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Related Rate - Rate of The Change in Distance of Two Ships

  1. Dec 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Ship A is 100 km west of ship B. Ship A is moving at 35 km/h (west), ship B is moving at 25 km/h (north). Find the rate of change of the distance between the two ships after 4 hours.


    2. Relevant equations



    3. The attempt at a solution

    the distance between the two ships is shown by

    [itex]a^{2}+b^{2}=c^{2}[/itex]

    Am I right to assume:

    [itex][\frac{ds_{a}}{dt}]^{2}+[\frac{ds_{b}}{dt}]^{2}=[\frac{ds_{c}}{dt}]^{2}[/itex]

    So, I need to find the rate of change of C? After 4 hours, the ships are 260 km apart.

    I need a helping hint.
     
    Last edited: Dec 23, 2012
  2. jcsd
  3. Dec 23, 2012 #2

    Dick

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    Can you write the distance apart as a function of the time t? Your derivative formula isn't correct.
     
  4. Dec 23, 2012 #3
    You mean like d=vt? I'm highly confused. I'm not sure how to relate the total distance C with d=vt.
     
  5. Dec 23, 2012 #4

    Dick

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    Pick coordinates. Suppose A is at (0,0) at time t=0 and B is at (100,0). Where are they at another time? You figured out the distance at t=4. Just do it for any t.
     
  6. Dec 23, 2012 #5
    at time t, they're at [itex]d_{a}=35t[/itex]

    [itex]d_{b}=25t[/itex].
     
  7. Dec 23, 2012 #6

    Dick

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    You can't describe locations in the plane with a single number. Use coordinates.
     
  8. Dec 23, 2012 #7
    Ok, I'm not following. Let me try another example.

    A pile of gravel is being poured from a belt at a rate of 20 m cubed per min. The base will always be half the height. How fast the the pile changing when h=10m?

    So, you take the equation for the volume of a cone. take the derivative with respect to h. And we also have the change in volume with respect to time. How do we relate all this to time?

    We have [itex]\frac{dV}{dh}=\frac{1}{4}\pi h^{2}[/itex]

    And: [itex]\frac{dV}{dt}=20m^{3}/s[/itex]

    Now what? We have these two derivatives. This is the part I don't understand.
     
  9. Dec 23, 2012 #8

    Dick

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    Ok. How about using [itex]\frac{dV}{dh} \frac{dh}{dt}=\frac{dV}{dt}[/itex]? That's the chain rule. You can use the same approach for the first one if you clearly define what a, b and c are.
     
  10. Dec 24, 2012 #9

    symbolipoint

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    Assume at the start that ship B is at a reference point.
    At any time t, let a be distance of ship A from the reference point, and let b equal the distance of ship B from the reference point.

    [itex]\[
    a = 35t + 100,\quad b = 25t
    \][/itex]

    Now, let h be the distance between the two ships at any time t, so that, by pythagorean theorem,
    [tex] \[
    \begin{array}{l}
    h^2 = a^2 + b^2 \\
    h^2 = \left( {35t + 100} \right)^2 + \left( {25t} \right)^2 \\
    \end{array}
    \]
    [/tex]
    and using suitable steps,

    [tex]
    \[
    h = 5\sqrt {74t^2 + 280t + 400}
    \]
    [/tex]


    You should be able to understand, and then continue and finish the exercise from this.
     
  11. Dec 24, 2012 #10
    [itex]\frac{dV}{dh} \frac{dh}{dt}=\frac{dV}{dt}[/itex] = [itex]\frac{1}{4}\pi h^{2}\frac{dh}{dt}[/itex] = 20

    At this point we plug in 10 for h? so that means [itex]\frac{dh}{dt}[/itex]=.255?
     
  12. Dec 24, 2012 #11

    Dick

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    Sure.
     
  13. Dec 24, 2012 #12
    So that begs the question, WHY is this correct? I can't find anything to explain it clearly. I don't know why you just plug in dh/dt
     
  14. Dec 24, 2012 #13

    symbolipoint

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    Astrum, what are you asking in post #7? How fast is height, h changing when it is 10 meters?

    Let me try.

    b is base, r is radius, given base is half of height h. Radius is half of b.
    [tex]
    \[
    \begin{array}{l}
    {\rm }b = {\textstyle{1 \over 2}}h \\
    {\rm }r = {\textstyle{1 \over 2}}b \\
    \end{array}
    \]

    [/tex]

    [tex]
    \[
    V = \frac{\pi }{{3 \cdot 8}}h^3
    \]
    [/tex]
    V and h are functions of time. V is a function of h, but both of them are functions of time. That is why you use Chain Rule when you find derivative of the right-hand side.

    note: may need to check my V result carefully in case arithmetic mistakes; but I believe they're good.
    EDIT: Now the formatting in the post is not working. I'll try in the next post.
     
    Last edited: Dec 24, 2012
  15. Dec 24, 2012 #14

    symbolipoint

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    base b is half of height h,
    radius is half of b,

    [tex]\[
    \begin{array}{l}
    b = \frac{1}{2}h \\
    r = \frac{1}{2}b \\
    \end{array}
    \]
    [/tex]

    [tex] \[
    V = \frac{\pi }{{3 \cdot 8}}h^3
    \]
    [/tex]
     
  16. Dec 24, 2012 #15
    Yes. I'm just confused as to how you can get dh/dt in there. My confusion comes form not understand implicit differentiation 100%.
     
  17. Dec 24, 2012 #16

    symbolipoint

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    Implicit differentiation is unnecessary. You can start (or continue) with the formula for volume of the cone, and this is a function of height, h. You need a bit of substitution as I showed to put the expression in the form as I have.

    THEN,
    1. Differentiate V with regard to t,
    2. Use Chain Rule to differentiate h with regard to t, because h is a function of t.
     
  18. Dec 24, 2012 #17

    Dick

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    Like I was saying in post 8, implicit differentiation is really just using the chain rule. Do you understand the chain rule?
     
  19. Dec 24, 2012 #18

    symbolipoint

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    I am possibly unclear about what is and is not implicit differentiation. I have not reviewed it specifically, so I have learned (or relearned) how to find derivative without deciding if I am doing so "implicitly" or not.

    Simply, V and h both are function of t; so one differentiates both sides of the formula with respect to t.
     
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