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ta123

  • 4 years ago

At noon, ship A is 150 km west of ship B. ship A is sailing east at 35 km/h and ship B is sailing north at 25km/h. How fast is the distance between the ships changing at 4:00PM?

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  1. ta123
    • 4 years ago
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    Can someone explain how to do this problem step by step, please

  2. eseidl
    • 4 years ago
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    We need a function describing distance between the ships as a function of the time. The initial distance between the ships is 150km, which occurs at what I will call time t=0. I will call the East direction my positive x and North my positive y. Lets refer everything from the point of view ship A which we will call at rest. Then ship B is approaching A with a horizontal component and moving away from A with a vertical component. Calling d=distance between the ships, we have: \[d=\sqrt{(150-35t)^2+(25t)^2}\]

  3. ta123
    • 4 years ago
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    is there any way you could draw it?

  4. eseidl
    • 4 years ago
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    The draw function on the site is tough

  5. ta123
    • 4 years ago
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    also what would plug into the t variable, time, to get rid of the t

  6. ta123
    • 4 years ago
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    so you can't draw the figure?

  7. eseidl
    • 4 years ago
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    Now that we have our function we need to know how fast it changes with time. t=0 is noon here and t=4 would be 4:00pm. We find the derivative of this function (with respect to time) and then plug in t=4 to find out how fast the distance between the ships is changing at 4:00pm.

  8. eseidl
    • 4 years ago
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    i'll try to draw it...it's going to look brutal though hehe

  9. eseidl
    • 4 years ago
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    |dw:1318178380389:dw|

  10. ta123
    • 4 years ago
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    so we have to take the derivative of d=((150-35t)^2+(25t)^2)^(1/2), then plug in 4 for t, right?

  11. eseidl
    • 4 years ago
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    The rectangle on the left is ship A and the rectangle on the right is ship B. d=distance between them (which has contributions from the horizontal distance,x and the vertical distance, y) x=(150-35t) y=25t

  12. eseidl
    • 4 years ago
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    Yep

  13. eseidl
    • 4 years ago
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    let me know what you get

  14. eseidl
    • 4 years ago
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    I get a rate of change of about +21.4km/h at time t=4 (4:00pm). i.e., the distance between the ships is increasing by this amount at 4:00pm

  15. ta123
    • 4 years ago
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    well I got 1.379, can you show your step on how you got that number

  16. ta123
    • 4 years ago
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    maybe I didn't take the derivative right because those squares inside the square root was kinda confusing on how to take the derivative of those properly.

  17. eseidl
    • 4 years ago
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    Yeah, expand them first, then take derivative.

  18. eseidl
    • 4 years ago
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    You should get:\[\frac{d/dt(150-35t)^2+(25t)^2)}{2\sqrt{(150-35t)^2+(25t)^2)}}\]

  19. eseidl
    • 4 years ago
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    After you expand and take the derivative of the numerator you have you function of how the distance changes with time. Plug in t=4 to find the value at 4:00pm

  20. ta123
    • 4 years ago
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    wait did you use natural log to expand it?

  21. eseidl
    • 4 years ago
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    No. Just expand as usual and you get a polynomial on top. Take the derivative of this polynomial.

  22. ta123
    • 4 years ago
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    is that rule or something because I haven't seen that before except its look similar to natural log

  23. ta123
    • 4 years ago
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    like for example ln x=1/x

  24. eseidl
    • 4 years ago
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    The numerator should be:\[d/dt(22500-2(150)(35)t+1225t^2+625t^2)\]

  25. eseidl
    • 4 years ago
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    okay...going back to the distance function\[d=\sqrt{w}\]where w=all the stuff under the root sign Take the derivative with respect to t:\[=\frac{d/dt(w)}{2\sqrt{w}}\]

  26. eseidl
    • 4 years ago
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    I'm just using the chain rule here

  27. eseidl
    • 4 years ago
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    and the normal rule for taking derivatives of roots

  28. ta123
    • 4 years ago
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    if your using the chain for the numerator should you do outside of derivative first which wil be 2(150-35t). Then inside which would be 2(150-35t)(35)

  29. eseidl
    • 4 years ago
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    You dont need the chain rule for the numerator (although you can use it). expand it first then take derivative.

  30. eseidl
    • 4 years ago
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    consider the function\[y=\sqrt{x^3}\]To take the derivative you use substitution rule: let u=x^3 \[y=\sqrt{u}\]Now take derivative wrt x:\[dy/dx=(dy/du)(du/dx)=(\frac{1}{2\sqrt{u}})(3x^2)\]No get rid of u, using u=x^3:\[dy/dx=\frac{3x^2}{2\sqrt{x^3}}\]Your problem is analogous to this.

  31. eseidl
    • 4 years ago
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    gotta go. good luck. hope I earned a medal ;)

  32. ta123
    • 4 years ago
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    lastly did you use (a-b)^(2)=a^2-2ab+b^2 to expand the numerator?

  33. ta123
    • 4 years ago
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    I still didn't get your answer

  34. ta123
    • 4 years ago
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    when you comeback could you please show the format of your final polynomial function after taking the derivative of the numerator? mine was -10500+3700t/2((150-35t)^(2)+(25t)^(2))^(1/2), then I plug in 4 and got 200.99

  35. eseidl
    • 4 years ago
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    \[\frac{d/dt(22500-10500t+1225t^2+625t^2)}{2\sqrt{(150-35t)^2+(25t)^2}}\]

  36. eseidl
    • 4 years ago
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    This gives:\[\frac{(0-10500+3700t)}{2\sqrt{(150-35t)^2+(25t)^2}}\]

  37. eseidl
    • 4 years ago
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    Now plug in t=4:\[\frac{14800-10500}{2\sqrt{(10)^2+(100)^2}}\]Giving,\[\frac{4300}{200.99}\approx21.4\]

  38. ta123
    • 4 years ago
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    Thanks!

  39. eseidl
    • 4 years ago
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    np :)

  40. ta123
    • 4 years ago
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    hey esidl could you help me with one more problem?

  41. eseidl
    • 4 years ago
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    sure

  42. ta123
    • 4 years ago
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    A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate 1 m/s, how fast is the boat approaching the dock when it is 8 m from dock?

  43. eseidl
    • 4 years ago
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    Ok. We have another triangle here. The hypotenuse is shrinking at a rate of 1 m/s. The height of the triangle is constant at 1m. We want to know how fast the length, x, of the triangle is shrinking when x=8m.

  44. ta123
    • 4 years ago
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    ok

  45. eseidl
    • 4 years ago
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    \[c^2=x^2+y^2\]But y=constant=1. So,\[c=\sqrt{x^2-1}\]Differentiating:\[dc/dt=\frac{(2x)(dx/dt)}{\sqrt{x^2-1}}\]Okay, we have a lot of stuf here now. Keeps your knowns and unknowns in mind. dc/dt is known, it is the rate the hypotenuse is shrinking i.e., 1m/s. We want dx/dt. What about x? Well, we want dx/dt when x=8 so this is known. We have:\[1=\frac{8(dx/dt)}{\sqrt{63}}\]So,\[dx/dt=\sqrt{63}/8\]

  46. eseidl
    • 4 years ago
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    oooops I made a boo boo. Forgot the factor of 2 in the denominator when I differentiated. But the answer take this into account, just a typo

  47. ta123
    • 4 years ago
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    its ok, so its suppose dx/dt=squrt(63)/16 on the denominator? Also could you draw a figure again.

  48. eseidl
    • 4 years ago
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    no the answer is still dx/dt=sqrt63/8 I just didn't type the 2 in the denominator

  49. ta123
    • 4 years ago
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    2X8=18

  50. ta123
    • 4 years ago
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    oops! I meant 16

  51. eseidl
    • 4 years ago
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    |dw:1318192994858:dw|

  52. eseidl
    • 4 years ago
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    rectangle is the boat, circle is the pulley, 1m above the level of the boat. distance is the horizontal line

  53. eseidl
    • 4 years ago
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    again, dc/dt should have been:\[dc/dt=\frac{2x(dx/dt)}{2\sqrt{x^2-1}}\]cancel the 2 on top with the one on the bottom. everything else works out. answer is still the one given.

  54. ta123
    • 4 years ago
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    ok thanks so much

  55. eseidl
    • 4 years ago
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    no worries...hope is correct (should be!)

  56. ta123
    • 4 years ago
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    when you take the derivative of inside of square root should be 2x on the denominator

  57. ta123
    • 4 years ago
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    unless that 2 on denominator was suppose to represent the 1/2 and the numerator represented taking the derivative of the inside of the squrt(x^2-1) in the denominator which would be 2X

  58. eseidl
    • 4 years ago
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    I made an error above; c^2=x^2+y^2=x^2+1 I put c^2=x^2-1 for some unknown reason lol answer should be sqrt(65)/8

  59. ta123
    • 4 years ago
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    ok, Thanks again

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