Change of axis in case of Horizontal tangent

Discussion in 'Homework Help' started by zulfi100, Dec 14, 2013.

  1. zulfi100

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    Jun 7, 2012
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  2. studiot

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    It tells you the answer in 8.36, just before 8.37.

    What does it say about dy/dx in 8.36?
     
  3. zulfi100

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    Jun 7, 2012
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    Hi,
    Thanks for your attention. Plz explain me how you say that answer is in 8.36. In my view answer is in the statement:
    which is stated in 8.37.
    Plz guide me about your understanding.

    Zulfi.
     
  4. studiot

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    Well I did ask you a very specific question about 8.36.

    What is your answer?
     
  5. zulfi100

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    Jun 7, 2012
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    Hi,
    Thanks for your consistency in my problem. I didnt consider it significant. It says that dy/dx does not exist at origin. Actually i didnt consider that 8.36 is related to my prob. However i have answered you, now plz explain me why you think its related to answer of my question being asked in this post.

    Zulfi.
     
  6. studiot

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    The whole section of your text is about calculating the length of a curve between some point where x=a to some point where x=b

    On page 417 of your text it defines the length, L, to be

    L = \int_a^b {\sqrt {\left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)} } dx

    It says this requires a 'smooth function of x'

    On page 418 there is a simple worked example using this definition (example2).


    Page 418 then goes on to discuss what happens when the function is not smooth ("vertical tangents, corners and cusps")

    and in example 3 shows what happens when dy/dx does not exist because the tangent is vertical so cannot be input to the formula.

    You need to understand that a vertical tangent means that dy/dx is infinite so cannot be used in a normal formula.

    It shows that one way round this problem (you should check that you really cannot input any values for dy/dx into the formulae for example 2) is to use the second formula they state. That is rearranging the equation for the curve so instead of y in terms of x they use x in terms of y, for the same curve.

    So we rotate the curve through 90 degrees and a vertical tangent become a horizontal one dx/dy which does exist and equals 0.
    This can be input to the second formula the gives you for arc length on page 417, namely


    L = \int_c^d {\sqrt {\left( {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} \right)} } dy

    Does this help?

    Don't forget that the point x=a is now the point y=c and the point x=b is now the popint y=d. That is you need new limits of integration for the same length of curve.
     
    Last edited: Dec 16, 2013
  7. zulfi100

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    Jun 7, 2012
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    Hi,
    Thanks for your answer. You mean that we have a function y=x to the power 1/3. It represents a vertical tangent. If we find dy/dx then it would be
    1/(3x to the power 2/3). At origin x= 0 so it would be infinity. In order to avoid this we do a 90 degree rotation so instead of y=x to the power 1/3 , we have
    x= y3 (horizontal tangent) & now we will find dx/dy which would be 3y2 so it would be zero at origin.

    Thanks again for your help.

    Zulfi.
     
  8. studiot

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    Yup , you got it.

    :):D
     
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