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Related Rates rectangle area

  1. Oct 20, 2007 #1
    1. The problem statement, all variables and given/known data


    A rectangle is expanding so that its length is always twice its width. The perimeter of the rectangle is increasing at a rate of 6cm/min. Find the rate of increase of the area of the renctangle when the perimeter is 40 cm.



    2. Relevant equations



    3. The attempt at a solution



    p = 6x
    dp/dt = 6 dx/dt
    6 = 6 dx/dt
    dx/dt = 1

    40 = 6x
    x = 40/6

    a = 2x^2
    da/dt = 4x^3 dx/dt
    da/dt = 4(40/3)^3 (1)
    da/dt = 32000/27

    However, this is not the right answer as the correct answer is 80/3 cm^2/min.

    Can anyone please help me solve this question? I have been stuck on it for the past half hour. :cry:
     
    Last edited: Oct 20, 2007
  2. jcsd
  3. Oct 20, 2007 #2
    [tex]A=2x^{2}[/tex]

    [tex]\frac{dA}{dt}=4x\frac{dx}{dt}[/tex]
     
  4. Oct 20, 2007 #3
    Thank you so much! It worked!
    Sometimes, small mistakes can make big differences.
     
  5. Oct 20, 2007 #4
    You are on the right track. Keep in mind what you have and use that for the solution,
    i.e. the derivative of the perimeter is known.
    So in order to know the change in area with respect to time find the relation between the area and the perimeter.
    Taking the derivative of this relation with respect to time gives you the equation you are looking for.
     
  6. Oct 20, 2007 #5
    Strange, dx/dt is not given. (although dA/dt=4x*dx/dt is correct)
    A=p^2/18, from which:
    dA/dt=p/9*dp/dt
    This is thus dA/dt=40cm/9*6cm/min=80/3cm^2/min
     
  7. Oct 20, 2007 #6
    you can solve for it, but it's been a while since i've seen Related Rates.

    [tex]P=6w[/tex]

    [tex]\frac{dP}{dt}=6\frac{dW}{dt}[/tex]

    dP\dT given

    [tex]6\frac{cm}{min}=6\frac{dW}{dt}[/tex]

    then i differentiated Area and plugged in and got 80/3.
     
    Last edited: Oct 20, 2007
  8. Oct 20, 2007 #7
    OK, it's clear how you did it. It is a way bit longer to derive, but perfectly valid. Nice.
     
  9. Oct 20, 2007 #8
    Thanks everyone.

    I have another question on related rates.

    At 9am, ship A is 50km east of ship B. Ship A is sailing north at 40km/h and ship B is sailing south at 30km/h. How fast is the distance between them changing at noon?

    I have drawn a diagram with two triangles connected but don't know how to get the distance between the two by differentiating a^2 + b^2 = c^2 with respect to t. Answer for this is 68 km/h. Any help would be GREATLY appreciated!
     
  10. Oct 20, 2007 #9
    Which calc class are you in, meaning which college? If you are in college

    Edit: Have any work for this btw?
     
    Last edited: Oct 20, 2007
  11. Oct 20, 2007 #10
    I'm not in college yet; grade 12 in high school.

    Yes, I have homework on related rates and am struggling big time!

    Any help on the above question? :cry:
     
  12. Oct 20, 2007 #11
    I'm sorry it's been a while since I've done related rates so its taking me some time to get the answer you have
     
  13. Oct 20, 2007 #12

    arildno

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    Let B's inital position be (0,0), and A's position (0,50), measured in km.
    Now, as a function of time, A's position A(t)=(40t,50), B's position B(t)=(-30t,0)

    What is now the expression for D(t), the distance between the boats as a function of time?
     
  14. Oct 20, 2007 #13
    starting pos. is at 9am, you're trying to determine the change at noon. how many hours have passed and how far have you gone with respects to each ship?
     
  15. Oct 20, 2007 #14
    The way to do this is to look for the equation stating the distance between the ships as a function of time. This can be done by considering the triangles you described.
    The lengths are depending on time which is nothing more than the velocity times time.
    L(t)=sqrt(4900t^2+2500)
    This function needs to be differentiated with respect to time. The derivative is standard, a square root and a power. After this put in the numbers and there should your solution be.

    In any case you should have 1470/sqrt(466)km/h, which is the correct one. It is approximately equal to 68km/h.
     
  16. Oct 20, 2007 #15
    The answer is right but I'm still not confused as to how you got it.
    Where did you get the equation for L(t)?
     
  17. Oct 20, 2007 #16

    arildno

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    Have a look at my previous post.
     
  18. Oct 20, 2007 #17
    It may be helpful for you to draw one big triangle to see the distance between the 2 boats per hour.

    Edit: The boats are moving at opposite directions at different rates so you can som those to get the distance per hour, and you are given that the ships started out 50km apart. Get it now
     
    Last edited: Oct 20, 2007
  19. Oct 20, 2007 #18
    The equation for the distance between the ships can be derived by considering the triangle formed by them. L=sqrt(a^2+b^2). What are the numbers a and b?
     
  20. Oct 20, 2007 #19
    I have been doing this question by deriving a^2 + b^2 separately for each ship.

    Once for the south ship so for a I have 90km and da/dt is 30 km, b = 50 and db/dt = 0.

    The second time I derive it, I derive it for the north ship so a = 120km and da/dt is 40km and b=50 and db/dt=0.

    Where am I going wrong? :cry:
     
  21. Oct 20, 2007 #20
    you need the "hypotenuse"
     
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