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Simple partial derivative question

  1. Jun 11, 2013 #1
    Hello, just a quick question about interpreting the partial derivative as a rate of change.

    My example is the area of a parallelogram: A = absinθ, with a and b being the adjecent sides with θ being the angle between them.

    We found the rate of change of the area A with respect to the side a by holding b and θ constant:

    (for a = 10, b = 20, θ = ∏/6)

    ∂A/∂a = bsinθ

    ∂A/∂a = 20sin(∏/6) = 10.

    Does this mean for every unit increment of a, A increases 10 times that, ie. increasing a by 2 increases A by 20?

    Thanks for reading.

    Lee
     
  2. jcsd
  3. Jun 11, 2013 #2

    SteamKing

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    You can check this with the example in the OP. If a = 11, b = 20 and theta = pi/6, what is the area of your P-gram?
     
  4. Jun 11, 2013 #3

    HallsofIvy

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    Here, your original function, [itex]A= ab sin(\theta)[/itex], is linear in both variables so, yes, you can think of the partial derivative as a "slope"- if a or b doubles, A also doubles.

    If it were not linear, that would not be true. Take the example of a rectanguar solid with square cross sections- two perpendicular edges of length x and the third of length y. The volume is [itex]V= x^2y[/itex] so the two partial derivatives are [itex]V_x= 2xy[/itex] and [itex]V_y= 2x^2[/itex]. If x= 2, y= 2 then V=(4)(2)= 8. If x= 2, y= 4 (so x remains the same while y doubles), V= (4)(4)= 16. Yes, the volume has doubled. But if x= 4, y= 2 (so y remains the same while x doubles), V= (16)(2)= 32. The volume has been multiplied by [itex]2^2= 4[/itex].
     
  5. Jun 11, 2013 #4
    Thanks for the replies.... can't believe I didn't think to just plug in some numbers and see it for myself.

    Lee
     
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