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When a cold drink is taken from a refrigerator, its temperature is $5^{\circ} \mathrm{C}$ . After 25 minutes in a $20^{\circ} \mathrm{C}$ room its temperature has increased to $10^{\circ} \mathrm{C}$ .(a) What is the temperature of the drink after 50 minutes?(b) When will its temperature be $15^{\circ} \mathrm{C} ?$

a) $$13 . \overline{3}^{\circ} \mathrm{C}$$b) $$67.74~minutes$$

Calculus 1 / AB

Chapter 3

INVERSE FUNCTIONS

Section 4

Exponential Growth and Decay

Derivatives

Differentiation

Applications of the Derivative

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Hey, guys, welcome back. It's probably another Newton's law of cooling equation problem. We'll use our equation Changing temperature over change in time. Physical tok temperature of the object Mice, temperature, surroundings. We know the temperature of the surroundings are 20 degrees Celsius. Yes, we can relate this to say, change in temperature over change in time. Okay. T minus 20. We also can then say we want to let why equal toe t minus 20. And if you differentiate both sides of that equation, you get that. Why? Prime is simply equal Teoh t prime. Then if we were to substitute into this equation right here said every and e t weaken right d y the wise equal to DT. Yeah, change. And why over the change in time is equal to K times. Why now? That's more like what we're used to seeing this exceptional growth chapter, we can integrate both sides and we're left with fact that why little to the initial y value times he raised the Katie. We know that initially, at times equal to zero temperature of the drink is five degrees. So at times equal to zero Capital T is equal to five degrees C. I would say that why is equal to T minus 20 is equal to five minus 20 the negative 15. So we could say that why zero is night of 15. It will be our initial condition. We can then also say that after 25 minutes, temperature of the drink is 10 degrees Celsius, so at time is equal to 25 minutes. Temperature of the drink is now 10 degrees Celsius. Still y Z T minus 20 equal to 10 minus 20 native 10. We can say that why? Of 25 equal to native 10. Now we can plug both of those values and this equation and Saul for K. So, you know, wise native 10. You know that our initial Y value is negative. 15. He raised okay and our T valued at the negative 10 was 25 minutes. We can divide both sides by get a 15 left with positive. 2/3 is equal to you to the 25 k We want to take the natural longer. Both sides were left with natural log 2/3 equal to 25 k Okay. Is able to Ellen 2/3 over 25 And if you do them algebra on that, that comes out to roughly negative 0.0 1 16 Center equation becomes Why sickle to our initial value, we said, was negative. 15 if he raised the negative 0.162 T and the first asked orders temperature of the drink after 50 minutes. So what is the temperature after 50 minutes? Yes, we can plug in 50 for D. That why is equal to negative 15 erased the negative point 0162 multiplied by 50. And when you that algebra we get a Y value of native six point 667 But that is not a temperature. Remember, we still have this equation over here where we let wise to capital T minus 20 with wise you'll to capital t minus 20 or tea? Well, two y plus 20 the native six, 667 plus 20. We had 22 that I'm gonna value of 13.13. So after 50 minutes, temperature after 50 minutes, see below 13 point 33 degrees. Our second part of the problem rest to figure out when the temperature becomes 15 degrees. Sorry, I don't wanna temporarily 15 degrees we're looking for. When is capital t equal to 15 degrees? Well, we know that why is able to t minus 20? He told a 15 minus 20 negative five. So we're trying to figure out when why he was thinking of five so we can plug in negative five into our equation. Negative five is equal to negative 15 times e raised a negative 150.0 162 times t now divide both sides by negative 15. We're left with positive 1/3 equal to E based negative 0.162 We're not even after a long of both sides. We're left with the natural log of 1/3 is equal to negative 0.16 to t and this T is equal to the natural log of 1/3 divided by a negative 0.162 And when you that equation do the algebra, you're left with a value 67 0.74 So thus it takes about 67 minutes just a little over for the drink to reach 15 degrees Celsius. And that concludes this problem. Thanks for watching

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