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Math Help - Implicit Differentiation

  1. #1
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    Red face Implicit Differentiation

    Hi

    Could someone please show me how to do question 12?
    Attached Thumbnails Attached Thumbnails Implicit Differentiation-scan.jpg  
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    Hi

    Could someone please show me how to do question 12?
    Dear xwrathbringerx,

    By differentiation with respect to x,

    3x^2+3y^{2}\frac{dy}{dx}=3\left[y+x\frac{dy}{dx}\right]

    [3y^{2}-3x]\frac{dy}{dx}=3y-3x^2

    [y^2-x]\frac{dy}{dx}=y-x^2

    \frac{dy}{dx}=\frac{y-x^2}{y^2-x}

    Can you do it from here?? If you need more help don't hesitate to ask.

    Hope this helps you.
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  3. #3
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    Hmm how exactly do I know that the tangents are horizontal and vertical?

    Also could you please show me how to do (b) and (c)...I have no idea how to do them
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  4. #4
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    A horizontal line has slope 0, a vertical line does not have a slope.

    For (b) you are to show that \frac{1+ (y/x)^3}{y/x}= 3/x.

    Starting from the original equation, x^3+ y^3= 3xy, divide both sides by x^3 to get 1+ (y/x)^3= 3y/x^2= 3(y/x)/x. Now multiply both sides by y/x.

    For (c) you need to know that x^3+ y^3= (x+ y)(x^2- xy+ y^2)= (x+y)(xy)(x/y- 1+ y/x).
    Last edited by mr fantastic; April 14th 2010 at 12:44 AM. Reason: Fixed latex error.
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  5. #5
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    hmmm for (b), when x = infinity, shouldnt y/x = infinity too? (since y/x = x/3 + y^2/3x^2?)
    Last edited by xwrathbringerx; April 3rd 2010 at 05:26 AM.
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  6. #6
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    Quote Originally Posted by xwrathbringerx View Post
    hmmm for (b), when x = infinity, shouldnt y/x = infinity too? (since y/x = x/3 + y^2/3x^2?)
    Dear xwrathbringerx,

    How did you get, \frac{y}{x}=\frac{x}{3}+\frac{y^2}{3x^2}~??

    This is obviously a wrong equation since, when multiplied by 3x^2 you get, 3xy=x^3+y^2 which contradicts the original equation.
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  7. #7
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    the 1st part of (b) says "show ...". I cross multiplied this to obtain it with y/x =
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  8. #8
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    Quote Originally Posted by xwrathbringerx View Post
    the 1st part of (b) says "show ...". I cross multiplied this to obtain it with y/x =
    Dear xwrathbringerx,

    Seems your cross multiplication is wrong. If you cross multiply you should obtain, \frac{y}{x}=\frac{x}{3}+\frac{y^3}{3x^2}
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  9. #9
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    Exclamation

    Ooops

    Bur Sudharaka, doesnt that still give infinity?
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  10. #10
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    Quote Originally Posted by xwrathbringerx View Post
    Ooops

    Bur Sudharaka, doesnt that still give infinity?
    Dear xwrathbringer,

    Suppose when x\rightarrow{\infty}\Rightarrow{y\rightarrow{-\infty}}. Therefore generally we cannot say that \frac{y}{x}\rightarrow{\infty} since we don't know what happens to y when x goes to infinity. Does this solve your problem??
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  11. #11
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    OH

    But I dont get it.Why did the question say y/x --> -1?
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  12. #12
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    Quote Originally Posted by xwrathbringerx View Post
    OH

    But I dont get it.Why did the question say y/x --> -1?
    Now consider, \frac{1+\left(\frac{y}{x}\right)^{3}}{\frac{y}{x}}  =\frac{3}{x}

    When, \frac{y}{x}\rightarrow{-1}\Rightarrow{}\frac{1+\left(\frac{y}{x}\right)^{3  }}{\frac{y}{x}}\rightarrow{0}\Rightarrow{\frac{3}{  x}\rightarrow{0}\Rightarrow{x\rightarrow{\infty}}}

    Get the idea??? If you have any questions please don't hesitate to ask. I'll try my best to answer them.
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