Calculus Help

Discussion in 'Homework Help' started by davidh93, Dec 10, 2010.

Calculus Help
  1. Unread #1 - Dec 10, 2010 at 11:01 PM
  2. davidh93
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    Calculus Help

    hey guys,

    I need some help on my calculus homework. There are some problems that I don't understand:

    1. liquid is pouring into a vat. after x hours there are 5x-x^(1/2) gallons in the vat. at what instantaneous rate of change is the liquid flowing into the tank after 4 hours? what was the average rate of change in the liquid over the first four hours?

    2. find the first derivative of y=(sin2x)(2cosx)
    I cant seem to find the chart and I dont have a calc book

    3. find the first derivative of y=(lnx)/(e^x)

    4. if the distance is given by 8x^3 - (1/2)x^2 + 3x then find the acceleration at x=2

    thanks in advance :D
     
  3. Unread #2 - Dec 11, 2010 at 12:04 AM
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    Calculus Help

    1. Let y=5x-x^(1/2). y'=5-(x^(-1/2))/2. y'(4)=4+3/4
    If you really need me to, I can work out the average rate of change, but I don't feel like dropping the time to do it right now.

    2. Use the product rule here, no tables needed. y=(sin2x)(2cosx). y'=2cos2x*2cosx+sin2x*-2sinx

    3. Use quotient rule here. So f=lnx, g=e^x, f'=1/x, g'=e^x. y=(lnx)/(e^x). y'=[(e^x)/x-(e^x)*lnx]/[(e^x)^2]

    4. Take the second derivative. y''=48x-1. y''(2)=95
     
  5. Unread #3 - Dec 11, 2010 at 12:15 AM
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    Calculus Help

    Adding to video's explanation of #1, the average rate of change is this:

    f(0) = 5(0)-(0)^(1/2) = 0
    f(4) = 5(4)-(4)^(1/2) = 20-2 = 18

    Average = (f_final - f_initial)/(t_final - t_initial)
    = 18 - 0 / 4
    = 18 / 4 = 4.5 gal/hr
     
  7. Unread #4 - Dec 12, 2010 at 12:26 AM
  8. davidh93
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    Calculus Help

    :O thanks guys! helped a lot, my calculus teacher is horrible :(. prob gonna fail my final XD
     
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