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mathcalculus Group Title

HELP :( LINEARIZATION: Please use h in place of \Delta x and dx I. If y= f(x)= 8 x^2+ 31 x + 12 find a. dy =16xh+31 b. \Delta y = 8*(x+h)^2+31*(x+h)-8x^2-31x II. Evaluate each of these quantities if x=10 and h=0.04 a. dy= 37.4 b. \Delta y = 7.6528

  • one year ago
  • one year ago

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  1. mathcalculus Group Title
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    why are my answers wrong? :(

    • one year ago
  2. phi Group Title
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    for part a) you should have dy =16xh+31h

    • one year ago
  3. mathcalculus Group Title
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    why have 31 h?

    • one year ago
  4. phi Group Title
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    if you start with y= x d (y = x) is dy = dx or using h instead of dx dy = h for your problem \[ d (y = 8 x^2+ 31 x + 12 )\\ dy = 8 dx^2 + 31 dx \\ dy = 16x dx +31 dx \\ dy = 16x \ h + 31 \ h \]

    • one year ago
  5. mathcalculus Group Title
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    i thought we leave the constant alone?

    • one year ago
  6. mathcalculus Group Title
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    oh because it was an x because finding the derivative?

    • one year ago
  7. phi Group Title
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    compared with \[ d (y = 31 x ) \\ dy = 31\ dx \]

    • one year ago
  8. mathcalculus Group Title
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    yes it ='s to 1

    • one year ago
  9. phi Group Title
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    \[ \frac{d}{dx} (y = 31 x) \\ \frac{d}{dx}y = 31 \frac{d}{dx}x\\ \frac{dy}{dx}= 31\frac{dx}{dx}\\ \frac{dy}{dx}= 31\cdot 1=31 \]

    • one year ago
  10. mathcalculus Group Title
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    right.

    • one year ago
  11. mathcalculus Group Title
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    but i was confused why i couldn't leave 31 alone

    • one year ago
  12. mathcalculus Group Title
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    and why i had to add h in the end..

    • one year ago
  13. phi Group Title
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    maybe a better way to think of it is \[ \Delta y = \frac{dy}{dx} \Delta x \] where dy/dx is the derivative of your function \[ \frac{dy}{dx}= 16x + 31\] and \[ \Delta y = \frac{dy}{dx} \Delta x =(16x + 31)\Delta x= (16x + 31)h \]

    • one year ago
  14. mathcalculus Group Title
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    oh okay, got cha.

    • one year ago
  15. mathcalculus Group Title
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    what if it had like an x and a y inside

    • one year ago
  16. mathcalculus Group Title
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    could i still apply this?

    • one year ago
  17. phi Group Title
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    it would be the same idea, but you use "partial derivatives"

    • one year ago
  18. mathcalculus Group Title
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    ex: dy/dx

    • one year ago
  19. mathcalculus Group Title
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    so (d/dx+dy/dx)h ?

    • one year ago
  20. phi Group Title
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    do you have a specific question?

    • one year ago
  21. phi Group Title
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    it is a new topic to talk about functions of two variables f(x,y) = x^2 + y^2 for example you won't see those until calculus II

    • one year ago
  22. mathcalculus Group Title
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    well i know that we have to do this: 2x*y(dy/dx) +2y*(d/dx).. right?

    • one year ago
  23. phi Group Title
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    are you asking about f(x,y) = x^2 + y^2 or z = x^2 + y^2 ? we find the partial derivative with respect to x (that means treat y as a constant) fx (means partial derivative with respect to x) = 2x fy = 2y we then say \[ \Delta z = 2x \Delta x + 2y \Delta y \]

    • one year ago
  24. phi Group Title
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    But you won't see those type of problems.

    • one year ago
  25. mathcalculus Group Title
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    oh okay, thank you so much! :)

    • one year ago
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