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Differentiation rule ?

  1. Oct 7, 2004 #1
    f(t) = 400t+70/t+1 represents the temp. in an oven and f(t) is in F degrees

    first i used the quotient rule and got 330/(t+1)^2 = f'(t)

    A) the rate of change in temp of the ove with respect to the time of 2 minutes after turnig the oven on and at 10 minutes

    i pluged 2 and 10 in for t in the equation above

    330/(2+1)^2 = = 36.67 F degrees per min

    330/(10+1)^2 == 2.73 F degrees per min

    B) fined the rate of change of temp when the oven is 350 degrees

    i plugged in 350 where f(t) is

    and got t = 5 min

    thanks joe
  2. jcsd
  3. Oct 7, 2004 #2


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    Hmm. That's odd. I get a different answer for [tex]f'(t)[/tex]. You might want to check your work. (Even if I use [tex]f(t)=400t + \frac{70}{t+1}[/tex] rather than [tex]f(t)=400t + \frac{70}{t} + 1[/tex].)

    For part B:
    [tex]f(5)=400 \times 5 + \frac{70}{5} + 1=2000+12+1=2013[/tex]
    So, clearly t=5 is not the correct time.
  4. Oct 7, 2004 #3
    I guess he was saying f(t)=(400t+70)/(t+1)
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