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Math Help - Variables and unknown constants?

  1. #1
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    Variables and unknown constants?

    Hi,



    Through a maths course I have been learning about differentiating inverse trig functions. I just learnt the rule:
    $y = {\sin ^{ - 1}}\left( {{x \over a}} \right)$
    \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {{a^2} - {x^2}} }}




    I understand how to use the rule and how to prove the rule.




    I was recently shown a question y = {\sin ^{ - 1}}\left( {\frac{x}{{{a^2}}}} \right) and to solve it they did
    \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {{{\left( {{a^2}} \right)}^2} - {x^2}} }}
     = \frac{1}{{\sqrt {{a^4} - {x^2}} }}

    And then continued making it more elegant




    But what I dont understand is why were they able to use the same rule and just sub it into the end?




    I mean if you got
    y = {\sin ^{ - 1}}\left( {\frac{{{x^2}}}{a}} \right)
    You couldnt just say that
    \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {{a^2} - {{\left( {{x^2}} \right)}^2}} }}





    Because that would be wrong because you need to use chain rule.




    So why can a be pretty much anything but x must remain as x for you to be able to use the rule?



    I tried to ask my teacher and he said it is because a is a constant. That confused me even more because I am not sure what the difference is between a variable and an unknown constant. Seeing as they both can take on any value.



    Do constants have some sort of property where you can replace them for anything in a rule and the rule still works while variables cant be replaced without creating a new rule?



    Sorry if this is worded badly I am having a bit of trouble trying to explain why I am confused. I would really appreciate any help.



    Thank you
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Re: Variables and unknown constants?

    Quote Originally Posted by deltasalt View Post
    Hi,



    Through a maths course I have been learning about differentiating inverse trig functions. I just learnt the rule:
    $y = {\sin ^{ - 1}}\left( {{x \over a}} \right)$
    \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {{a^2} - {x^2}} }}




    I understand how to use the rule and how to prove the rule.




    I was recently shown a question y = {\sin ^{ - 1}}\left( {\frac{x}{{{a^2}}}} \right) and to solve it they did
    \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {{{\left( {{a^2}} \right)}^2} - {x^2}} }}
     = \frac{1}{{\sqrt {{a^4} - {x^2}} }}

    And then continued making it more elegant




    But what I dont understand is why were they able to use the same rule and just sub it into the end?




    I mean if you got
    y = {\sin ^{ - 1}}\left( {\frac{{{x^2}}}{a}} \right)
    You couldnt just say that
    \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {{a^2} - {{\left( {{x^2}} \right)}^2}} }}





    Because that would be wrong because you need to use chain rule.




    So why can a be pretty much anything but x must remain as x for you to be able to use the rule?



    I tried to ask my teacher and he said it is because a is a constant. That confused me even more because I am not sure what the difference is between a variable and an unknown constant. Seeing as they both can take on any value.



    Do constants have some sort of property where you can replace them for anything in a rule and the rule still works while variables cant be replaced without creating a new rule?



    Sorry if this is worded badly I am having a bit of trouble trying to explain why I am confused. I would really appreciate any help.



    Thank you
    I hope this doesn't confuse you, but let's generalize this even more to cover all cases.

    Let us consider the function y=\sin^{-1}\left(\frac{u(x)}{c}\right) where u(x) is any arbitrary function of x and c is any arbitrary nonzero constant.

    We now proceed to determine the derivative formula. We first note that

    y=\sin^{-1}\left(\frac{u(x)}{c}\right)\implies \sin y=\frac{u(x)}{c}

    We now differentiate both sides with respect to x to get:

    \frac{d}{dx}\left[\sin y\right]=\frac{d}{dx}\left[\frac{u(x)}{c}\right]\implies \cos y\frac{\,dy}{\,dx}=\frac{1}{c}\frac{\,du}{\,dx}

    Note that here, I let u^{\prime}(x)=\frac{\,du}{\,dx}. Thus, we now right

    \frac{\,dy}{\,dx}=\frac{1}{c\cos y}\frac{\,du}{\,dx}.

    Now, note that the range of y=\sin^{-1}x is y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]. Thus, \cos y is positive over such values, and thus we can take the square root of \cos^2y=1-\sin^2y to get \cos y=\sqrt{1-\sin^2y}.

    But this implies that \cos y = \sqrt{1-\sin^2 y} = \sqrt{1-\left(\frac{u(x)}{c}\right)^2} since \sin y=\frac{u(x)}{c}. Therefore, the derivative formula becomes the following:

    \frac{\,dy}{\,dx}=\frac{1}{c\sqrt{1-\left(\frac{u(x)}{c}\right)^2}}\frac{\,du}{\,dx}.

    But we continue to rewrite this as follows, using the fact that c=\sqrt{c^2}:

    \begin{aligned}\frac{\,dy}{\,dx}&=\frac{1}{c\sqrt{  1-\left(\frac{u(x)}{c}\right)^2}}\frac{\,du}{\,dx}\\  &=\frac{1}{\sqrt{c^2}\sqrt{1-\left(\frac{u(x)}{c}\right)^2}}\frac{\,du}{\,dx}\\  &=\frac{1}{\sqrt{c^2-u^2}}\frac{\,du}{\,dx}\end{aligned}

    When you use this formula, keep in mind that c can be any constant! So going back to your remark about taking the derivative of y=\sin^{-1}\left(\frac{x}{a^2}\right), note that a^2 is just another constant (think of it like c=a^2). So when you use the derivative formula, you always square the constant (as we already proved). The fact that we're working with an arbitrary constant in the formula gives us this flexibility. In short, whenever we have some constant appearing within the inverse sine function, it always ends up getting squared in its derivative.

    I'm a little too tired to discuss the issue about arbitrary constant vs variable, so I'll leave it for someone else who wants to address that.

    I hope this helps clarifies things.
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  3. #3
    Member
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    Jul 2011
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    Re: Variables and unknown constants?

    Quote Originally Posted by deltasalt View Post
    Hi,



    Through a maths course I have been learning about differentiating inverse trig functions. I just learnt the rule:
    $y = {\sin ^{ - 1}}\left( {{x \over a}} \right)$
    \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {{a^2} - {x^2}} }}




    I understand how to use the rule and how to prove the rule.




    I was recently shown a question y = {\sin ^{ - 1}}\left( {\frac{x}{{{a^2}}}} \right) and to solve it they did
    \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {{{\left( {{a^2}} \right)}^2} - {x^2}} }}
     = \frac{1}{{\sqrt {{a^4} - {x^2}} }}

    And then continued making it more elegant




    But what I dont understand is why were they able to use the same rule and just sub it into the end?




    I mean if you got
    y = {\sin ^{ - 1}}\left( {\frac{{{x^2}}}{a}} \right)
    You couldnt just say that
    \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {{a^2} - {{\left( {{x^2}} \right)}^2}} }}





    Because that would be wrong because you need to use chain rule.




    So why can a be pretty much anything but x must remain as x for you to be able to use the rule?



    I tried to ask my teacher and he said it is because a is a constant. That confused me even more because I am not sure what the difference is between a variable and an unknown constant. Seeing as they both can take on any value.



    Do constants have some sort of property where you can replace them for anything in a rule and the rule still works while variables cant be replaced without creating a new rule?



    Sorry if this is worded badly I am having a bit of trouble trying to explain why I am confused. I would really appreciate any help.



    Thank you
    A variable (for example in real Nos) can take any value in real Nos ,while a constant can take only one particular value in the set of real Nos
    Note a variable is allowed to take only one value at a time and not two or more
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  4. #4
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    Jul 2010
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    Re: Variables and unknown constants?

    Thank you both so much for your help I understand it now!
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