anonymous
  • anonymous
A freshly brewed cup of coffee has temperature 95°C in a 20°C room. When its temperature is 67°C, it is cooling at a rate of 1°C per minute. When does this occur? Approximate to three decimal places.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
let's say the temperature at time t is C(t). at t = 0, C(t) = 95 the rate of cooling dC(t)/dt is proportional to the difference in temperature between the coffee and the room. so, dC(t)/dt is proportional to C(t)-20 at C(t) = 67, dC(t)/dt = 1 = k*67-20or, k = 1/47
anonymous
  • anonymous
why is everyone asking questions in the wrong sections? is it casue no one checks the others?
anonymous
  • anonymous
since it is cooling down, k should be negative, i.e k is -1/47

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anonymous
  • anonymous
the final answer the i got was 23.37 but it says its wrong
anonymous
  • anonymous
how did you get your answer?
anonymous
  • anonymous
i started out subtracting 67-20
anonymous
  • anonymous
then i got -1/47 like u did
anonymous
  • anonymous
then?
dumbcow
  • dumbcow
another way of looking at it is; f(0) = 95 f(t) = 67 f'(t) =-1 *the rate of change at time t using the definition of the derivative f'(t) =f(t) -f(0)/t -1 = (67-95)/t solve for t
dumbcow
  • dumbcow
another way of looking at it is; f(0) = 95 f(t) = 67 f'(t) =-1 *the rate of change at time t using the definition of the derivative f'(t) =f(t) -f(0)/t -1 = (67-95)/t solve for t
anonymous
  • anonymous
ok my work is all over the place..but i ended up with t=-50ln(47/75)
anonymous
  • anonymous
i got my answer from 67=20+75e^(-t/50)
anonymous
  • anonymous
you need to find the rate of change of the rate of change. I am pretty sure a double derivative is needed here, but my calculus is a bit rusty.
anonymous
  • anonymous
its ok ill just ask my teacher about it
anonymous
  • anonymous
ok
dumbcow
  • dumbcow
i think you're making it more complicated than it needs to be. t=28
anonymous
  • anonymous
how did you get that?
dumbcow
  • dumbcow
look at my previous post
anonymous
  • anonymous
it said that answer is wrong
anonymous
  • anonymous
i tried submitting that in my h.w and it didnt work
dumbcow
  • dumbcow
oh ok sorry i guess i was wrong
anonymous
  • anonymous
lol its ok
anonymous
  • anonymous
dumbcow your answer is wrong. The rate of cooling is higher initially and it slows down as the temperature of the coffee approaches the room temperature. Thats why I said you need to find the rate of change of the rate of change.
dumbcow
  • dumbcow
ah gotcha thank you. i was assuming the cooling was constant. sorry
anonymous
  • anonymous
you are welcome.
dumbcow
  • dumbcow
hey if you're still on i think i figured it out. check my answer t = 21.914 min
dumbcow
  • dumbcow
hey if you're still on i think i figured it out. check my answer t = 21.914 min
dumbcow
  • dumbcow
anyway here is my solution: assume a decay function starting at 95 and going down and approaching 20 over time f(x) = a*b^x + 20 f(0) = a+20=95 => a = 75 f(x) = 75b^x+20 = 67 => b^x = 47/75 f'(x) = 75ln(b)b^x = -1 => ln(b)(75)(47/75)=-1 =>ln(b) = -1/47 thus b = e^(-1/47) so b^x = e^(-x/47) = 47/75 take ln both sides -x/47 = ln(47/75) => x = -47*ln(47/75) = 21.965
dumbcow
  • dumbcow
anyway here is my solution: assume a decay function starting at 95 and going down and approaching 20 over time f(x) = a*b^x + 20 f(0) = a+20=95 => a = 75 f(x) = 75b^x+20 = 67 => b^x = 47/75 f'(x) = 75ln(b)b^x = -1 => ln(b)(75)(47/75)=-1 =>ln(b) = -1/47 thus b = e^(-1/47) so b^x = e^(-x/47) = 47/75 take ln both sides -x/47 = ln(47/75) => x = -47*ln(47/75) = 21.965

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