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Math Help - instantaneous rate of change problem?

  1. #1
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    instantaneous rate of change problem?

    Hello everyone having some troubles with maths.

    I am a stationary object 150m below the waters surface.
    I observe an object that is moving towards the waters surface directly above me.
    I observe that the distance between me and the object is decreasing at 5m/s when it is 250m away from me.
    The objects speed is constant.
    Find the objects speed.

    I cannot figure out how to solve this problem. I have looked around but always get stuck. If possible I would really appreciate it if someone can show me how to solve problems of this type. You don't have to give me answers to my question specifically. I'm more concerned with the method.
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  2. #2
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    Re: instantaneous rate of change problem?

    This doesn't really make sense. If the object is approaching the surface of the water at 5m/s and its speed is constant then why are you asked to find its speed? The inclusion of water in the problem is also unnecessary unless you are dealing with the refractive index and the bending of light which makes the apparent speed 5 m/s but the actual speed different.
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  3. #3
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    Re: instantaneous rate of change problem?

    Hello,

    Sorry I should've said that I found out how to do it a little while ago (It involved implicit differentiation).

    However the object was not approaching the water's surface at 5m/s but me, who is 150m below the water's surface. Basically it's velocity in the X-direction was 5m/s (When the distance between me and the object was 250m) but it was not travelling strictly in the X direction but also in the Y direction and hence it's actual speed is different.

    my attempt at an image
    instantaneous rate of change problem?-maths-prob.png
    Last edited by ldwilson; March 21st 2014 at 03:31 AM.
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: instantaneous rate of change problem?

    OK, so the object is at the same depth as yuo, and is headed for a point that on teh surface and is directly above you. Hence the ratio of the object's vertical motion to its horizontal motion is 150/250 = 3/5, and thus its vertical velocity is 3 m/s. Its total velocity is therefore  \sqrt {v_x^2 + v_y^2} = \sqrt {5^2 + 3^2} = \sqrt {34} m/s.
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