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Related Rates problem involving triangle

  1. Mar 12, 2012 #1
    1. The problem statement, all variables and given/known data

    "At a given instant the legs of a right triangle are 8in. and 6in., respectively. The first leg decreases at 1in/min and the second increases at 2in/min. At what rate is the area increasing after 2 minutes?"


    2. Relevant equations

    A=[itex]\frac{1}{2}[/itex]bh

    [itex]\frac{db}{dt}[/itex]=-1

    [itex]\frac{dh}{dt}[/itex]=2



    3. The attempt at a solution

    A=[itex]\frac{1}{2}[/itex]bh

    [itex]\frac{dA}{dt}[/itex]=[itex]\frac{1}{2}[/itex]([itex]\frac{db}{dt}[/itex]*h+b*[itex]\frac{dh}{dt}[/itex])

    [itex]\frac{dA}{dt}[/itex]=[itex]\frac{1}{2}[/itex](-1*10+6*2)

    [itex]\frac{dA}{dt}[/itex]=[itex]\frac{1}{2}[/itex](2)=1

    Therefore the area is increasing at a rate of [itex]\frac{1in^{2}}{min}[/itex] after 2 minutes. Is my reasoning sound (I'm pretty sure my answer is correct but I want to be sure that my work is logical)?
     
    Last edited by a moderator: Mar 12, 2012
  2. jcsd
  3. Mar 12, 2012 #2
    I see no problem with this, looks nice!
     
  4. Mar 12, 2012 #3

    Mark44

    Staff: Mentor

    Your work is logical and mostly correct, but you have a small mistake. The two legs are 8" and 6", not 10" and 6" as you show in your work. The hypotenuse is 10", but it doesn't enter into this problem.
     
    Last edited: Mar 12, 2012
  5. Mar 12, 2012 #4
    I thought that you're supposed input the length of the two legs after 2 minutes (and according to the derivatives for both legs this would be 6 and 10 after a 2 minute interval)?

    Your supposed to input the original lengths?
     
  6. Mar 12, 2012 #5

    Mark44

    Staff: Mentor

    Sorry, I missed that "after 2 minutes" part in the first post. Your work is fine.
     
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