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Thread: Differentiate ε = 4x10^−5t^4−0.0024t^3+0.0487t^2−0.3428t+1.4532

  1. #1

    Differentiate ε = 4x10^−5t^4−0.0024t^3+0.0487t^2−0.3428t+1.4532

    Hi guys,

    The following equation was derived from a theoretical test, where t is equal to time:

    ε = 4x10^−5t^4−0.0024t^3+0.0487t^2−0.3428t+1.4532

    I was told I need to differentiate to dε/dt

    I have got the following as my answer:

    (4t^3−180t^2+2435t−8570) / 25000

    Does that look correct?

    I was then asked what the answer would be if the time was 15 hours, which I would then put in to the above equation - if it is correct?



    Thanks for any pointer.
    Last edited by mcarthyryan; 08-09-2017 at 04:33 PM.

  2. #2
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    Well, I'm assuming the superscript exponents didn't carry over from when you copied-and-pasted, and you actually meant:

    [tex]\epsilon = 4 \cdot 10 - 5t^4 - 0.0024t^3 + 0.0487t^2 - 0.3428t + 1.4532[/tex]

    [tex]\dfrac{d\epsilon}{dt} = \dfrac{4t^3 - 180t^2 + 2435t - 8570}{25000}[/tex]

    If that's the case, your derivation is almost correct, except for the 4t3 term. Recall the power rule tells us that [tex]\dfrac{d}{dt} \left( t^\alpha \right) = \alpha \cdot t^{\alpha - 1}[/tex] and the constant multiple rule tells us that [tex]\dfrac{d}{dt} \left( C \cdot f(t) \right) = C \cdot \dfrac{d}{dt} \left( f(t) \right)[/tex]. Combining these two rules, what does that make [tex]\dfrac{d}{dt} \left( 5t^4 \right)[/tex]? (Hint: It's not 4t3/25000 )

    As for your second query, assuming that the declaration of the variable was something like "t is the time, measured in hours, since...", then yes, you're trying to find the derivative at the point t = 15. Since the derivative of a function is itself a function, you can simply plug in the desired value and evaluate.

  3. #3
    Hi, Thank you for the reply. Yes, I don't think the equation copied properly. The original equation was:

    . . .[tex]\epsilon\, =\, 4\cdot 10^{-5}\,t^4\, -\, 0.0024\, t^3\, +\, 0.0487\, t^2\, -\, 0.3428\, t\, +\, 1.4532[/tex]

    My derivative was:

    . . .[tex]\dfrac{d\epsilon}{dt}\, =\, \dfrac{4t^3\, -\, 180t^2\, +\, 2435t\, -\, 8570}{25000}[/tex]

    Hopefully this shows OK for what I was trying to explain?
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    Last edited by stapel; 08-09-2017 at 05:27 PM. Reason: Typing out the text in the graphics.

  4. #4
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    Ah, okay. Armed with this correction, your derivative is absolutely correct. What did you get for the next step, when you plug in t = 15?

  5. #5
    Quote Originally Posted by ksdhart2 View Post
    Ah, okay. Armed with this correction, your derivative is absolutely correct. What did you get for the next step, when you plug in t = 15?
    Thanks for that

    I got 0.0382 as my answer.

  6. #6
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    That "/25000" form is a bit misleading. It gives an aura of exactness that doesn't actually exist. If you start with four decimal places, you probably should stay with the decimal places, rather than converting to a cleaner-looking picture.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  7. #7
    Quote Originally Posted by tkhunny View Post
    That "/25000" form is a bit misleading. It gives an aura of exactness that doesn't actually exist. If you start with four decimal places, you probably should stay with the decimal places, rather than converting to a cleaner-looking picture.
    Hi tkhunny,

    Sorry for the late reply.

    If I didn't include the /25000, what would be the best way to present the equation? Apologies, I don't understand.



    Thank you
    Ryan

  8. #8
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    Quote Originally Posted by mcarthyryan View Post
    Hi guys,

    The following equation was derived from a theoretical test, where t is equal to time (in what unit? seconds? minutes? hours? days?.:

    ε = 4x10^−5t^4−0.0024t^3+0.0487t^2−0.3428t+1.4532

    I was told I need to differentiate to dε/dt

    I have got the following as my answer:

    (4t^3−180t^2+2435t−8570) / 25000

    Does that look correct?

    I was then asked what the answer would be if the time was 15 hours, which I would then put in to the above equation - if it is correct? (answer to this question will depend on answer to that question)



    Thanks for any pointer.
    .
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  9. #9
    Quote Originally Posted by Subhotosh Khan View Post
    .
    Hi,

    in hours.

  10. #10
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by mcarthyryan View Post
    If I didn't include the /25000, what would be the best way to present the equation?
    I'm not sure there's an issue, with your form.

    If we change all decimal numbers to their Rational form, and we combine everything into a single ratio, then 25000 is the common denominator.

    Perhaps, when rounding your final result (when t=15), you don't have enough precision from the givens to justify reporting to four decimal places (see Significant Figures). You'll need to inquire, whether that's a concern.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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