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Math Help - Quotient rule differentiation problem..

  1. #1
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    Quotient rule differentiation problem..

    This is the question:

    The value, V, in dollars, of an antique solid wood dining set t years after it is purchased can be modelled by the function



    Determine the rate of change of the value of the dining set after t years.

    So I think that just means find the derivative of the function?

    If so this is what I did:

    V'(t) = (18t^2)(0.002t^2 +1)^1/2 - (5500 +6t^3)(1/2)(0.002t^2 +1)^-1/2 (0.004t) / [(0.002t^2 +1)^1/2]^2

    = (18t^2)(0.002t^2 +1)^1/2 - (5500 +6t^3)(0.002t^2 +1)^-1/2 (0.002t) / (0.002t^2 +1)

    = (18t^2)(0.002t^2 +1)^1/2 - (22t +0.024t^4)(0.002t^2 +1)^-1/2 / (0.002t^2 +1)

    = (0.002t^2 +1)^-1/2 [(18t^2)(0.002t^2 +1) - (22t +0.024t^4)] / (0.002t^2 +1)

    = (0.002t^2 +1)^-1/2 [0.036t^4 +18t^2 -22t -0.024t^4) / (0.002t^2 +1)

    = (0.002t^2 +1)^-1/2 [0.06t^4 +18t^2 -22t] / (0.002t^2 +1)

    = 0.06t^4 +18t^2 -22t / (0.002t^2 +1)(0.002t^2 +1)^1/2

    = t(0.06t^3 +18t -22) / (0.002t^2 +1)^3/2

    Where did I go wrong? The answer is supposed to be:

    Last edited by kmjt; April 2nd 2010 at 12:50 PM.
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  2. #2
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    I didn't look through the whole problem, but shouldn't it be 18t^2 instead of 18t on your very first line?
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  3. #3
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    Fixed that.
    Last edited by kmjt; April 2nd 2010 at 12:50 PM. Reason: corrected
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  4. #4
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    Does anyone see an error in my differientiation? I don't see how the bottom would be (5t^2 +2500)^3/2
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  5. #5
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    Quote Originally Posted by kmjt View Post
    This is the question:

    The value, V, in dollars, of an antique solid wood dining set t years after it is purchased can be modelled by the function



    Determine the rate of change of the value of the dining set after t years.

    So I think that just means find the derivative of the function?

    If so this is what I did:

    V'(t) = (18t^2)(0.002t^2 +1)^1/2 - (5500 +6t^3)(1/2)(0.002t^2 +1)^-1/2 (0.004t) / [(0.002t^2 +1)^1/2]^2

    = (18t^2)(0.002t^2 +1)^1/2 - (5500 +6t^3)(0.002t^2 +1)^-1/2 (0.002t) / (0.002t^2 +1)

    = (18t^2)(0.002t^2 +1)^1/2 - (22t +0.024t^4)(0.002t^2 +1)^-1/2 / (0.002t^2 +1)

    = (0.002t^2 +1)^-1/2 [(18t^2)(0.002t^2 +1) - (22t +0.024t^4)] / (0.002t^2 +1)

    = (0.002t^2 +1)^-1/2 [0.036t^4 +18t^2 -22t -0.024t^4) / (0.002t^2 +1)

    = (0.002t^2 +1)^-1/2 [0.06t^4 +18t^2 -22t] / (0.002t^2 +1)

    = 0.06t^4 +18t^2 -22t / (0.002t^2 +1)(0.002t^2 +1)^1/2

    = t(0.06t^3 +18t -22) / (0.002t^2 +1)^3/2

    Where did I go wrong? The answer is supposed to be:

    = (18t^2)(0.002t^2 +1)^1/2 - (5500 +6t^3)(0.002t^2 +1)^-1/2 (0.002t) / (0.002t^2 +1)

    = (18t^2)(0.002t^2 +1)^1/2 - (22t +0.024t^4)(0.002t^2 +1)^-1/2 / (0.002t^2 +1)

    -> There appears to be an error between these 2 steps
    (5500+6t^3)(.002t)=11t+.012t^4

    -> Continuing with these values I get
    (.024t^4+18t^2-11t)/(.002t^2+1)^3/2

    -> Multiply by 125000/125000 *given the form of the desired result, let's just distribute 125 in the numerator and keep 1000 factored out
    1000(3t^4+2250t^2-1375t)/[125000(.002t^2+1)^3/2]

    ->take note 125000=2500^3/2
    so [125000(.002t^2+1)^3/2] = [2500(.002t^2+1)]^3/2
    and distribute in the denominator also factor t out of the numerator


    1000t(3t^3+2250t-1375)/(5t^2+2500)^3/2
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  6. #6
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    Ok so I worked it to:

    =[(0.036t^4 +18t^2) - (11t +0.012t^4)] / (0.002t^2+1)^3/2

    =(0.036t^4 -0.012t^4 +18t^2 -11t) / (0.002t^2+1)^3/2

    =(0.024t^4 +18t^2 -11t) / (0.002t^2+1)^3/2

    Multiply by 125000/125000 *given the form of the desired result, let's just distribute 125 in the numerator and keep 1000 factored out
    1000(3t^4+2250t^2-1375t)/[125000(.002t^2+1)^3/2]
    I'm not entirely sure why you are multiplying by 125,000/125,000. And how did you know you would have to multiply by 125,000/125,000? It just seems like a random way to simplify further. Is it really necessary to simplify this far? 125000/125000 seems like a random number to me, how do we know thats what we have to multiply by?
    Last edited by kmjt; April 3rd 2010 at 02:08 PM.
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  7. #7
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    In all honesty, I multiplied by 125000/125000 to produce the answer that you had given. It is apparently the best way to get a result that has all whole integers. I don't personally think that it is intuitive or necessary.
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  8. #8
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    I'm going to assume I multiplied wrong. What did I do wrong?
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  9. #9
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    = 125000(0.024t^4 +18t^2 -11t) / 125000(0.002t^2 +1)^3/2

    = 3000t^4 +2250000t^2 -1375000t / ???

    = 1000t(3t^3 +2250t -1365) / ???

    So I now get how the top part worked out.. but i'm lost with the bottom. do I also multiply 125000 by (0.002t^2 +1)^3/2? I'm not sure if your aloud to do that with a weird exponent like that.. and if I did how would I work it out to the bottom like the book answer?
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  10. #10
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    Quote Originally Posted by kmjt View Post
    So I now get how the top part worked out.. but i'm lost with the bottom. do I also multiply 125000 by (0.002t^2 +1)^3/2? I'm not sure if your aloud to do that with a weird exponent like that.. and if I did how would I work it out to the bottom like the book answer?
    125000 \cdot (0.002t^2+1)^{3/2}

    = \left(125000^{2/3}\right)^{3/2} \cdot (0.002t^2+1)^{3/2}

    = 2500^{3/2}  \cdot (0.002t^2+1)^{3/2}

    = \left[2500 \cdot (0.002t^2+1)\right]^{3/2}

    = (5t^2+2500)^{3/2}
    Last edited by drumist; April 3rd 2010 at 11:52 PM.
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  11. #11
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    The book solution probably was originally arrived at by factoring the problem before taking the derivative.

    \frac{5500+6t^3}{\sqrt{0.002t^2+1}}

    = \frac{5500+6t^3}{\sqrt{0.002t^2+1}} \cdot \frac{50}{50}

    = \frac{50(5500+6t^3)}{\sqrt{2500(0.002t^2+1)}}

    = \frac{275000+300t^3}{\sqrt{5t^2+2500}}

    If you perform the derivative here, you should arrive directly to the same solution as the book gave you. Keep in mind your solution is not wrong by any means (unless you were asked to make all coefficients into integers).

    You may ask why did we choose to multiply by 50? The answer is because it is the smallest number you can multiply that would make all the coefficients into integers.
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  12. #12
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    Nvm got it thanks!
    Last edited by kmjt; April 4th 2010 at 09:58 AM.
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