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Using dx/dy to find a y-parallel tangent

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  1. May 24, 2017 #1
    1. The problem statement, all variables and given/known data
    The curve ##C## has equation
    $$y=x^2+0.2sin(x+y)$$
    Show that ##C## has no tangent(no point where ##dy/dx=∞##), that is parallel to the y axis.

    Attempt
    $$1=2x\frac{dx}{dy}+0.2cos(x+y)(1+\frac{dx}{dy})$$

    For a tangent to be parallel to y-axis,

    $$\frac{dx}{dy}=0$$

    $$1=0.2cos(x+y)$$

    $$cos(x+y)=5$$

    No value of ##(x,y)## exist for which ##cos(x+y)>1##, hence no y-axis parallel tangent.

    **Confusion**
    Never been taught or encountered ##\frac{dx}{dy}## and thus I am having doubts over the validity of the solution.

    Can somebody please verify it?
     
  2. jcsd
  3. May 24, 2017 #2

    BvU

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    What about ##dx\over dy## if ##{dy\over dx} < Z## where you can show that Z is finite ?
     
  4. May 24, 2017 #3

    Mark44

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    Thread moved. @Faiq, in the future, please post questions about derivatives in the Calculus & Beyond section, not the Precalc section.
     
  5. May 24, 2017 #4

    Mark44

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    Why not just find dy/dx, and show that it is defined for all real x?
     
  6. May 24, 2017 #5

    BvU

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    OK, ' is finite for all finite x ' -- thanks, Mark !
     
  7. May 24, 2017 #6
    Not sure what you're telling. Can you please elaborate?
     
  8. May 24, 2017 #7
    I am just concerned whether my solution is correct or not.
     
  9. May 24, 2017 #8

    Mark44

    Staff: Mentor

    Your solution looks fine to me.
     
  10. May 24, 2017 #9

    I guess it is correct because by inverse function theorem ##D_x (y) = 1/D_y(x)##. So if ##D_y(x) \to \infty##, ## D_x(y) \to 0##.
     
  11. May 24, 2017 #10

    SammyS

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    In that case, you may need to show how it is that requiring dx/dy = 0 gets you from
    ##\displaystyle 1=2x\frac{dx}{dy}+0.2cos(x+y)(1+\frac{dx}{dy}) ##​
    to
    ##\displaystyle 1=0.2cos(x+y) ##​
    .
     
  12. May 24, 2017 #11
    Okay thanks to all of you very much.
     
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