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Thread: Related Rate Problem

  1. #1 Related Rate Problem 
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    Problem: As a man walks away from a 12-ft lamppost, the tip of his shadow moves twice as fast as he does. How tall is the man?

    I get 8ft. tall for the man. However, I took out a ruler and pencil, did a rough sketch, and now I doubt my answer. I also doubt the method I used to do this problem. I just picked two arbitrary numbers to describe the ROC of the man's speed, and the ROC of the shadows speed. I made the ROC of the shadow's speed double the other ROC, and I get 8 ft., but I doubt that's correct.

    Also, how would I verify my answer with just a simple sketch or two.


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  3. #2  
    Forum Radioactive Isotope MagiMaster's Avatar
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    Just in my head, and using similar triangles, I get a different answer.

    I can't say how you'd verify it, since I don't know what you've tried.


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  4. #3  
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    This is how I am solving the problem.

    The pole is 12 feet tall. Call the distance from the pole to the man x, and call the distance from the man to the tip of his shadow k.

    By similar triangles

    12/(x + k) = h/k (h is man's height)

    Cross multiply

    12k =hx + hk

    Differentiate

    12(dk/dt) + h(dx/dt) + h(dk/dt)

    After this point I am picking an arbitrary value for (dx/dt) and (dk/dt), but making sure that the value I pick for dk/dt is twice as big as dx/dt. (his shadow moves twice as fast as he does)

    I get 8 ft. for the mans height after this, but I honestly think that it should be 6 feet. What kind of textbook would have an 8ft. tall man as the answer to a question. The answers are usually realistic.
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  5. #4  
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    You don't even need calculus for this one. At t=0 the man of height h is under the lamp and his shadow is length 0. At t=1 he is at distance x and his shadow is at 2x, since it was moving twice as fast as he was for the whole time t. By similar triangles h/(2x-x)=12/2x
    h=12/2=6
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  6. #5  
    Forum Radioactive Isotope MagiMaster's Avatar
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    Quote Originally Posted by SteveC
    This is how I am solving the problem.

    The pole is 12 feet tall. Call the distance from the pole to the man x, and call the distance from the man to the tip of his shadow k.

    By similar triangles

    12/(x + k) = h/k (h is man's height)

    Cross multiply

    12k =hx + hk

    Differentiate

    12(dk/dt) + h(dx/dt) + h(dk/dt)

    After this point I am picking an arbitrary value for (dx/dt) and (dk/dt), but making sure that the value I pick for dk/dt is twice as big as dx/dt. (his shadow moves twice as fast as he does)

    I get 8 ft. for the mans height after this, but I honestly think that it should be 6 feet. What kind of textbook would have an 8ft. tall man as the answer to a question. The answers are usually realistic.
    Your problem here is that x and k (and dx/dt and dk/dt) will be equal, since the distance from the pole to the tip of the shadow will be twice the distance from the pole to the man. Think about what would happen if they were moving the same speed. The shadow would always be right under him.
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  7. #6  
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    Thanks Harold, I understand what you wrote.


    Your problem here is that x and k (and dx/dt and dk/dt) will be equal, since the distance from the pole to the tip of the shadow will be twice the distance from the pole to the man. Think about what would happen if they were moving the same speed. The shadow would always be right under him.
    Well, I am still slightly confused about solving the problem with calculus, because assigning the same value to dx/dt and dk/dt is not making sense to me. If the man is moving at 2m/s, then his shadow needs to move at 4m/s. Thats why I'm making one twice as large as the other.

    However, when I assign the same value to both I do get a 6ft. tall man, which I definitely thought was the right answer.
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  8. #7  
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    Remember you defined k as the distance from the man to his shadow.
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