solve-the-given-problems-by-finding-the-appropriate-derivative-a-metal-bar-is-heated-and-then-allo-3

Question

Solve the given problems by finding the appropriate derivative.
A metal bar is heated, and then allowed to cool. Its temperature $T$ $\left(	ext { in }^{\circ} \mathrm{C}ight)$ is found to be $T=15+75 e^{-0.25 t},$ where $t$ (in min) is the time of cooling. Find the time rate of change of temperature after 5.0 min.

Stanley Enemuo · Accepted Answer

We need to find the time rate of change of temple for me to embark after five minutes. If the temperature is expressed by T And it&#x27;s equal to 13. Look, count to five. Mhm. To the zero 0.25 T. So the time rate of change of temperature can be obtained by different change given equation of our temperature. So the team T. C. That&#x27;s equal through- 18.75. Some speech and negative. They&#x27;re .25. Now we can find a time really change of temperature after five minutes but 50 tune and C. Is equal to five. Enter the equation and that gives you -5.4 and that&#x27;s the time we will change the temperature. The negative sign indicates that the temperature decreases with time.

Amrita Bhasin · Answer

this question asks for the temperature of the bar at the end of one hour and the average value of the temperature. What we know is that the end of one hour we have 20 plus 9 80 Eat the negatives. Your 0.1 terms 60 which leaves us with 22.43 degrees Celsius report be. We know we have the average value of 1/60 times the integral from 0 60 of the function. I&#x27;m just gonna write off of T function is in the problem. This is equivalent to 1 60 times 1200 because we have 20 times 60 minus 980 divide by 0.1 times E to the negative six minus one, which is equivalent to 182.93 degrees. And then the unit is in Celsius Now for part C. What? We know that the average value of the temperature over the first hour is going to be smaller than the average value of the temperature at the beginning and at the end, because we know the graph of f of X is corn cave up. No,

Stephen Hobbs · Answer

in this Newton&#x27;s law of cooling problem. We&#x27;re waiting to see how long a possible boiling water must set out for it to reach a temperature of 30 degrees. So we could start with, um, the original temperature equation as a function of time, Little t is gonna equal C you to be negative, Katie. Now, one thing that we will have to solve for is see, the next thing we&#x27;re gonna have to Seoul for is K. And so we&#x27;ll start. And this is also gonna have ah, bless A as well, um, we&#x27;re going to start with this initial temperature. It&#x27;s always good to start with the time of zero, cause it&#x27;ll knock out the exponential so we&#x27;ll replace the left side of the equation with that 100 degree temperature equals C E to the in this case, time zero and then plus the ambient temperature, which is the temperature of the room just 10 degrees so we can see right away. Uh, the temperature is going to be ah or C. Rather, it&#x27;s becomes won&#x27;t be to the zero becomes one. So C is going to equal 90. We&#x27;ll use that in the next part, so we&#x27;ll take this same equation. But next, we&#x27;re going to use the information that now that what if the temperature is 80 we&#x27;ll replace the left side of the equation with 80 which is equal to see. We just found C was 90 agrees, and then times e so the negative K, Except we also know that the time in this situation would it&#x27;s 80 degrees. Time is thirties. What could say, like negative, but the okay or negative K 30 with whichever and then plus a, um and A you could just go ahead and place of with the temperature of the room here, which in this case is 10. So plus 10. Okay, will subtract that 10 over to the left side. Then we&#x27;ll get 70 and then we&#x27;re gonna divide both sides by 90. I&#x27;m doing so the right side of the equation is just gonna remain then e to the negative 30. Okay. But of course, in order to solve for K, we need to take the natural log of both sides, will take the Ln of the left side and we&#x27;ll take the Ellen of the right side because this will allow us to then cancel the natural log with E. So Ellen and E will both cancel and then we&#x27;ll get that. Ah, the only thing left on the right. His negative 30 k So if we divide this now by negative 30 and this side by a negative 30 we could find I find out what Value K is so that would get us a K value of should be positive. That looks like and it is that will give us zero 0.0 zero eight and then it depends on how many decimals you want to use. We&#x27;ll just included anymore, whether or not they&#x27;re significant or not. 0.837 It&#x27;s only when you know C and K that you can then use Thies this information to then solve or an unknown time given a temperature. Finally, we are given a temperature of 30 degrees Celsius and we want to find out how long this will take So you could go at any news you calls See, which is 90 e to the this time negative since even though we solved K is a positive, it&#x27;s ah negative in this equation since the temperatures decreasing, exponential decay, eat and then, like we said, if you go and include 37 you care to, um, Times T and plus 10. Okay, so this will be the last equation that we saw on this. T value will tell us how long this will take. Subtracting 10 from both sides. We will get 20 from the left, divide both sides by 90. 20/90. And that would then leave us on the right side. Just negative. Zero point you. I should be a deer. Okay. The seven density. We could take the natural log of both sites. No one and natural will come right. Like it&#x27;s sons. Elena. Meter in verses that gets them to cancel. Finally, you could divide both sides by negative 0.837 Divide that on the left and the right. And finally, we&#x27;ll get a T value of positive 100 and 80 minutes approximately. And you could, um, point this last step in if you need a little bit more specific than to the nearest minute time on that one

Stephen Hobbs · Answer

in this problem. We&#x27;re using Newton&#x27;s law of cooling to find the temperature reached by a glass of milk, Um, at 90% the ambient temperature and on the right side, you can see that calculation there, so 0.9 times 20 would get a temperature of 18 degrees. So that&#x27;s our goal is to find out how long it takes to reach that. We&#x27;re going to start with most Newton&#x27;s law of cooling problems with the temperature as a function of time Or just again, uh, the textbook. It&#x27;s the shortcuts to the differential equation, and so we&#x27;ll go ahead and use it. So we&#x27;ve got C E to the It could be positive or negative k. It will actually work out no matter what. So we&#x27;ll go ahead and use positive to switch it up. It&#x27;s time, times T and then plus the ambient temperature, which in this case is 20 degrees plus 20. Next, we&#x27;re going to use the initial condition, which is when time is zero. We&#x27;ll plug in the temperature and degrees on the left side. Five degrees is the temperature after zero minutes, which is nice, because that then knocks out the exponential function we&#x27;re gonna get E to the zero. Anything to the power zero is one which is just see, in this case and then plus 20. So this allows us at least to find out what the see value it&#x27;s. And we can subtract the 20 over and we&#x27;ll go ahead and write it. Ah, negative. 15. Because our see value. Okay, we then need another value besides the initial to continue from here. And so you&#x27;re going to use this equation and the subsequent information that after one minute, the glass of water has cooled to seven degrees. So we&#x27;ll replace the temperature on the left side again. This time with seven equals. C is negative 15 as we just solved previously times e to the Katie. But we do not know K, which is fine, because actually, we do know time, right? It&#x27;s over here. After one minute, the temperature is seven degrees. So time is one. And then plus 20. So far, so good. So we&#x27;re gonna isolate K. We&#x27;re gonna subtract 20 over, and that&#x27;ll get us. Give us negative three, and I&#x27;m sorry. Negative 13. And then we&#x27;re gonna divide by negative 15. We&#x27;ve got our negative 15. And once we do that, subtracted the 20 and then divided by a negative 15 on the right side, it&#x27;s actually just gonna leave us with E to the K. So we&#x27;re very close to solving this. In fact, the only thing we have to do at this point is take the natural log of both sides. And the reason we do this is whenever we have an exponential on one side of the equation and we want to get rid of it. And so because Ellen and you&#x27;re in verses of each other, they cancel Leave us with K. And in this particular situation, we find out that K is going to equal. That&#x27;s all right. Que is going to equal negative 1.1431 So that&#x27;s the result of Ellen of 13 divided by 15. And then finally we can jump to the last part. Remember, we wanted to find out when the temperature was 90% of the ambient. 90% of 20 was 18. So in the end, we want to stop. Find out, um, how much time it takes to reach a temperature of 18 degrees Celsius set this equal to, um see, we found out with his negative 15 times the exponential e. We did find out K just a second ago, 0.1 or 31 and then times t again, That&#x27;s the thing that we don&#x27;t know, Plus the ambient temperature of the room so less that 20 very similar process to what we just did to solve for K. It&#x27;s just we&#x27;re solving for t Subtract the 20 over will get negative, too. So it&#x27;s Subtract that from both sides. Dubai Negative 15 by both sides that will get rid of it on the right side. And that will just leave us with at this point, um, e to the negative 0.1 for 31 They could probably see what we&#x27;re gonna do in the last two steps here. We&#x27;re gonna take the natural log of both sides and again, that really only helps us because it gets rid of the exponential on the right side, since that they are opposites and versus so those cancel. And then we could just divide both sides by negative 0.1431 And that would be the last step towards isolating, um t. So divide that side by mega do one 4.1 for 31 and we will get a time a very reasonable time of about 14 minutes. Um, if you get something that&#x27;s about two hours, you know, just from your own experience that, um, a glass cools very quickly in an ambient room, and so it will take a very short amount of time, so 14 seems very reasonable.

Regina Hays · Answer

Okay. What we want to do is we want to find the instantaneous rate of change. Well, my brain is going faster than my writing rate have changed for the function g of T equal to two money&#x27;s t squared, Um, at T equal to negative five. And so we know that the instantaneous rate of change is given by the limit process. So this is gonna be the limit as H goes to zero of by function evaluated at that tea value plus h minus. My function evaluated at that tea value divide by age. So we get the limit as h goose zero of two minus the negative five plus h squared minus two minus. The native five squared. And this is all gonna be over H. And so when I, um, multiply out that parentheses, we get on the limit as H goes to zero of two, um, minus 25 plus 10 h minus h squared. And then we have a minus two, um, plus 25 and this is all over H. And so we knows the twos add up to zero. The native 20 five&#x27;s add up to zero, and then what we can do is the limit as h goose zero. I&#x27;m a factor at an H from both of my terms in the numerator. And so the H and the denominator in the H in the numerator reduced down toe one. And so we have the instantaneous rate of change. AT T equal to negative five is 10.

Nick Johnson · Answer

Okay, so we know that according to the law of cooling the temperature oven ab of an object, um, at time T is well governed by this differential equation. Um, DT, DT Capital T DT is equal Jew. Well, negative k times t minus t sub em. Okay, so it is given here the temperature of the room t&#x27;s mm is equal chew 72 F. So we substitute 72 into our equation for T&#x27;s Obama, and we get well ah, DT The TV is equal to, um negative K times. Well, t minus 72. Okay, so now, um, we can do is rewrite our equation by separating the variables to get one over T minus 72. DT is equal to negative K times. Well, negative k d t. So that we can integrate both sides are equation. You have the camera if he&#x27;s on one side and, uh, tees or the DTs on the other side. So just integrate both sides and the left side. Well, that is just equal to the natural log. So the natural log of the absolute value of T minus Sepp t minus 72 and that is equal to Well, we just have negative, Katie. Plus a constant. Okay, um, so now we can exponentially eight both sides to find t of tea. So t capital t of t is equal. Chew. Well, 72 plus. See someone times e to the negative K t. Okay, I basically taking, you know, e to both sides e to the natural log of something just gives you back. Um, something. Right. So what we got here is while the temperature of the object after one minute, um, so t someone All right, this is TFT, so tee or tee off one is, um, is 1 50 So t of one is 1 50 So therefore, we&#x27;ve got 1 50 is equal to while, 72 plus. See someone times e to the negative. K t. Um, maybe a few apps messes. Kind of. It&#x27;s a little sloppy. Saris. We just ate que looks. Okay, let&#x27;s see. All right, So, um, some facts. 72 here. What we got? We got 78 is equal to see someone times e to the negative k. Okay, So the rape, um, the rate of change off temperature is 20 F, so we can substitute in negative. 20. So we&#x27;re gonna substitute, um, negative 20 for t t t. Why? Well, because the rate of change is 20 F and the rate of change that is the derivative there. So in this substitute, negative 24 the derivative T T T T um, and 1 50 and 1 50 um, degrees Fahrenheit for tea for tea. Um, in our equation, the T V t equals negative K times t minus 72. So what we end up getting here is Well, we&#x27;ve got negative. 20 is equal to negative K. Times 1 50 minus 72. Okay, um, so we&#x27;ve got native 20 is equal to Well, this just becomes negative. 78 k and then and stuff. It&#x27;s okay. We&#x27;ve got dividing through running of 78. We&#x27;ve got K is equal to zero point 256 Okay, so, um, 0.256 is equal to Kate Now. We can substitute the value for K in our equation. 78 equals C suborn times eats negative K. So what we&#x27;ve got is 78 is equal to well, cease of one times e raised to the, uh, negative K So times e raised to the negative. 0.256 Okay, so therefore, we&#x27;ve been selfless thing for cease of one is to buy through by each of that power. And where we got we got 100. Um, point 76 is equal to ours. See somewhat. Okay. Now, in order to find the initial temperature of the object weaken, substitute zero for tea. Um, and well, then all the other known values. And what do we get you get? Um, so we got our equation, right? Is t off T is equal to while, 72 plus, uh, plus See someone Times e raised to the negative K t. Okay, this is a plus sign here. Okay, so we&#x27;re gonna find out t of zero is equal to while 72 plus we found on C support is ah, 100. Um, points 76 times eat. Raised to the negative. R K. Value is what, zero 0.256 Um, and then our tea is times zero. OK, but notice that we have something. Time. Zero, This is gonna be e 20 which is just one. Okay, Just whole thing just becomes t of zero is just equal to write this e to this power. It&#x27;s something Time. Zero. So it&#x27;s this is gonna be eating zero and any Internet 01 So this is just one. So this just becomes 72. Um, plus ah, 100 0.76 That whole e term is gone because this just becomes he did a zero, which is one. Okay, so this just becomes, well, 172 0.76 So there we go. Get the initial temperature as 172.76 degrees Fahrenheit. Okay, there is our initial our initial temperature. Okay, Now, let us find the temperature of the object after 10 minutes. So after 10 minutes, what we&#x27;re gonna find gonna find, well t of 10. Okay, So t of 10 is equal to 72 plus 100 0.76 times e ah, to the negative. 0.256 times 10. Okay, so, um, this is gonna equal while 72. Plus, where we get here, it just gets 7.79 you know? So, plus 7.79 And that is equal to while 70 nine, um, 79.79 Okay, so, um, now we can substitute 79.79 40 and 72. 40 and 0.256 for K in our equation. Right. The guy whips we&#x27;ve got, um, into when substitute these values into the equation. Oops. The TV, too. The T e t is equal to negative K times t minus 72. Okay, so what do we get? We get d t d t is equal to Well, negative, Kate. So negative. 0.256 times she which we got a 79.79 and then minus 72. Okay, so this, um, what is, um, the you can t t Here were all this is gonna be approximately equal to, um, negative, too. So the rate of the rate of change in the rate of temperature after 10 minutes is, um is to he agrees. Very night. All right, take care.

Problem

Solve the given problems by finding the appropria…

Problem 27 Easy Difficulty

Answer

$-5.4^{\circ} \mathrm{C} / \mathrm{min}$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

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Heather Z.

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$-5.4^{\circ} \mathrm{C} / \mathrm{min}$

Basic Technical Mathematics with Calculus

Anna Marie V.

Heather Z.

Caleb E.

Joseph L.

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.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus