Results 1 to 4 of 4

Thread: Problem with trigonomic derivative (pretty simple)

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    11

    Problem with trigonomic derivative (pretty simple)

    I know how to derivate sinx, cosx and standard equations like

    y = x^4 + sqrt.(x)
    y' = 4x^3 + x^-1/2

    y = sinx + cosx
    y' = cosx - sinx

    and so on. But now they want me to "decide an equation for the tangent to the curve":

    a) y = cosx (pi/2, 0)

    Answer: y = -x + pi/2

    b) y = 2x - 5sinx (pi, 2pi)

    Answer: y = 7x -5pi

    I'm new to derivatives with sinx, and they have almost no examples in the book unfortunetly. I have almost no idea how to solve this, except that I guess pi would stand for 180 degrees, so then pi/2 would be x = 90 degrees. After that I'm lost!

    Appreciate help!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by vane505 View Post
    I know how to derivate sinx, cosx and standard equations like

    y = x^4 + sqrt.(x)
    y' = 4x^3 + x^-1/2

    y = sinx + cosx
    y' = cosx - sinx

    and so on. But now they want me to "decide an equation for the tangent to the curve":

    a) y = cosx (pi/2, 0)

    Answer: y = -x + pi/2

    b) y = 2x - 5sinx (pi, 2pi)

    Answer: y = 7x -5pi

    I'm new to derivatives with sinx, and they have almost no examples in the book unfortunetly. I have almost no idea how to solve this, except that I guess pi would stand for 180 degrees, so then pi/2 would be x = 90 degrees. After that I'm lost!

    Appreciate help!!
    I take it that the numbers in the brackets are the co-ordinates of the point the tangent should touch...

    a) $\displaystyle \frac{dy}{dx} = -\sin{x}$.

    At the point $\displaystyle x = \frac{\pi}{2}, \frac{dy}{dx} = -\sin{\frac{\pi}{2}} = -1$.

    Remember that the equation for a line is given by $\displaystyle y = mx + c$ where $\displaystyle m$ is the slope and $\displaystyle c$ is the $\displaystyle y$-intercept. Since we have the slope of the tangent line and a point that lies on the line, we can solve for $\displaystyle c$.

    $\displaystyle y = mx + c$

    $\displaystyle 0 = -1\left(\frac{\pi}{2}\right) + c$

    $\displaystyle 0 = -\frac{\pi}{2} + c$

    $\displaystyle c = \frac{\pi}{2}$.

    So the equation of the tangent line to the curve at the point given is

    $\displaystyle y = -x + \frac{\pi}{2}$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by vane505 View Post
    I know how to derivate sinx, cosx and standard equations like

    y = x^4 + sqrt.(x)
    y' = 4x^3 + x^-1/2

    y = sinx + cosx
    y' = cosx - sinx

    and so on. But now they want me to "decide an equation for the tangent to the curve":

    a) y = cosx (pi/2, 0)

    Answer: y = -x + pi/2

    b) y = 2x - 5sinx (pi, 2pi)

    Answer: y = 7x -5pi

    I'm new to derivatives with sinx, and they have almost no examples in the book unfortunetly. I have almost no idea how to solve this, except that I guess pi would stand for 180 degrees, so then pi/2 would be x = 90 degrees. After that I'm lost!

    Appreciate help!!
    b) $\displaystyle y = 2x - 5\sin{x}$ so $\displaystyle \frac{dy}{dx} = 2 - 5\cos{x}$.

    At the point $\displaystyle x = \pi, \frac{dy}{dx} = 2 - 5\cos{\pi} = 2 - 5(-1) = 7$. This is $\displaystyle m$.

    So now we put $\displaystyle x, y, m$ into $\displaystyle y = mx + c$ and solve for $\displaystyle c$.

    $\displaystyle y = mx + c$

    $\displaystyle 2\pi = 7\pi + c$

    $\displaystyle c = -5\pi$.


    So the equation of the tangent line at the point given is

    $\displaystyle y = 7x - 5\pi$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2008
    Posts
    11
    Great! Thank you very much for your help and very clear explenation, I get it now
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Mar 5th 2010, 09:25 PM
  2. Pretty simple trigonometry but need help
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Feb 20th 2010, 12:25 PM
  3. pretty simple algebra problem (i think)
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Aug 30th 2009, 10:45 AM
  4. Pretty simple question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 25th 2009, 02:27 PM
  5. pretty simple abstract value problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Dec 31st 2008, 09:51 AM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum