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Question fromRates of Change the Natural and Social SciencesThe quantity of charge Q in coulombs (C) that has passed through point in wire up to time (measured in s...

Question

Question fromRates of Change the Natural and Social SciencesThe quantity of charge Q in coulombs (C) that has passed through point in wire up to time (measured in seconds) is given by Q(t) 7_2t2 4t + 3 [See this example_ The unit of current is an ampere A = Cls ] Find the current when 0.6Find the current when t =At what time the current the lowest?

Question from Rates of Change the Natural and Social Sciences The quantity of charge Q in coulombs (C) that has passed through point in wire up to time (measured in seconds) is given by Q(t) 7_2t2 4t + 3 [See this example_ The unit of current is an ampere A = Cls ] Find the current when 0.6 Find the current when t = At what time the current the lowest?



Answers

The quantity of charge of $ Q $ in coulombs (C) that has passed through a point in a wire up to time $ t $ (measured in seconds) is given by $ Q(t) = t^3 - 2t^2 + 6t + 2. $ Find the current when (a) $ t = 0.5 s $ and (b) $ t = 1 s. $ [See Example 3. The unit of current is an ampere ( 1 A = 1 C/s).] At what time is the current lowest?

And this question were given the charge that goes through a wire at a certain time and we want to find the current at specific times and we want to figure out when the currents are lowest. So we need to remember first that the we need to remember that the time derivative of the charge of the charge flow is current. So that states that we should first differentiate the function of charge that were given. So we know that the we know that this time is we take the derivative just normally and this is equal to this is the current that's flowing through Now. If we want to find the values at specific points, just plug in the number in this case we want to have Yeah. Oh yeah. We just plug in the number into the current and we get that we have that. This is equal to 4.75 amperes current is measured in amperes. And we do the same with just one with the number one. Okay, Let me get that. This is equal to five amperes. Yeah. To find out when the current is the lowest. Yeah. Consider that the current is just a problem that's facing downwards. Yeah, it's facing upwards. It's actually facing upwards. So we can find the vertex Yeah. The vertex of this Parabola. Yeah. Which the specific time we want is equal to negative B over two A. With a jet With a generic of course using a generic polynomial. In which case we have got a negative 4/2 times three Equals -2/3. Okay. Oh yeah. Right. So we know that the lowest current is that T equals two third second. Do you guys hate basketball? And that's how I do this question, Yeah.

So to find a rate that's the change in columns over the change in time. And we're interested in that rate at three seconds. So we'll take the limit As each approaches zero of our function evaluated at three plus H minus or function evaluated at three. All divided by age. So plugging in three plus H we get one third three plus H queued plus three plus H. Well then subtract our function evaluated at three. Everything all divided by H. So probably the hardest part of this problem is just expanding three plus H cube. So let's do that just as a side side step here three plus H times three plus ages nine plus six H plus H squared. Now we need to multiply that try no meal by the binomial three plus H. We get 27 18 H. Three H squared and then multiplying by H. We get nine H. Six H. Squared and H cube. So combining like terms we get 27, 27 H nine H Squared in H. So let's go back to our problem. We have the limit as H approaches zero. We need to take a third of those quantities. So we have nine nine H. Three H squared and a third of H cube. Then we have plus three plus H. In the next set of brackets we have three cubed which is 27 times. The third is nine plus three is 12. So let's see nine plus three. He would notice his 12 minus 12. Those canceled and all the terms left having each in common. So we can factor out that H and divide. So we get nine plus three H plus one third H squared plus one. As a church tends to zero, these terms would disappear And we're left. What's 10? So what is the current in Amperes? It's 10 columns Oops 10 km/s. Or we could just say 10 amperes Yeah. In the follow up question where asked when will a 20 ampere fuse in the line blow. So what we need to do is figure out The women is h approaches zero of one third, T plus H cube plus T plus H minus 1/3, T. Plus T. All divided by age. So what I'm doing now is exactly what we did before. We just don't know what the time T. It's so we'll expand this before T plus H cubed. You'll see as T cubed plus three T squared H plus three th squared plus H cube. So we'll have the limit as H approaches zero. A third of those terms would be a third T cubed plus t squared H plus th squared plus a third of h cube. Well then add T plus H. So that's the first set of brackets -1 3rd. TQ Busty. Yeah. All divided by H. We see that a third he cube and T cancel all the terms left have each of them So we can divide by one as a church tends to zero these cancel away and generally we get T squared plus one. And note when he is +33 squared plus one would be the 10 that we already found. So when will a 20 ampere fuse in the line blow? We can now set this equal to 20 and solve for T by square rooting. We get the squared of 19 And we know the squared of 16 is four. So the squared of 19 should be a bit more. This works out to about 4.359 seconds. So at that time we know we have 20 amperes in the line and thus the line would blow

For the problem given to us here, we see that the electric charge and a circuit at any given point is given by Q. Which we're going to call fx is equal to x times the second of the square root of 0.2 X squared. That's one. And this is going to be angle measurement. So when we take the derivative, we see that we would have something um of the form just this because when we take the derivative, this is the product rule. So we take the derivative X. It's just one and then it's gonna be plus X times uh the second of this, mm times the tangent of this. And then we have to use chain rule again. So we have to do chain role within chain role. Um But instead of doing this, another option that we have is to evaluate f prime of X At X equals 0.8 seconds. So we end up getting about 2.5 um as our value. So that's going to be 2.5 amps because we took the change in the charge with respect to time, that would give us 2.5 Amps.

Were given a charge function. Q. Of T. And we want to determine what is the current current capital. I of T is the first derivative of the charge function with respect to times Q Prime of tea. And for this it's going to get a little bit complicated. We want to be careful. We're gonna use the product rule and the chain role. So let's define FFT as e to the negative to t It's a FFT equals E to the negative to t and then f prime of tea use the chain role and say, Well, the first derivative of E to the negative to t would be either the negative to tee times the first derivative of negative to T, which is negative two. So we'll get negative too e to the negative to t that are other function GFT That's gonna be this part right here. It's a GFT is going to be co sign of three T minus two times the sign of three T, and the first derivative of that again will have to use a little bit of the chain role. First derivative of co sign is negative sign and the first derivative of three. Tia's three. So we're gonna end up with negative three times the sign of three T and for our second term, the first derivative of Sinus co side in the first derivative of three. T is three, so we'd have three co sign of three tee times. Negative two is negative six times the co sign of three T and our product rule tells us that if you have f times G and you want to take the first derivative, you'd say times g prime plus g times f prime. So let's go ahead and do that F That was each of the negative to t times g prime. That was negative three times the sign. But keep that need. It's gonna be hard enough not to make any mistakes. Alright, negative three times the sign of three T minus six times the co sign of three t and we're gonna add to that g Times f prime. So we'll put it in the other order. Well put f prime first. So it's going to be negative, too no to the negative to t Times G, which is co sign of three T minus to sign of three T Now we could clean this stop, and there are several ways to go about it. But let's go ahead and do it sort of step by step so we don't mess anything up, so we'll distribute those eat of the negative to tease. We eat with the negative to t times Negative three Sign. Excuse me. Negative three sign of three t so we'll have negative three e to the minus two T sign of three T That's our first term. Minus six e to the negative to T co sign of three t What's that term minus two e to the negative to t times the co sign of three T that's that term negative. Two times negative two is positive four plus four e to the minus two T times this sign of three T Now that we can go ahead and combine up our like terms so we'll go ahead and take our to sign terms and combine them and we get for you to the minus two T minus three e to the minus. Two tea leaves us e to the minus two T times the sign of three T, and they will also combine up our our co sign of three teas. We get negative six feet of the minus two T minus two. So that's minus eight E to the minus two t times the co sign of three t We're almost there. One more step. It's factor out that e to the negative to t So we end up with E to the negative to t times the sign of three T minus eight times the co sign of three t and then we take a quick moment to pat ourselves on the back and be super proud cause we got it right.


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