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Thread: interception

  1. #1
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    interception

    a cruiser is 12 km due west of a destroyer . the cruiser is travelling in a direction 60 degrees N of E at 30 km hr^-1 and the destroyer is travelling at 15 km hr^-1. find the direction the destroyer must travel in order to get as close as possible to the cruiser

    not sure how to go about this.
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  2. #2
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    Re: interception

    Let's use vectors. The cruiser's initial position is $0\hat{i}+0\hat{j}$. The destroyer's original position is $12\hat{i}+0\hat{j}$.

    The cruiser's position after $t$ hours will be:

    $30t\cos 60^\circ \hat{i} + 30t\sin 60^\circ \hat{j}$

    The destroyer's position after $t$ hours will be:

    $12+15t\cos \theta \hat{i} + 15t\sin \theta \hat{j}$

    The distance between them will be:

    $\sqrt{(12+15t\cos \theta - 15t)^2+(15t\sin \theta - 15t\sqrt{3})^2}$

    You are minimizing with respect to two variables.
    Last edited by SlipEternal; Jul 5th 2017 at 02:10 PM.
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  3. #3
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    Re: interception

    I believe that the destroyer should travel at 60 degrees North of West. I found that by taking the partial derivative with respect to $t$, setting it equal to zero, solving for $t$, plugging it back into the distance formula, then since I had a function of one variable, I minimized with respect to $\theta$ and got $\theta = \dfrac{2\pi}{3}$, which is $120^\circ$, or 60 degrees North of West. However, this is not a guaranteed minimum. This may not even be a local minimum. There are certain conditions that must be met in order for this to be a minimum. I get that $t=24\text{ mins}$ gives the minimum by this method. That would be 6km apart after 24 minutes. I am not sure if you can get them any closer together, but I certainly did not prove that this was, indeed, the minimum.
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  4. #4
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    Re: interception

    that is the right answer but im not sure how you get it. when i multiply out the expression you gave for the distance between them i get
    √{ 225t^2(cosa)+t^2*(225cos^2 a)+225t^2+(360cosa)*t -360t+144 +225t^2(sin^2 a)-450√3*t^2(sina)+675t^2 }
    I am not sure how to differentiate this as there is 2 variables

    how did you do it in your answer?
    thanks
    Last edited by markosheehan; Jul 6th 2017 at 12:50 AM.
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  5. #5
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    Re: interception

    I treated $\theta$ as a constant and differentiated WRT $t $.
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  6. #6
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    Re: interception

    Did I multiply it out correctly? Should I use the chain rule and not multiply it out.?
    I still not working out for me
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  7. #7
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    Re: interception

    I used Wolframalpha to find derivatives. The way I did it makes it a very nontrivial problem. Perhaps there is a geometric interpretation that simplifies it, but it did not jump out at me.
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  8. #8
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    Re: interception

    does any one else know how to do this problem?
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  9. #9
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    Re: interception

    Do not multiply out. Just use the chain rule.

    $D_t = \dfrac{1}{2}\left(2(12+15t \cos \theta - 15t)(15\cos \theta - 15)+2(15t\sin \theta - 15t\sqrt{3})(15\sin \theta - 15\sqrt{3})\right)\left((12+15t\cos \theta - 15t)^2+(15t\sin \theta - 15t\sqrt{3})^2\right)^{-1/2} = 0$

    $(12+15t\cos \theta - 15t)(15\cos \theta - 15) + (15t\sin \theta - 15t\sqrt{3})(15\sin \theta-15\sqrt{3}) = 0$

    $(12+15t\cos \theta - 15t)(15\cos \theta - 15)+t(15\sin \theta - 15\sqrt{3})^2 = 0$

    $12(15\cos \theta-15)+t(15\cos \theta-15)^2+t(15\sin \theta-15\sqrt{3})^2 = 0$

    $t = \dfrac{12(15-15\cos \theta)}{(15\cos \theta-15)^2+(15\sin \theta-15\sqrt{3})^2} = \dfrac{4}{5}\dfrac{1-\cos \theta}{(\cos \theta-1)^2+(\sin \theta-\sqrt{3})^2}$

    Plug that in for $t$ in the original equation for the distance and then minimize with respect to $\theta$.
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  10. #10
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    Re: interception

    You get

    $12\sqrt{\left(1-\dfrac{\sin^2 \theta}{(1-\cos \theta)^2+(\sqrt{3}-\sin \theta)^2}\right)^2+\left(\dfrac{(\sin \theta-\sqrt{3})(1-\cos \theta)}{(1-\cos \theta)^2+(\sqrt{3}-\sin \theta)^2}\right)^2}$

    Differentiate that with respect to $\theta$ and set it equal to zero. You should get $\theta = \dfrac{2\pi}{3}$.
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  11. #11
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    Re: interception

    Does not work with partial derivatives...
    Last edited by SlipEternal; Jul 6th 2017 at 08:39 AM.
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    Re: interception

    thanks for doing this work but my knowledge of calculus is not that advanced
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  13. #13
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    Re: interception

    Then there is probably something with circles and triangles that will yield the same result. I yield the thread to those more knowledgeable in such things.
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  14. #14
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    Re: interception

    Had to recall plotting relative motion on a maneuvering board from my Navy days ...

    Cruiser is at the origin, relative motion of the destroyer is the vector $\color{red}{\Delta r}$ in the attached diagram.
    Attached Thumbnails Attached Thumbnails interception-cpa_vector.jpg  
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  15. #15
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    Re: interception

    ive tired to solve it another way. i should of mentioned its a question from a relative velocity chapter.

    Vc =15sinai+15sinaj

    Vd=15i+15√3j


    Vcd =(15cosa-15)i+(15sina-15√3)
    if you look at skeeter diagram this is this represents the line labelled △r

    in my diagram we want x to be a minimum so tan theta should be a minimum

    to minimise tana : d(tana)/d theta =0

    so i use the chain rule

    tan a = (15sina-15√3)/(15cosa-15)

    so i simply this and in the chain rule let u=3sina-3√3 and v=3cosa-3

    so eventually i get cosa+√3sina=1

    i am doing something wrong but i know this method does work as i have seen it work on similar questions.
    where am i going wrong or could someone do out the question in the way i am trying to do it correctly?
    Last edited by markosheehan; Jul 7th 2017 at 12:51 PM.
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