Results 1 to 2 of 2

Math Help - relation rate word problem

  1. #1
    Member
    Joined
    Jun 2007
    Posts
    79

    relation rate word problem

    2. A ladder 13 ft. long rest against a vertical wall and is sliding down the wall at the rate of 3ft/s at the instant the foot of the ladder is 5 ft. from the base of the wall. At this instant how fast is the foot( or base) of the ladder moving away from the wall? Side note, this problem looks like a pythag. theorem: x^+y^2=r^2. the answer comes out to 7.2 ft./s. though I know what formula to use for the certain situation, I really do not know the steps in solving this problem.

    And also, I tried isolating dr/dt and it gave 2x+2y-2r=dr/dt and getting the value for x and y(which i know does not make any sense).
    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 driver327 View Post
    2. A ladder 13 ft. long rest against a vertical wall and is sliding down the wall at the rate of 3ft/s at the instant the foot of the ladder is 5 ft. from the base of the wall. At this instant how fast is the foot( or base) of the ladder moving away from the wall? Side note, this problem looks like a pythag. theorem: x^+y^2=r^2. the answer comes out to 7.2 ft./s. though I know what formula to use for the certain situation, I really do not know the steps in solving this problem.
    Yes! it is using Pythagoras. i hope you drew a diagram to see what's going on (i won't, i usually do, but not this time ).

    Introducing r gets you in trouble, we only need x and y here. r is a constant, we don't need to replace it with a variable, since it does not change! the ladder will always be 13 ft long

    Let x be the distance the foot of the ladder is away from the wall.
    Let y be the vertical distance from the foot of the wall to the top of the ladder
    Let x' be the rate at which the horizontal distance is increasing
    Let y' be the rate at which the vertical distance is decreasing

    We want x' when y' = -3 and x = 5

    By Pythagoras, we have

    x^2 + y^2 = 13^2 ..............(1)

    from the above, when x = 5 , we have y = 12

    going back to equation (1), we proceed using implicit differentiation

    \Rightarrow 2x~x' + 2y~y' = 0

    \Rightarrow x' = - \frac {y~y'}{x}

    We care about the instant where x = 5, y' = -3, y = 12

    Thus we have:

    x' = - \frac {(12)(-3)}{5} \implies \boxed { x' = 7.2 }


    Note, x' = \frac {dx}{dt} and y' = \frac {dy}{dt}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rate of Change Word Problem
    Posted in the Calculus Forum
    Replies: 11
    Last Post: November 7th 2011, 10:30 AM
  2. quadratic relation word problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: July 24th 2011, 03:30 PM
  3. Inverse relation word problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: August 28th 2010, 01:08 AM
  4. Rate word problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 11th 2010, 04:15 PM
  5. Word problem/rate
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 14th 2006, 03:29 AM

Search Tags


/mathhelpforum @mathhelpforum