Results 1 to 3 of 3
Like Tree2Thanks
  • 1 Post By MarkFL
  • 1 Post By Bradyns

Math Help - Derivatives and other fancy stuff

  1. #1
    Newbie
    Joined
    Apr 2013
    From
    Melbourne, Australia
    Posts
    15

    Derivatives and other fancy stuff

    Having trouble with a couple of questions

    The first one:
    The height of a projectile at a particular time (measured in seconds) is given by h(t)=10t(1-t)m. Find the rate of change of height from first principles (using the definition of the rate of change).

    - How do I find the rate of change of height and I'm a bit confused on what 'first principles' are.

    The second: Consider the function f(x)=x(4-x). Find the equation of the tangent to the graph of f(x) at x=2. [NOTE: when calculating f'(2), use first principles.]

    - Again I'm not too sure on the 'first principles' thing.

    Any help is much appreciated,

    moob
    Last edited by moob; April 20th 2013 at 09:16 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Derivatives and other fancy stuff

    Computing a derivative by first principles means using the definition:

    f'(x)\equiv\lim_{h\to0}\frac{f(x+h)-f(x)}{h}
    Thanks from moob
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member Bradyns's Avatar
    Joined
    Apr 2013
    From
    Maitland, NSW, Australia
    Posts
    39
    Thanks
    12

    Re: Derivatives and other fancy stuff

    First one:
    h(t)=10t(1-t)

    \Rightarrow h(t)= 10t-10t^2

    h'(t) = \lim_{\delta\rightarrow 0} \frac{h(t+\delta)-h(t)}{\delta}

    I've used delta so as not to confuse the function with the limiting value.

    h'(t) = \lim_{\delta\rightarrow 0} \frac{10(t+\delta)-10(t+\delta)^2-(10t-10t^2)}{\delta}

    You just need to expand some terms.. cancel like terms, factor out a 'h' from the numerator and cancel out the bottom 'h', take the limit and you're done.

    Second one:


    \Rightarrow f(x)=4x-x^2

    Same deal with first principles, this time we have an 'x' value.

    f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}

    f'(2) = \lim_{h\rightarrow 0} \frac{f(2+h)-f(2)}{h}


    f'(2) = \lim_{h\rightarrow 0} \frac{4(2+h)-(2+h)^2-(4(2)-(2)^2)}{h}

    Same deal as above..
    You just need to expand some terms.. cancel like terms, factor out a 'h' from the numerator and cancel out the bottom 'h', take the limit and you're done.

    Once we have that equation, that solution with be your gradient (m).
    You then use the point gradient formula:

    y - y_1 = m(x - x_1)

    Where
    x_1 = 2
    y_1 = f(x)= 4(2)-(2)^2
    m = derivative at 2.

    Rearrange to get it into the form y = mx + b.
    Thanks from moob
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Fancy Functions
    Posted in the Algebra Forum
    Replies: 0
    Last Post: April 1st 2013, 01:58 AM
  2. An integral. Nothing too fancy... or not?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 13th 2011, 12:47 PM
  3. fancy some toast?
    Posted in the Statistics Forum
    Replies: 3
    Last Post: August 6th 2009, 05:53 AM
  4. Fancy working something out for me?
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: December 8th 2008, 01:07 PM
  5. [SOLVED] fancy entropy exercise?help needed
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: October 27th 2007, 04:50 AM

Search Tags


/mathhelpforum @mathhelpforum