taking the natural loggers of both sides. We have little Q max over. Thank you. It&#x27;s equal to negative. All right, t over hell. So the time it takes to reach this state of affairs, it&#x27;s going to be negative to well over our times a log of Hugh Max over Q, which this is one over the square root of two. Um, so negative log of one over the square root of two is the same as the log of the square root of two. Because this is the properties of logarithms, and then two times the log of this period of two is the log of the square root of two squared or two. And so this is the amount of time that it&#x27;s going to take to, uh, it&#x27;s hell over our lock, too. So that&#x27;s the amount of time it&#x27;ll take two. Reach half the initial energy. Um, no. Well, and, uh, this is basically the identity we used. Uh, So having a negative to out here means that you invert the argument and then swear it. That&#x27;s how we got our two from this very door to one over this weary of two down here

Robert Daugherty · Answer

session 67. Problem number 27. So they give us a formula that approximates the surgeon energy in energy and they give us the rate. So the rate per unit of time, which is in ours, is ours. He go to t e to the minus a t now where they want us to figure out is the total energy. So the total energy so are this represents, right, That is a first derivative. So to figure out the total energy, I&#x27;m gonna need to take the anti derivative of that and they want us to find the energy used in the first t ours. So that&#x27;s gonna be the integral of tea you to the minus 80 DT. So this formula is going to give me the total energy that gets used. So in order to do that first, I&#x27;m gonna need to figure out howto I integrate t e to the minus a t d t. Well, you let you equal t then d&#x27;you is equal to d t. Then Devi is equal to e to the minus a t d t, which means that V is equal to minus one over a e to the minus a t so this definite indefinite integral t e to the minus a t d t is going to be equal to UV, So minus t over a you to the minus a tea. So you ve minus the inner role of V d. U. And so this becomes minus T over a E to the minus a t and then minus one over a squared E to the minus a t plus a constant of integration. And so this becomes minus one over a E to the minus a tea, and then you&#x27;re left with t plus one over a plus a constant of integration. Now, what I can do if I don&#x27;t like that one over a I could just write this as minus one over a squared E to the minus a t. And then I could write it as a t um so 80 plus one. Now back to what we needed. We need the definite integral from zero to capital T so hee is equal to the interval from zero to t of t e to the minus a t d t. So that is going to be minus one over a squared you to the minus a t 80 plus one. All of this evaluated from zero to capital T. So let&#x27;s do all of those evaluations And what will we come up with? Okay, so your answer is going to be minus one over a squared, and then when you substitute the capital T end, you get e to the minus a capital T, and then you get a T plus one minus e to the zero, and then you get zero plus one. So this is minus one over a squared, and this becomes right. Yeah. So let&#x27;s just do minus one a squared and multiply by the first term. And that&#x27;s going to be, um, e to the minus a t Time&#x27;s a t. And then the last term is just simply going to be, um so when you look at e to the zero, that&#x27;s going to be, um, minus one. And so you&#x27;re gonna have minus one over a squared, Um, so minus and minus says she&#x27;s simply gonna be plus one over a squared. Sorry, I skipped the term. So what I did, I was multiplying distributing with the first term and had to distribute across that one uh, So you&#x27;re going tohave minus one over a squared, Um, E to the minus a tea. And then when you distribute in this last term here, you&#x27;re gonna have plus one over a square, and we see then what? We can simplify to to go further here. So now it looks like I can factor out that minus one over a squared. Um, and so you&#x27;re gonna have the first term simplifies to You&#x27;re gonna have a simplification here with the A&#x27;s right there. So we go to a new page, you&#x27;re gonna have end up with minus one over. Hey, so minus one over a t e to the minus a t. You know, And then you&#x27;re left with plus one over a squared. So the one over a squared, and then you&#x27;re gonna be left with a one minus e to the minus 80 one, minus e to the minus A t. Okay, so that is my total energy from zero to time T So again. So the total energy, it&#x27;s the interval from zero to t of t E to the minus a t t t. And that could be written as minus one over a t e to the minus a t plus one over a squared one minus e to the minus A t. So the integration by parts and doing a definite interval Now what they ask us in the next part in part B, They&#x27;re asking us to look at what is the limit as t approaches infinity of e which is the limit as t approaches infinity off everything that we just said, They&#x27;re so copying down this answer so minus one over a t e to the minus a t plus one over a square and and one minus E to the minus 80. So as T goes to infinity so as t goes affinity, have to look and see what happens. Okay, this is a linear term approaching infinity, and this would be an exponential term approaching zero. So the exponential term approaching, um, approaching zero or the linear term approaching figure, this is gonna win out and be zero. So this is or you could look at it if you if you didn&#x27;t see it that way from the rates, um, you could look at and say What is the limit? As t approaches infinity of tea over e to the A T. That&#x27;s going to be the same as the limit as T approaches infinity of one over a e to the A t, which is going to be zero. So this limit right here this approaches zero and then e to the A t that approaches zero. So this just turns out to be one over a squared. So this limit as T approaches zero is just simply one over hey squared.

Katie Mcalpine · Answer

Hey, So for this one, we have a battery resistor and Dr usual characters. And then we want we&#x27;re given the inductive time. Constant is 37 milliseconds will go ahead and write that down Hoops all call that town on T 37 times 10 to the minus three seconds and, yeah, let&#x27;s see what else is going on. And we want to know when the energy dissipated the resistor equal to the energies rate at which the energy is stored in the in doctors magnetic field. Um, lots of words. Just some calculus, though. So, um yeah, so the power. So it basically the power through the resistor. The energy didn&#x27;t energy dissipated. The resistor is the power. Um ah. Through the resistor and that&#x27;s I squared of times are, um so we could say p his I squared r and then we want to set this sequel to the energy through the derivative of the energy through the in Dr Cell. Kind of Put this up here. Um, the energy stored in dined on in un in doctor is ally squared. And if you take a derivative of this with respect to time, if l stays cost and I just changes. So then you have You want us to be setting that equal to l I D I d t great. And, um, then kind of doing some algebra canceling these eyes? Um oh, yeah, so that we need to take us. Have the current going through the this, and so you just double check the problem. Okay, So time equals zero. You&#x27;re connecting it. So we can say that I Is this, um, on second V over Al, isn&#x27;t this supposed to be over our stubble? Checking my notes? Um, let me just pull up the equations. I feel comfortable writing all of this. Yeah, So have you over our times. One minus you to the hotel one minus e to the minus. T he over tell. Wait, and then you can I&#x27;ll just kind of do this on the side. So this problem is kind of just organized into some little boxes, so d i d t equals. So this term goes away. And so then you just have to be over our, um, and then also looks Then Tao is, uh, over our So anyway, it&#x27;s view over our and then the minuses. Cancel and then you have one over towel on the outside. And so the inverse of towers are over. How and then we still have this exponential factor to you over towel, and then these are is canceled. And you do stuff with you if you were out. So if we plant these two into, um, here, in here, in here, um, doing out the algebra I got I&#x27;m not gonna fully right out the algebra, But you basically just put all that and and then you get one equals two e to the minus t over Tao. Um And then algebraic was solving that I got t is equal to Tao time. See Elena to And when I put that into a calculator, I got 25.6 milliseconds.

Farhanul Hasan · Answer

all right. First of all, current is equal to I not times e to the minus r t over. Oh, um and so and power is equalled. I squared Times are so therefore power at a given time T is equal to I not squared are times e to the minus two rt over. Oh, that&#x27;s your expression. Next we integrate. And so power has dimensions of energy over time. And so you integrate that over time you get dimensions of energy. So you just in a girl from zero to infinity time goes from zero to infinity off power function of time over t So the ice square I not squared are his time independent comes out of the integral, and you&#x27;re left with, um, exponents minus two rt over l d t. Okay. And so, uh, the equal here is such that the minus two are over. L comes out, uh, of the bottom here in the denominator and e to the minus two r t over our stays on the numerator, and you integrate. So this is your, uh, indefinite in a growth which you integrate from which you, in great from t equals of and so you give the values T equals zero and t equals infinity as your limits. All right, um, so you as I not squared, are times one over minus two are over. L. So this is, uh, independent. Off to a T equals zero. This is one A T equals infinity. This 00 minus one. Right. Um, so this is l Over to our That&#x27;s you get I not squared are times l over to our the arse. Cancel your left. His one half l A times I not squared for your, um, and you.

Problem

Solve the given problems by finding the appropria…

Problem 35 Easy Difficulty

Solve the given problems by finding the appropriate derivative.
The energy $E$ (in $\mathrm{J}$ ) dissipated by a certain resistor after $t$ seconds is given by $E=\ln (t+1)-0.25 t .$ At what time is the energy dissipated the greatest?

Answer

$$3.0 \mathrm{s}$$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

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Precalculus Review - Intro

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$$3.0 \mathrm{s}$$

Basic Technical Mathematics with Calculus

Grace H.

Kristen K.

Samuel H.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

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

Grace H.

Kristen K.

Samuel H.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus