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For the following problems, set up and solve the differential equations.You boil water to make tea. When you pour the water into your teapot, the temperature is $100^{\circ} \mathrm{C}$ . After 5 minutes in your $15^{\circ} \mathrm{C}$ room, the temperature of the tea is $85^{\circ} \mathrm{C}$ . Solve the equation to determine the temperatures of the tea at time $t .$ How long must you wait until the tea is at a drinkable temperature $\left(72^{\circ} \mathrm{C}\right) ?$

$T(t)=15+85 e^{-k t}$$t=10.3$ minutes

Calculus 2 / BC

Chapter 4

Introduction to Differential Equations

Section 5

First-order Linear Equations

Differential Equations

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work with a Newton's law of cooling problem, where we have a liquid at 90 degree Celsius currently in a room as an ambient temperature of 25 year resells. Yes, we also know, after a minute it's 85 degrees Celsius, and I want to know when will. That would be 30 degree Celsius. So a lot of these problems there are long derivations to get where we want to get. But there's also that derivation designed to do you a quick kind of result. And as you can see in your text, they work that their vision out and a nice little shortcut toe or condensed version of the Newton's Law of cooling. It looks like this temperature with respect to T or in terms of time in minutes is equal to some constant. That's not Celsius, by the way Times E raised to the negative. Katie, where K is another constant plus your ambient temperature. Now, this is a This is a decreasing exponential function because the negative Katie and we know is gonna be that horizontal last month. So it should make sense, by the way, that we have essentially as one toe at a in our liquid doing something along the lines of this in terms of temperature cooling and approaching that value. So, like in a lot of other problems that have differentials involved, which this does in the background, um, way could kind of go ahead and start solving for a lot of our constants, given what we know. So I'm gonna first use our initial value of TV. Here was 90 under ambient temperature, and we're going to go and find a seat once we have to see, we're gonna find K, and then we'll have our general equation that we can go forward to figure out when will the liquid end of being 30 degrees Celsius. So I'm gonna go ahead right? 90 Over here, ID equals C times the race to the zero there zero hour because you're going Negative K, which you don't know times zero plus 25. That's nice, cousin. And right now we just get 65 equals C. So now we know that our see value is 65 degrees Celsius. Now, what I'll do is I'll use my second initial value, and all right, 85 equals 65. Since we know that now, at times e to the negative K pits in the negative K times one since we're looking at one minute later plus 25. So I tracked 25. Get 60 by Divine by 65 I get 12. 13th equals e to the negative K. Once I go ahead and natural log both sides to get rid of our exponential, we get the natural log of 12. 13th equals negative case. How? But I say negative natural longer. 12 13th. Okay, And this is gonna be a roughly 0.8 So what I have to do is in writing out my general equation is I like to write it as 65 e now since isn't raised to the negative, K T and R K happens to be negative. That's a negative. Negative. So we're gonna end up just having the natural log of 12 13th T and then plus 25. Okay, Some people like to run like this. Okay, which is fine, as long as you recognize that point away is a rounded value. And if you're gonna do any further calculations you're not using point away, you're using the exact value. Your calculator. So now we're gonna say, Hey, what is this thing you called? A 30 degree Celsius? I'm gonna write like this natural log of 12 13th T plus 25. I must attract 25. That's gonna be a six or five. Apartment five, divided by 65 1/13. Similar Bulat with 1 13 equals E to the natural log 12 13th t I will have to natural log again. No, that the's not cancel out because of this tea here she left with the natural log of 1 13 equals the natural log of 12 13th tee. So the very final step would be to take the natural log of 1/13 divided by the natural law of 12 13th notice have used any round that valued yet. And you get the value of 32.4 and so forth. So it may be your time is gonna be roughly around 32 minutes for that quickly to cool to 30 degrees

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