solve-the-given-problems-by-integration-find-the-root-mean-square-current-in-a-circuit-from-t0-s-t-2

Question

Solve the given problems by integration.
Find the root-mean-square current in a circuit from $t=0$ s to $t=0.50 \mathrm{s}$ if $i=i_{0} \sin t \sqrt{\cos t}$

Will Erickson · Accepted Answer

hi there. And this problem were asked to compute this definite integral. Now it&#x27;s the definite integral of vector valued function. So from the end of this chapter, let&#x27;s remember our definition of the integral of a vector valued function, uh is we&#x27;re gonna have three separate into girls. We can just add together. So it will take the definite integral of this first component which, let&#x27;s rewrite that as Tito the 1/3 power. As long as we&#x27;re at it, that will be our I component. Plus again definite integral from 0 to 1 of our function, one over t plus one that were J component and finally definitely a girl from 0 to 1 of e to the minus t. That&#x27;ll be RK component. So we just to each of these separately, let&#x27;s begin with this 1st 1 here. Let&#x27;s find an anti derivative. In order to use the fundamental theorem of calculus of the anti derivative of tea to the 1/3 we bumped power up by one to get 4/3 and then divide by that in front. So we have 3/4 t of the 4/3 we&#x27;ll plug in one and zero, and that will be our I coordinate next for the J Cordant uh, anti derivative one over T plus one. Now, since it&#x27;s one over a function, we suspect the natural log of that bottom function. And there&#x27;s no need to use U substitution here, since a derivative of T plus one is just one. So the anti derivative truly is a natural log of t plus one on again. We will plug in one end zero Jay here and finally for K component this last one here, anti derivative eat of the minus t is minus you. The minus T. We&#x27;ll plug in zero on one. That will be our key component. Okay, so let&#x27;s compute these now. So if you plug in 11 of the 4/3 power is just one. So plugging in one gives us just 3/4 times one, which is one plugging zero clearly gives us zero. So that&#x27;s quick for the I component. Now let&#x27;s move on to the J component. Ah, natural log of one plus one says natural. Log of to minus. I was plugging zero for tea. Natural log of zero plus one. This natural log of one, the natural log of one is zero on any log of one is just zero. So that&#x27;s our K component. Is there a J component now for K? I was plugging one for so minus eat of the minus one and minus. That&#x27;s a minus minus, so we end up with plus and eat. Of the zero is just one. There&#x27;s RK component. So we&#x27;re almost done. Just we&#x27;ll clean this up a little bit. The end up with 3/4 I That&#x27;s it for the I component plus natural lot of to Jay, that&#x27;s RJ component. And finally for K. Let&#x27;s see, we can rewrite this a little queen. Or maybe that&#x27;s right. The one first and then eat of the minus one is the same as one over E. So that&#x27;s RK component and we are done

J Hardin · Answer

here we have a definite integral from zero to pie over two of co sign of tea divided by the square root of one plus sign square. Let&#x27;s do it. Use up here and let&#x27;s take you to be sci fi Then do you is cosign TT Since this is a definite no girl, we should switch these limits of integration So the new lower limit will become using this formula. Pierre. So we plug in t equals zero and we have u equals sign zero, which is zero and then for the upper limit. Piper too. So we have upper limit you equals signed by over two, which is one and then by Are you sub cosign t dt is just do you and then the denominator we have one plus you square inside the radical. So now for this new integral we should do entrance up. So let&#x27;s take you equals Tan Daito Then do you a sequence where data and once again because we have Ah, definitely, girl. We should go out and switch these limits. So originally we had zero for you also Wait. One more observation since this is a definite no girl. Oh, when you&#x27;re used when your tricks up is of the form eight and data the requirement is that date is between negative five for two and powerful too. So plugging in you equal zero to get a new lower limit, we have zero equals tan data. And if they does in this interval appear the only time tangent could be zero is a data equals zero. So that&#x27;s our new lower limit for the upper limit. We have one equals tan data and the only time this happens in our honorable negative Piper tutto perverts Who is that power for some running on room here. So let me go to the next page. So we had that was the general. So now, after using our tricks up, we have sealed up IRA for do you with sequins. Where? Data. And in the bottom we have tan squared plus one. So using the Pythagorean identity for Tangent Square, we have seeking squared inside the radical and then the square root of seek and square as you see again so we could cross off one of the sequence Sierra Pi over four seek and data and we know this integral to be natural log, absolute value seek and data plus tan data. So this is a trick in a girl and there end points or zero power before. So if we plug in pi over four first, seeking a pi over four is rude, too. Tangent on power for is one and then seek in of zero is one. Tension of zero is zero and then natural log of one zero so we could ignore the second term and their final answer. Natural log one plus radical too. And you could lose the absolute value because one plus radical to his positive number.

Vipender Yadav · Answer

here in this problem, we have to evaluate the integral zero to pi by two Hussein d divide by under it off one place sine squared e Dr DT Oh, here. I&#x27;m going to substitute us sine squared d Also the same So from here we can say that Do you will be cools to good sign t dot on duty And when de is zero, we get Tita as zero So when he is, you know, forget to us zero And when de is by way to we get you as one So the integration becomes 0 to 1 it is Do you divide by a nerd off one Plus you square no, again I&#x27;m substituting. You asked danger, Tito. So you get do you as seeking schoolteacher Door Detoyato And when you is zero, we get t does zero And when the U is one, we get t desk by bio full So the indication we gums zero to buy. Therefore, do you is seeking square Tito dot de Tito Divide by under Rudolf one Bless you School is one Palestinian square Tito So we know that underdog opinion school teacher is seeking schoolteacher so zero to buy were full Seacon Square, Tito deter divide by second Tito on the second day of unticketed of cancer and we&#x27;re left with zero to buy before this is seeking Teeter data. The answer of this integrationists second data is natural lock Cygan Tita Pless Danger Tedo on the limit is from zero to pi. Therefore simply were deployed the limits, so it will be called to natural Look sickened by by four is roach 22nd and changing by before his one silicate route to plus one minus the lower limit. So live in natural seeking zero is one antennae in 00 to get natural log one which is equals to natural log route two plus one. So this is an answer for the Given Integral which waas co sign t divide by unsettled off one plus sine squared D. Norton Duty on the limit was from zero to buy by two. I hope you got the problem. Thank you.

Amrita Bhasin · Answer

We know that you is one plus co sign t which means that do you There&#x27;s never a good sign T d t Which means that native do you is sign she did you? Which means we have negative times the ventricle of you to the 1/2 to you which gives us knighted to you the three over to divide by three plus C because you&#x27;re using a powerful which means the exploding increases by one You divide by the new exponents you plug in, you&#x27;re back. Substitute you is one Posco Santy.

Foster Wisusik · Answer

Okay. This question wants us to find the anti derivative of this function. So usually with this, we always pick you to be the arguments of our trick function. So let&#x27;s have you equals the square root of T plus three and taking the derivative here. D&#x27;you is equal to 1/2 squared of tea d t. But we only have one squared of tea in our integral. So let&#x27;s multiply both sides by two To get to do you is equal to one over square to t DT. And now we see that that&#x27;s something we have exactly are integral. So now we can do our substitution. So we have the integral of co sign of you times too. Do you or two times the integral of co sign you? Do you and the integral of co sign is just signed So we get to sign of you plus c and looking back in for you, we get to sign of Well, we said you was square it of tea plus three and then close our integration constant. And that&#x27;s our anti derivative

Problem

Solve the given problems by integration. In the s…

Problem 46 Easy Difficulty

Solve the given problems by integration.
Find the root-mean-square current in a circuit from $t=0$ s to $t=0.50 \mathrm{s}$ if $i=i_{0} \sin t \sqrt{\cos t}$

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

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus