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On a cold winter day $\left(10^{\circ} \mathrm{F}\right),$ an object is brought outside from a $70^{\circ} \mathrm{F}$ room. If it takes 40 minutes for the object to cool from $70^{\circ} \mathrm{F}$ to $30^{\circ} \mathrm{F}$, did it take more or less than 20 minutes for the object to reach $50^{\circ} \mathrm{F}$ ? Use Newton's law of cooling to explain your answer.

$44.64^{\circ}$Less than 20 minutesdecrease to $50^{\circ}$

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 1

Differential Equations Everywhere

Differential Equations

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in this problem. We're told that T at T is the temperature at time t in minutes after team minutes, it's been removed from an up a room. We'd like to know in this problem what the temperature will be 20 minutes later. And for this problem, using the formula for Newton's law of cooling temperature at Time T is equal to the surrounding temperature. T sabb em my a c times e to the negative. Katie. So what we need to do next is hold off on evaluating t 20 and first go through all the initial conditions that have been given. Our objective is to determine what the constant Tisa BM is C as well as K or to determine Tisa. Bam. We can just do an immediate substitution that t sabb em is equal to 10 F will obtain temperature at Time T is equal to 10 minus See Time's each of the negative. Katie, the next constant that we'd like to determine is see itself. We can do that from the very first initial condition. The starting temperature or th zero is 70 degrees. Let's use this last differential equation solution that we obtained to make that substitution. Well, right. T s zero is equal to first. It's 10 minus C times e to the negative K time zero. But we also know that each of the zero result in one. So this factor cancels out immediately. The next substitution will make is that TF zero itself is 70 degrees. So now we have an equation. 70 equals 10 minus c. Subtracting 10 from both sides obtains 60 is equal to negative c So negative 60 degrees itself is that constant? C Let's and put that back into the differential equation. Now we have two negatives to deal with. So temperature at time T will be 10 with a positive 60 times E to the negative. Katie, we have one last differential equation. Initial condition to consider TF 40 is 30 degrees or the temperature 40 minutes later decreased to 30 degrees, and that could be used to solve four K. So as we did before, let's use this last step of this equation. We've been building a substitute in 40 for Time T. If we do that, then T at 40 will be equal to 10 plus 60 times e to the negative of 40 times k. Ah Further substitution tells us that T A 40 itself is 30 degrees, so that equation becomes 30 equals 10 plus 60 times each of the power negative 40 k We'll solve this equation for K to do that vs subtract 10 from both sides we obtain 20 is 60 times each of the negative 40 k If we divide 60 into both sides will get to six after reduction or 1/3. So now we have 1/3 equals e to the power of negative 40 k and this can be solved for K by first taking the natural logger them of both sides. We obtained that natural auger them of 1/3 is equal to negative 40 k So finally, after dividing both sides by negative 40 que is now equal to negative 1/40 times the natural log of 1/3 and that will be substituted back into this portion off that differential equation solution. We just have to be mindful of the double negatives. So our next step is that temperature at Time T is equal to 10 plus 60 times the base exponential to the power of 1 48th times the natural log of 1/3 times t. Finally, we can address what is the temperature 20 minutes after this up, object has been removed from the room we can write Temperature at 20 minutes later is equal to buy substitution from the last step of our equation. 10 plus 60 each of the power of 1/40 times the natural log of 1/3 times 20 for a place of tea. If we put this all into, a calculator will obtain that this is approximately equal to 44.64 degrees, and then what this tells us is it takes less than 20 minutes for the temperature to decrease to 50 F.

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