Results 1 to 6 of 6

Math Help - Another rate of change question

  1. #1
    Junior Member
    Joined
    Apr 2006
    Posts
    29

    Another rate of change question

    A dog instantly spots a squirrel sitting on the ground 15m away. She approaches the squirrel at a constant rate of 2m/s and the squirrel darts straight up the tree at a constant rate of 4m/s. Assuming the ground is smooth and flat and the tree is perfectly vertical:
    a) At what rate is the distance between the dog and the squirrel changing at the moment when they are the closest together? (i.e. at the minimum distance apart?)
    b) Determin the time at which the distance is at its minimum
    c) Calculate the minimum distance between the dog and the squirrel.

    I know that this problem involves the dog and squirrel forming a triangle, but there doesnt seem to be enough information given to solve the question, please help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    The thing to do first is to find the time when they are closest.

    Since the dog is moving at 2 m/s, her distance is (15-2t).

    The squirrels distance is 4t

    Therefore the time they're closest together is by ol' Pythagoras.

    D(t)=(15-2t)^{2}+(4t)^{2}

    D'(t)=40t-60

    t=\frac{3}{2} \;\ sec

    After 3/2 seconds they are closest together.

    If the dog is moving at 2 m/s for 3/2 seconds then she moves 3 feet.

    Therefore, she is 15-2(3/2)=12 feet from the tree.

    The squirrel is 4(3/2)=6 feet up the tree,

    The distance between them along the hypoteneuse is

    \sqrt{12^{2}+6^{2}}=6\sqrt{5}\approx{13.42} \;\ meters
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2006
    Posts
    29
    I just have one question, when you used the pythagorean theorem for the D(t) equation, why dont you have to take the squareroot of the right side?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,678
    Thanks
    609
    Hello, sugar_babee!

    When you used the pythagorean theorem for the D(t) equation,
    why don't you have to take the squareroot of the right side?
    Galactus used a very handy "trick" . . .


    You are correct . . . the distance is: . d \;=\;\sqrt{(15-2t)^2 + (4t)^2}

    . . Square both sides: . d^2 \;=\;(15-2t)^2 + (4t)^2

    Now let D = d^2\!:\;\;D \;=\;(15-2t)^2 + (4t)^2


    Can you see that if we minimize D, we are also minimizing d ?

    So we can solve: D' = 0 and avoid that one-half power.

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2006
    Posts
    29
    okay so how would I find the rate for part a?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    You have all the info now, use related rates to solve.

    Differentiate implicitly.

    D^{2}=x^{2}+y^{2}

    D\frac{dD}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: April 12th 2011, 09:51 AM
  2. Rate of change question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 10th 2010, 12:55 AM
  3. Rate of change question.
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 8th 2009, 02:46 PM
  4. Rate of Change Question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 4th 2009, 04:05 PM
  5. Rate of change question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 15th 2008, 02:24 PM

Search Tags


/mathhelpforum @mathhelpforum