Results 1 to 10 of 10

Math Help - Just another trivial problem

  1. #1
    Member Jones's Avatar
    Joined
    Apr 2006
    From
    Norway
    Posts
    170

    Just another trivial problem

    Hello, i hope you can help me with this trivial problem(i hope)

    Determine the the equation for the tangent to the curve y=tan x/x

    at the given point of ((pi/4),(4/pi))

    Is it just the orinary y-y1=k(x-x1) formulae we need to use or is it different this time?

    Thank you, Jones
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Jones
    Hello, i hope you can help me with this trivial problem(i hope)

    Determine the the equation for the tangent to the curve y=tan x/x

    at the given point of ((pi/4),(4/pi))

    Is it just the orinary y-y1=k(x-x1) formulae we need to use or is it different this time?

    Thank you, Jones
    The equation y-y_1=k(x-x_1) is the equation of the line through
    the point (x_1,y_1) with slope k. For it to be the equation
    of the tangent to the curve at the given point k must be the slope
    of the curve at the point. That is:

    <br />
k=\frac{d}{dx} \left\{\frac{\tan(x)}{x}\right \} \left \left \vert _{x=\pi /4}<br />
,

    and x_1=pi/4,\ y_1=4/pi.

    RonL
    Last edited by CaptainBlack; April 16th 2006 at 12:00 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Jones's Avatar
    Joined
    Apr 2006
    From
    Norway
    Posts
    170
    Don't we need to differentiate the function first?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Jones
    Don't we need to differentiate the function first?
    That appears to be what 5th line of my earlier post says

    RonL
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by CaptainBlack
    For it to be the equation
    of the tangent to the curve at the given point k must be the slope
    of the curve at the point. That is:

    <br />
k=\frac{d}{dx} \left\{\frac{\tan(x)}{x}\right \} \left \left \vert _{x=\pi /4}<br />
,

    <br />
\frac{d}{dx} \left\{\frac{\tan(x)}{x}\right \}= \frac{1}{x (\cos(x))^2}-\frac{\tan(x)}{x^2}<br />
.

    Therefore

    <br />
k=\frac{8}{\pi}-\frac{16}{\pi^2}\approx 0.92534<br />

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member Jones's Avatar
    Joined
    Apr 2006
    From
    Norway
    Posts
    170
    Quote Originally Posted by CaptainBlack
    <br /> <br />
-\frac{\tan(x)}{x^2}<br />
.
    where did this come from?

    Im a little slow you see, almost dumb. therefor you have to be very legible
    Last edited by Jones; April 25th 2006 at 02:02 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Jones
    where did this come from?

    Im a little slow you see, almost dumb. therefor you have to be very legible
    From the product rule for differentiation.

    <br />
\frac{d}{dx} f(x)g(x)= f'(x)g(x)+f(x)g'(x)<br />

    and here g(x)=1/x and f(x)=\tan(x) so:

    <br />
\frac{d}{dx} \left\{\frac{\tan(x)}{x}\right \}= \frac{1}{x (\cos(x))^2}-\frac{\tan(x)}{x^2}<br />

    I could have used the quotient rule, but since it is equivalent to this
    use of the product rule I cant be bothered to remember it (as it is
    redundant).

    RonL
    Last edited by CaptainBlack; April 25th 2006 at 02:11 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,857
    Thanks
    321
    Awards
    1
    Quote Originally Posted by CaptainBlack
    I could have used the quotient rule, but since it is equivalent to this use of the product rule I cant be bothered to remember it (as it is redundant).
    RonL
    I'm glad I'm not the only one with that "hang-up."

    -Dan
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by topsquark
    I'm glad I'm not the only one with that "hang-up."

    -Dan
    I'm not sure it is a "hang-up" I find it difficult to see why time and
    limited human memory resources are wasted on teaching the quotient
    rule.

    It seems to me that the teaching of the q-rule in the anomaly -
    one of those distrubing features of this reality which makes one suspect
    that one has slipped into a parallel universe without having noticed.

    RonL
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,857
    Thanks
    321
    Awards
    1
    Quote Originally Posted by CaptainBlack
    I'm not sure it is a "hang-up" I find it difficult to see why time and
    limited human memory resources are wasted on teaching the quotient
    rule.

    It seems to me that the teaching of the q-rule in the anomaly -
    one of those distrubing features of this reality which makes one suspect
    that one has slipped into a parallel universe without having noticed.

    RonL
    I think I live in a parallel universe. (If I only had a "brane!" sings the scarecrow.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trivial Problem...
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: October 8th 2010, 07:02 AM
  2. A trivial problem . . . and solution
    Posted in the LaTeX Help Forum
    Replies: 3
    Last Post: September 5th 2010, 03:29 PM
  3. [A Trivial Problem] adj(AB) = adj(A)adj(B)
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: June 26th 2010, 04:26 PM
  4. Need help with Trivial/Non trivial solution
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 6th 2010, 05:53 AM
  5. Trivial vector bundle problem
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: April 1st 2010, 05:56 AM

Search Tags


/mathhelpforum @mathhelpforum