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Related rates car problem

  1. Mar 13, 2007 #1
    2 cars start moving from the same point. one travels south at 60 mi/h and the other travelks west at 25 mi/h. at what rate is the distance between the cars increasing 2 hourse later.

    What do i do. I need help some serious help with this one.
  2. jcsd
  3. Mar 13, 2007 #2
    what equation do i use
  4. Mar 13, 2007 #3
    Pythagoras Theorem A^2+B^2=C^2
  5. Mar 13, 2007 #4
    ok, so i put 60^2+25^2=c^ which equals 4225
    then i square root 4225 and i get 65 which is the answer in my book, thanks man if this is the right way to do it
  6. Mar 13, 2007 #5
    NO you have to do the dy/dx to get the incresing speed!
  7. Mar 13, 2007 #6
    whats the dy/dx is it 2
  8. Mar 13, 2007 #7
    ok so i have to find the derivative of he P.T which is 2c=2a+2b
  9. Mar 13, 2007 #8
    And put the dy/dx on it dc/dt(2c) = da/dt(2a)+db/dt(2b)!
    but you have to calculate how much have they both travel in 2 hour and put into the equation and solve!
  10. Mar 13, 2007 #9
    ok so is this what it will look like c=2*60+2*25
  11. Mar 13, 2007 #10
    NO you have to calculate how much have they both travel in 2 hour

    dy/dx is the speed a b and c is distance~!
  12. Mar 13, 2007 #11
    so in 2 hours 60mi/h is 120 mi and 25 is 50 mi
  13. Mar 13, 2007 #12
    yep , and plug the speed and the distance into equation and solve and the third distance between you have to calculate that too. use the Pythagoras Theorem you have already found up there!
  14. Mar 13, 2007 #13
    so the distance is 170 mi
  15. Mar 13, 2007 #14
    so what would it look like
  16. Mar 13, 2007 #15
    would it look like this 2c=2*60+2*25

  17. Mar 14, 2007 #16


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    Science Advisor

    I am starting to wonder if you are serious! You cannot differentiate c2= a2+ b2 and get 2c= 2a+ 2b. For one thing, you have completely left out the question of which variable you are differentiating with respect to. Here the question is about how fast the distance between them, c, is increasing so you are trying to find dc/dt,, not c.
    As yaho8888 alrready told you, you need to differentiate both sides of c2= a2+ b2 with respect to x! Using the chain rule, that is 2c dc/dt= 2a da/dt+ 2b db/dt. Now what are a, b, c, da/dt, and db/dt after two hours?
    (And c, the distance between the cars after 2 hours, is NOT 170 miles.
  18. Mar 14, 2007 #17
    you have to solve for dc/dt~!
    The question ask you for speed! dx/dy means speed remember that.

    dc/dt(2c) = da/dt(2a)+db/dt(2b)
    first solve the distance between a and b in two hours use Pythagoras Theorem.
    (2*60mi/hr)^2+(2*25mi/hr)^2= c^2
    if you calculate that out c = 130miles

    you found the c plug into the equation~!
    dc/dt(2*130miles) = (60mi/hr)*(2*120mi) +(25mi/hr)*(2*50mi)
    then just solve for dc/dt~!
    which is 65mile/hr!
  19. Mar 15, 2007 #18
    oh, i thought i had to differentiate the pythagorean theorem.
  20. Mar 15, 2007 #19
    thanks man, how come when i plug in 25^+60^2=4225 square root of 4225 i get the same answer
  21. Mar 17, 2007 #20
    Man you have no Idea what calculus is!
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