Results 1 to 2 of 2

Math Help - Rates of change

  1. #1
    Newbie
    Joined
    Jul 2008
    Posts
    10

    Rates of change

    I need to find the derivative of;

    h(t) = 62 + 60sin ((5t-1)pi / 2)

    All help is appreciated.

    xoxo Lex.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Mercy99 View Post
    I need to find the derivative of;

    h(t) = 62 + 60sin ((5t-1)pi / 2)

    All help is appreciated.

    xoxo Lex.
    Again, keep in mind that \frac{d}{dt}\left[\sin(x)\right]=\cos(x).

    Proof:

    If f(x)=\sin(x), then f'(x)=\lim_{h\to{0}}\frac{f(x+h)-f(x)}{h}=\lim_{h\to{0}}\frac{\sin(x+h)-\sin(x)}{h}

    Note that \sin(\varphi+\vartheta)=\sin(\varphi)\cos(\varthet  a)+\cos(\varphi)\sin(\vartheta)

    Thus, \lim_{h\to{0}}\frac{\sin(x+h)-\sin(x)}{h}=\lim_{h\to{0}}\frac{\sin(x)\cos(h)+\co  s(x)\sin(h)-\sin(x)}{h}

    \implies \lim_{h\to{0}}\frac{\sin(x)\cos(h)-\sin(x)}{h}+\lim_{h\to{0}}\frac{\cos(x)\sin(h)}{h}

    \implies \lim_{h\to{0}}\sin(x)\frac{\cos(h)-1}{h}+\lim_{h\to{0}}\cos(x)\frac{\sin(h)}{h}

    We should know that \lim_{h\to{0}}\frac{\cos(h)-1}{h}=0 and \lim_{h\to{0}}\frac{\sin(h)}{h}=1

    Thus, \lim_{h\to{0}}\sin(x)\frac{\cos(h)-1}{h}+\lim_{h\to{0}}\cos(x)\frac{\sin(h)}{h}=\sin(  x)\cdot 0+\cos(x)\cdot 1=\color{red}\boxed{\cos(x)}




    We need to apply the chain rule as well : \frac{d}{dt}\left[f(g(t))\right]=f'(g(t))\cdot g'(t)

    Thus, \frac{dh}{dt}=60\cos\left(\frac{(5t-1)\pi}{2}\right)\cdot \frac{5\pi}{2}=\color{red}\boxed{150\pi\cos\left(\  frac{(5t-1)\pi}{2}\right)}

    Does this stuff seem to make more sense now?

    Feel free to ask more questions if you have them. I may not be on for much longer. There are others besides me who can answer your questions.

    --Chris
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rates of Change
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 23rd 2010, 03:15 PM
  2. Rates of Change
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 6th 2009, 05:40 PM
  3. Rates of change
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 25th 2009, 10:44 AM
  4. Rates of Change
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 19th 2009, 07:15 AM
  5. Rates of change
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 7th 2009, 04:24 AM

Search Tags


/mathhelpforum @mathhelpforum