Results 1 to 6 of 6

Math Help - Few hard questions

  1. #1
    Newbie
    Joined
    Sep 2010
    Posts
    3

    Few hard questions

    1) Use the definition of derivative to show that function y = g(x) implicitly defined by equation: y^{2/3} + x^{2/3} = 1 is not differentiable at point (0,1)

    2) Find all points at which the tangent lines to the curve are horizontal for equation 2(x^2 + y^2)^2 = x^2 - y^2

    3) Let f(x) = sin x / (1+cos x), Show that f'(x) = 1/(1+cos x)

    These questions are from my university's past exams. It won't be much problem for me, IF they give me more time to solve it... The first two problems are given like only about 10 minute to solve, and the last one is not even 5 minute (Given 180 min for 17 problems, but last one is like part a of the problem)

    So, it either means 1) I am doing it the long way through, or 2) the university examiners don't expect anyone to pass the exam... Therefore, I want to ask for opinions on how to solve these types of complex questions in a more time-efficient way.

    PS: I used 1-2 hours just to pull that lim h -> 0 of ((1-h^{2/3})^{3/2} - 1 )/h from the first problem
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,536
    Thanks
    778
    1) Use the definition of derivative to show that function y = g(x) implicitly defined by equation: y^{2/3} + x^{2/3} = 1 is not differentiable at point (0,1)
    As you wrote, y(x)=(1-x^{2/3})^{3/2}. Using Maclaurin series for (1+x)^\alpha, we get y(x)=1-3x^{2/3}/2+o(x^{2/3}). Therefore, (y(x)-y(0))/x=-3/(2x^{1/3})+o(1/x^{1/3}). This expression tends to -\infty when x\to0^+ and to \infty when x\to0^+, so there is no limit.

    It also helps to have a sketch of the graph of y(x). This function is symmetric w.r.t. the vertical x = 0 axis because x occurs squared. Also, the behavior of x^{2/3} near 0 is similar to that of \sqrt{x}, in particular, (x^{2/3})'\to\infty when x\to0^+. Therefore, 1-x^{2/3} has a sharp angle at (0,1), which is its maximum point. I believe that raising everything to power 3/2 does not fundamentally change this behavior.

    3) Let f(x) = \sin x / (1+\cos x), Show that f'(x) = 1/(1+\cos x)
    This is pretty straightforward using the formula for the derivative of a fraction and the fact that \sin^2x+\cos^2x=1.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2010
    Posts
    3
    I have never used Maclaurin series before. Therefore, I don't really understand the term o(x^{2/3}), so if you can explain me, then that would be great. Also I did use L'Hopital Rule for solving 1st question and I check limit left-hand = right-hand to prove it's not differentiable. However, when I apply the L'Hopital Rule, I get a constant as the limit, not infinity. Therefore, I'm not sure if I'm applying it in the right situation or not. For the 3rd problem, i think they want a proof? I'm not sure, but for me.. using the quotient derivative is kind of "too easy for an exam question".
    Last edited by biloon; September 24th 2010 at 09:50 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by biloon View Post
    [snip] For the 3rd problem, i think they want a proof? I'm not sure, but for me.. using the quotient derivative is kind of "too easy for an exam question".
    Nonsense. You're asked to find a derivative. The method is not prescribed. Therefore using the quotient rule is the obvious approach to take. A competent student will get it done easily in under 5 minutes.

    Quote Originally Posted by biloon View Post
    [snip]

    2) Find all points at which the tangent lines to the curve are horizontal for equation 2(x^2 + y^2)^2 = x^2 - y^2

    [snip]
    The clock starts ...... now:

    Using implicit differentiation:

    \displaystyle 4(x^2 + y^2) \left( 2x + 2y \frac{dy}{dx} \right) = 2x + 2y \frac{dy}{dx}

    \displaystyle \Rightarrow \frac{dy}{dx} = \frac{x(1 - 4x^2 - 4y^2)}{y(2x^2 + 2y^2 - 1)}.

    Solve \displaystyle \frac{dy}{dx} = 0 subject to 2(x^2 + y^2)^2 = x^2 - y^2:

    x(1 - 4x^2 - 4y^2) = 0 .... (A)

    2(x^2 + y^2)^2 = x^2 - y^2 .... (B)

    From (A):

    Case 1: x = 0. Substitute into (B): y = 0. However, dy/dx is indeterminant at (0, 0) so that point is probably a crunode. (Certainly a crunode if you recognise the given relation as defining a lemniscate).

    Case 2: 4x^2 + 4y^2 = 1. Substitute into (B) etc. You have 5 minutes left, plenty of time. Test that dy/dx is not indeterminant for the solutions. (Here's a short cut: solve 4x^2 + 4y^2 = 1 and 2(x^2 + y^2)^2 = x^2 - y^2 - Wolfram|Alpha)
    Last edited by mr fantastic; September 25th 2010 at 02:54 AM. Reason: Merged posts, Added the link.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,693
    Thanks
    1466
    A "crunode" is a point at which a graph intersects itself. (I had to look that up!)

    Also called a "double point". (Which I do recognize.)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,536
    Thanks
    778
    I have never used Maclaurin series before. Therefore, I don't really understand the term o(x^{2/3}), so if you can explain me, then that would be great.
    I suppose it would not help telling that Maclaurin series is Taylor series at x = 0. Little-o notation f(x)=o(g(x)) when x\to\alpha means that f(x) becomes negligible compared to g(x) when x\to\alpha.

    Also I did use L'Hopital Rule for solving 1st question and I check limit left-hand = right-hand to prove it's not differentiable. However, when I apply the L'Hopital Rule, I get a constant as the limit, not infinity. Therefore, I'm not sure if I'm applying it in the right situation or not.
    Using l'H˘pital's rule to find \displaystyle\lim_{x\to0}\frac{(1-x^{2/3})^{3/2}-1}{x} seems justified because the limits of the nominator and denominator are both 0. Further, \displaystyle[(1-x^{2/3})^{3/2}-1]'=3/2(1-x^{2/3})^{1/2}(-(2/3)x^{-1/3})=-\frac{(1-x^{2/3})^{1/2}}{x^{1/3}}. Now, the nominator tends to 1, so the limit is -\infty when x\to0^+, and it is +\infty when x\to0^-.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Hard Questions
    Posted in the Geometry Forum
    Replies: 3
    Last Post: October 7th 2009, 10:48 PM
  2. 4 Hard Questions!! Help Please!!
    Posted in the Calculus Forum
    Replies: 11
    Last Post: August 3rd 2009, 06:08 PM
  3. Hard questions are here...
    Posted in the Geometry Forum
    Replies: 0
    Last Post: January 23rd 2009, 08:17 AM
  4. Really hard questions...
    Posted in the Geometry Forum
    Replies: 2
    Last Post: June 2nd 2008, 10:13 PM
  5. [SOLVED] Help!! hard math questions
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 4th 2008, 08:52 PM

Search Tags


/mathhelpforum @mathhelpforum