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Homework Help: Finding exact point at this tanget line of a graph

  1. Feb 14, 2008 #1
    1. The problem statement, all variables and given/known data
    Given the equation: x - cos^2(x^2 - y^2) = xy - sin^2(x^2 - y^2)

    Here's a graph of this equation: http://img240.imageshack.us/my.php?image=graphtq8.png

    What is the exact point on the tangent line?
    And what is the slope of the tangent line?

    3. The attempt at a solution

    I know since x=0 at the point, I would need to find y when x=0. But this is confusing me a bit since when I plug in 0 I'm left with -cos(-y^2) = -sin^2(-y^2). And I don't really know when to go from there. I also figure im going to need to derive the equation by implicit differentiation to find the slope I assume. I'm just looking for some suggestions.
  2. jcsd
  3. Feb 14, 2008 #2
    well, now you need to find the y-intercept right.
    you are left with sth like this, after u plug in 0 for x.
    [tex]cos^{2}(-y^{2})=sin^{2}(-y^{2})[/tex] since sine is squared we can get rid of the minus sign, for cosine we know it is even so it does not matter

    [tex]cos^{2}(y^{2})=sin^{2}(y^{2})[/tex] devide by [tex]cos^{2}(y^{2})[/tex], but we lose solutions here for y^2=kpi , so
    [tex]tan^{2}(y^{2})=1[/tex] from here we get [tex]tan(y^{2})=+-1[/tex], but since y^2 can never be negative we get [tex]tan(y^{2})=1[/tex], now
    so du u see now?
  4. Feb 14, 2008 #3


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    How is the question stated? You said there is a tangent line, but it's equation is not given, only that it appears to be a tangent line at x=0.

    You can find the exact point on the graph at x=0 by simply plugging x=0 into the equation of graph. You'll get a trigo equation, which you must then solve for a general solution in terms of y. The exact point on the tangent line would then be the first instance where y is negative according to the general solution.

    You have to use implicit differentiation here. Just treat y as a function of x. Once you get [tex]-cos^{2}(-y^{2}) = -sin^2(-y^2)[/tex], which you already did, then note that sine is an odd function which means you can simplify RHS and solve for y to get the general solution.
  5. Feb 14, 2008 #4
    I do not think he will need to implicitly differentiate at all.-That is what i first thought though. because he just needs the y value for which x=0, so basically that will be
  6. Feb 14, 2008 #5


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    He'll need to differentiate it implicitly because he's also asking for the slope of the tangent line.
  7. Feb 14, 2008 #6
    I did give the equation on the first line. The equation of the tangent line isn't given, they just want to know the coordinates at that point on the graph and also the slope of the tangent line.

    Anyways, the sqrt of the arctan(1) gave me about -.8662269255, which seems reasonable by looking at the graph.
  8. Feb 14, 2008 #7


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    You shouldn't use the calculated value. Arctan 1 = pi/4 Square root that for the exact expression.
  9. Feb 14, 2008 #8
    Sorryyyyy! My bad, i did not look at the second question at all, i just missed it, that one about the slope. So you are right, he defenitely will need to differentiate implicitly.
  10. Feb 14, 2008 #9
    I see, thanks for the correction.

    Anyways, after using implicit on the equation I ended up with: y` = [tex]\frac{1}{xy - 2x + 2y}[/tex]

    So is it safe to say the slope is [tex]\frac{1}{2(-\sqrt{pi/4})}[/tex] ?
  11. Feb 14, 2008 #10
    For the op: after you differentiate it implicityly, you need to bring on one side dy/dx, while on the other side the rest. After that you can easily find the slope of the tangent line, just plug in x=0 and y=-sqrt(pi)/2
  12. Feb 14, 2008 #11
    Yep, if you did the calculations correctly when you differentiated implicitly, then i guess that is safe to say so!
  13. Feb 14, 2008 #12
    I have this feeling that you have done some mistakes somewhere, because by just looking at the graph you provided, one could tell that the slope of the tangent line at that point is positive. So, look somewhere if you haven't missed some minus sign somewhere.
  14. Feb 14, 2008 #13

    Edit: i guess it is not safe at this case!
  15. Feb 14, 2008 #14
    Hmm, this is the work i did for getting the derivative

    d/dx [ x - cos^2(x^2 - y^2) ] = d/dx [xy - sin^2(x^2 - y^2)]

    1 + sin^2(x^2 - y^2)*(2x-2y)*y` = x * y` * y - cos^2(x^2 - y^2)*(2x-2y)*y`

    1 = x * y` * y - cos^2(x^2 - y^2)*(2x-2y)*y` - sin^2(x^2 - y^2)*(2x-2y)*y`

    1 = y`[ xy - cos^2(x^2 - y^2)*(2x-2y) - sin^2(x^2 - y^2)*(2x-2y)]

    1 = y`[xy - (2x-2y)( cos^2(x^2 - y^2) + sin^2(x^2 - y^2) ) ]

    1 = y`[xy - (2x-2y)(1)]
    1 = y`[xy - 2x +2y]
    y` = (1)/(xy - 2x +2y)

    Can you spot my mistake by any chance?
  16. Feb 14, 2008 #15
    Here it is the first mistake i can spot, there might be others, but i will show you the idea

    well think of it this way say u=u(x) that is u is a dependent variable, then

    [cos^2(u)]'=[cos(u)cos(u)]'=[cos(u)]'cos(u)+cos(u)[cos(u)]'=-sin(u)u'cos(u)+cos(u)(-sin(u)u' =-2cos(u)sin(u)u'

    i hope this helps!
    By the way, your mistake is on the second line!
    Last edited: Feb 14, 2008
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