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Homework Help: Calculus: answer check please

  1. Mar 8, 2005 #1
    Just wondering if anyone here could check to see if I did this question right.

    An astroid has the equation [tex] x^\frac{2}{3} + y^\frac{2}{3}=1 [/tex]. The equation defines two continuous functions y=y(x) in the interval [tex] -1\leq x \leq 1 [/tex]

    a) Solve the equation for y to obtain formulas for the functions.
    b) Give formulas for the first derivatives of the functions for x cannot equal 0, ±1 by differentiating the equation of the curve implicitly and then substituting in the formulas for y from part (a).
    c) Compute the same derivatives by differentiating the formulas from part (a)

    a) [tex] y= (1-x^\frac{2}{3})^\frac{3}{2}[/tex]
    [tex] =\pm \sqrt{(1-x^\frac{2}{3})^3} [/tex]

    b) [tex] y'= \frac {-\sqrt[3]{y}}{\sqrt[3]{x}} [/tex]
    substituting y:

    c) [tex] y= (1-x^\frac{2}{3})^\frac{3}{2}) [/tex]
    [tex] y'= \frac {3}{2} (1-x^\frac{2}{3})^\frac{1}{2} (\frac{-2}{3}x^\frac{-1}{3}) [/tex]
    [tex] = \frac{-1}{\sqrt[3]{x}} (1-x^\frac{2}{3})^\frac{1}{2} [/tex]

    Thank you all in advance.

    Edit: Sorry if my work is incomplete, it's not letting me post the rest for some reason.
    Last edited: Mar 8, 2005
  2. jcsd
  3. Mar 8, 2005 #2


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    (a) looks good to me.

    (c) looks good, that part of it which I can see.

    I can't really follow (b). Unfortunately, not all of your equations turned out.

    Nice work though...

    Edit: I think you're in the middle of fixing it.
  4. Mar 8, 2005 #3
    Yes,for some reason it's not all showing up. Anyway, my final answer for b and c turned out to be:

    ± (1-x^(2/3))^(1/2) / x^(1/3)

    I'm hoping this is the right answer....
    Last edited: Mar 8, 2005
  5. Mar 8, 2005 #4


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    I'm pretty confident because, like I said, part (c) looks on track. If you got the same answer using two different methods, you should be pretty confident too. :smile:

    But why is there a +/- in front?

    Isn't this :

    [tex] = \frac{-1}{\sqrt[3]{x}} (1-x^\frac{2}{3})^\frac{1}{2} [/tex]

    (which I stole from your post) the final answer?
    Last edited: Mar 8, 2005
  6. Mar 8, 2005 #5
    It said formulas for part a so solving for y:

    [tex]\pm \sqrt{(1-x^\frac{2}{3})^3} [/tex]

    So calculating the derivative for the positive equation and the negative equation, wouldn't it give me ± (1-x^(2/3))^(1/2) / x^(1/3) ? Or perhaps I'm wrong?
  7. Mar 8, 2005 #6


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    Yeah, I think you're right. Well...sort of. I think that if you express y(x) entirely in terms of powers, then there is only one formula for y(x) (but it defines two functions). If you then differentiate, the expression is still in terms of powers, and again there is only one formula. Then if you express all your powers to the 1/2 as radicals, you get two separate expressions, one positive, one negative. But I guess what I'm trying to say is that:

    y = x^1/2 defines both possible functions, and it's derivative:

    y' = 1/2(x)^(-1/2) defines both derivatives (the derivative of each one).

    You only see explicitly that the expression allows for two possible functions when you write it in terms of square roots...then you get your +/-'s in front. At least I think so.

    So you could express your final answer either as:

    [tex] \frac{dy}{dx} = - \frac{1}{\sqrt[3]{x}} (1-x^\frac{2}{3})^\frac{1}{2} [/tex]

    or (equivalently) as

    [tex] = \frac{\mp \sqrt{1-x^\frac{2}{3}} }{\sqrt[3]{x}} [/tex]

    I hope that's right. somebody tell me if I'm way off base here...
    Last edited: Mar 8, 2005
  8. Mar 9, 2005 #7
    Thanks for the help cepheid :smile:
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