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A glass of water whose temperature is $50^{\circ} \mathrm{F}$ is taken outside at noon on a day whose temperature is constant at $70^{\circ} \mathrm{F}$. If the water's temperature is $55^{\circ} \mathrm{F}$ at $2 \mathrm{p} . \mathrm{m} .,$ do you expect the water's temperature to reach $60^{\circ} \mathrm{F}$ before $4 \mathrm{p} . \mathrm{m}$. or after $4 \mathrm{p} . \mathrm{m} . ?$ Use Newton's law of cooling to explain your answer.

after $4 \mathrm{pm}$

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 dealing with a glass of water and t at lower case T indicates the temperature of that water after noon. 12 PM were given that tia zero so directly at 12 PM the water starts out as 50 degrees. Two hours later, the temperatures increased to 55 degrees and the surrounding temperature is a nice 70 degrees itself. What we like to do is determine the solution and its form for this differential equation. They want to term decide whether the temperature will reach 60 degrees before or after four o'clock. So let's start by writing down the solution to this differential equation, we have t at Time T is equal to t sebum the surrounding temperature, plus a constant see times e to the power of negative Katie. First, we're given that T sebum is 70 degrees. So start by doing this is a direct substitution here. So the next form of our solution to the differential equation is that temperature of Time T is equal to 70 degrees plus the constant C times Either the naked of Katie. The next step in this process will be determined. The value of that constant C Let's look at the information we've used so far. We've used that the surrounding temperature of 70 degrees and we have to initial conditions. Let's start with this initial conditions here it says T s zero is 55 degrees, but we sub substitute zero into the differential equation we have so far we would obtain 70 degrees, plus the constant C times e to the power of negative K time. Zero. Then another Substitution tells us that temperature at time zero is 50 degrees itself. So we have the equation. 50 degrees is equal to 70 degrees, plus the constant C, since each of the power of zero invokes a cancellation immediately, then subtracting 70 degrees from both sides we know have negative 20 degrees. Is that constant? C? Let's input that back into our solution on the left. Now the equation reads temperature at Time T is 70 degrees minus 20 degrees times E to the power of negative Katie. The next step is to determine the value of that constant lower case K, and to do that will use the final initial condition that wasn't used temperature at time to is 55 degrees. Let's write that out on the right hand side, so temperature at time two hours from noon would be equal to 70 degrees minus 20 degrees times E to the negative K times two. Source substitution happens at the value for T. Now what we're left with is an equation that if we saw four K itself will be able to make a full substitution and know exactly what the solution would look like. So let's work on next temperature at time to is 55 degrees, which is equal to 70 degrees minus 20 degrees times each of the power of negative two K. So the next step in solving for the variable K will be to subtract 70 from both sides. The next equation is negative. 15 degrees equals negative 20 degrees each of the power of negative two K free divide bull sites. By negative 20 we would get positive. 3/4 equals e to the power of negative two K. Now let's take the natural log of both sides of this equation to obtain Natural log of 3/4 is the natural lock of the Basie exponential, but that gives us a cancellation to say the right hand site is now negative. Two k Once we divide both sides by negative, too. We obtained that that constant K is negative 1/2 times a natural log of 3/4. Let's substitute that back into our original solution to that differential equation Temperature at time T. After all, this work is equal to 70 degrees minus 20 degrees. Let's substitute next in for K, where we already have one negative sign to account for. So this will be eats the power of positive 1/2 natural log of 3/4 times T. So this is our full format for the solution to that differential equation. In a calculator, we can now check that temperature at time. Four. It's 70 degrees minus 20 degrees times E to the power of 1/2 times the natural log of 3/4 times. For if we can put all this information to a calculator, we would get 58 0.75 degrees four hours after the water has been left outside. Likewise, temperature at time five would be equal to 60.2 degrees after we make the substitution and then put the value to the calculator. And so this work tells us finally that the temperature will reach 60 degrees Fahrenheit after 4 p.m. Since we just discovered that at four PM were close to 60 degrees and we obtained just a little bit more than 60 degrees at five PM itself.

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