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Differntiating a circle

  1. Dec 14, 2008 #1
    How would I go about finding the derivative of x2 + y2 = 36.

    I know it is a circle with radius 6. Is there a better way to find the derivative then:

    y2 = 36 - x2
    y = (36 - x2)1/2
    y = 1/2(36 - x2)-1/2(-2x)
    y' = -x/(36-x2)1/2
     
  2. jcsd
  3. Dec 14, 2008 #2
    Usually with cases like this where it is inconvenient to differentiate explicitly you can use implicit differentiation.
     
  4. Dec 14, 2008 #3

    James R

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    Another way is implicit differentiation:

    [tex]x^2 + y^2 = 36[/tex]

    [tex]2x + 2y \frac{dy}{dx} = 0[/tex]

    [tex]\frac{dy}{dx} = -\frac{x}{y} = -\frac{x}{\sqrt{36 - x^2}}[/tex]

    Are you familiar with implicit differentiation?
     
  5. Dec 14, 2008 #4
    You may use implicit differentiation. If f(x) = g(x), then f'(x) = g'(x). Therefore, we can have f(x) = x2 + y(x)2 (where I have written y explicitly as a function of x) and g(x) = 36.

    Hence, we can differentiate both sides with respect to x without isolating y. Differentiating the left hand side, we get 2x + 2yy' using the chain rule. On the right hand side, differentiation 36, a constant function, just gives us 0.

    We can then solve for y': 2x + 2yy' = 0 so y' = -x/y. Notice that we have y' in terms of both x and y(x) instead of just in x; this is a hallmark of implicit differentiation. If you solve for y in terms of x and plug it in, you find that

    [tex]\frac{dy}{dx} = \frac{-x}{\pm\sqrt{36-x^2}}[/tex]​

    depending on whether y was positive or negative. This is a more complete version of the answer you found by solving for y first.
     
  6. Dec 14, 2008 #5
    Implicit differentiation. I have learned it, but obviously need to make more use of it. Thank you kind sirs.
     
  7. Dec 14, 2008 #6
    On a second note, would it be better to leave the final answer in terms of x and y. Or should I solve for y and find y' in terms of x only. I prefer x/y but if the question asks for y' how should I answer?
     
  8. Dec 15, 2008 #7

    HallsofIvy

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    1: You titled this "differentiation of a circle" which makes no sense. You cannot differentiate a geometric figure!

    2: You then wrote "find the derivative of x2 + y2 = 36" which also makes no sense. You can differentiate (both sides of) an equation but you have to specify with respect to what variable.

    3: Everyone here has assumed you really meant "find the derivative of y with respect to x, assuming that x2+ y2= 36".
     
  9. Dec 15, 2008 #8
    Sorry for the mix up. Yes I meant solve for dy/dx.
     
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