Results 1 to 4 of 4

Thread: tangent line

  1. #1
    Newbie
    Joined
    Feb 2008
    Posts
    14

    tangent line

    find the equation of the tangent line to y=(sin x)^4x at the point (pi/2, 1)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Bust2000 View Post
    find the equation of the tangent line to y=(sin x)^4x at the point (pi/2, 1)
    a line is of the form $\displaystyle y = mx + b$

    we will find the line by the point-slope form. that is, $\displaystyle y - y_1 = m(x - x_1)$, where $\displaystyle m$ is the slope (given by the derivative evaluated at $\displaystyle (x_1,y_1)$) and $\displaystyle (x_1,y_1)$ is a point the slope passes through (here it is (pi/2, 1)).

    can you continue?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,854
    Thanks
    138
    Quote Originally Posted by Bust2000 View Post
    find the equation of the tangent line to y=(sin x)^4x at the point (pi/2, 1)
    Do you really mean:

    $\displaystyle f(x) = \left(\sin(x)\right)^{4x}$ ...... ??

    If so re-write f to:

    $\displaystyle f(x) = e^{4x \cdot \ln(\sin(x))}$ ...... and use the chain rule to calculate the derivative:

    $\displaystyle f'(x) = \frac{e^{4x \cdot \ln(\sin(x))}}{\sin(x)} \cdot \left(4 \sin(x) \cdot \ln(\sin(x)) + 4x \cdot \cos(x) \right)$

    Now plug in $\displaystyle x = \frac{\pi}{2}$ to calculate the slope of the tangent.

    Since $\displaystyle \ln\left(\sin\left(\frac{\pi}2 \right) \right) = 0$ and $\displaystyle \cos \left(\frac{\pi}2 \right) = 0$ too you'll get:

    $\displaystyle f'\left(\frac{\pi}2 \right) = 0$ and therefore the equation of the tangent is

    $\displaystyle y = 0$

    EDIT: According to Jhevons reply the equation of the tangent is $\displaystyle y = 1$
    Last edited by earboth; Feb 11th 2008 at 12:17 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by earboth View Post
    Do you really mean:

    $\displaystyle f(x) = \left(\sin(x)\right)^{4x}$ ...... ??

    If so re-write f to:

    $\displaystyle f(x) = e^{4x \cdot \ln(\sin(x))}$ ...... and use the chain rule to calculate the derivative:

    $\displaystyle f'(x) = \frac{e^{4x \cdot \ln(\sin(x))}}{\sin(x)} \cdot \left(4 \sin(x) \cdot \ln(\sin(x)) + 4x \cdot \cos(x) \right)$

    Now plug in $\displaystyle x = \frac{\pi}{2}$ to calculate the slope of the tangent.

    Since $\displaystyle \ln\left(\sin\left(\frac{\pi}2 \right) \right) = 0$ and $\displaystyle \cos \left(\frac{\pi}2 \right) = 0$ too you'll get:

    $\displaystyle f'\left(\frac{\pi}2 \right) = 0$ and therefore the equation of the tangent is

    $\displaystyle y = 0$
    i believe it would be y = 1 if the slope is zero
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Nov 4th 2011, 06:42 PM
  2. Replies: 6
    Last Post: Jan 12th 2011, 02:38 PM
  3. Replies: 5
    Last Post: Nov 4th 2009, 06:49 PM
  4. tangent line & normal line question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Mar 15th 2009, 03:45 AM
  5. Replies: 4
    Last Post: Mar 17th 2008, 01:53 PM

Search Tags


/mathhelpforum @mathhelpforum