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  1. #1 higher derivatives 
    Forum Ph.D. Heinsbergrelatz's Avatar
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    can anyone help me DERIVE this question, not just a simple substitution, thank you

    what is the nearest point on the curve , to the point (3,0)?

    just by looking at it, you might say it is (1,1) and it is that answer, but how on earth would you derive it? by using higher derivatives. something tells me that you must derive through the distance equation on the Cartesian plane, but can anyone help me solve this fundamental yet, tricky question?


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  3. #2  
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    you are right. Just write the general distance from a point on the curve to the given point and derive this expression.

    You get a square root, but that doesn't matter here. Since you are just looking for the minimum, and you know a distance can only be positive, you can just look for the minimum of the function beneath the square root.

    The root of that expression is your answer. It's always a good idea to check whether it's indeed a minimum with the second derivative, but in this case it is easy to see that's the only possibility.


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  4. #3  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    would you be generous, and try and show me the derivation please?

    is the distance equation in this situation go like this?;

    how would you get the exact set of co-ordinate (1,1) with the derivative under the root?
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    As you can see, you can just ignore the square root, because it moves to the denominator anyway and the roots are found with the numerator. So you have to find x for which:
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  6. #5  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    oo i see, thank you for the help, really appreciate it
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  7. #6  
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    Quote Originally Posted by Heinsbergrelatz
    would you be generous, and try and show me the derivation please?

    is the distance equation in this situation go like this?;

    how would you get the exact set of co-ordinate (1,1) with the derivative under the root?
    It will be easier if you minimize the square of the distance. The minima of the square will occur at the same point as the minima of the distance itself.
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  8. #7  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    It will be easier if you minimize the square of the distance. The minima of the square will occur at the same point as the minima of the distance itself.
    can you elaborate? please
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  9. #8  
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    It's what I meant when I said to ignore the square root and derive only the function beneath it.

    The minimum of the distance will be at the same point as the minimum of the distance squared, but the latter is easier to calculate (although the former isn't really that difficult either).
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  10. #9  
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    Quote Originally Posted by Heinsbergrelatz
    It will be easier if you minimize the square of the distance. The minima of the square will occur at the same point as the minima of the distance itself.
    can you elaborate? please
    Take the distance and multiply it times itself -- that is what "squaring" is.
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  11. #10  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    okay.... umm not much of a difference on the scale of difficulty, but thanks anyway
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  12. #11  
    . DrRocket's Avatar
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    Quote Originally Posted by Heinsbergrelatz
    okay.... umm not much of a difference on the scale of difficulty, but thanks anyway
    Actually, it should be quite a bit simpler.
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  13. #12  
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    Listen to Dr Rocket. As a matter of course it is almost always better to think in terms of distance^2 rather than distance with regard to maximization and minimization problems. I'll grant that in this particular instance, it's not that important. But for more complicated problems, it can save you a lot of computational headaches.
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  14. #13  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    i never really said im not listening to him, yes i get his approach but of course everyone has different views, and when i approached both ways, it doesnt really make a significant difference in terms of duration. Yes you are right about that, you can reduce its difficulty, but i prefer doing it the maximization and the minimization way.
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  15. #14  
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    I'd recommend the following method for solving absolute maxima and minima problems:

    First draw a diagram of the problem, labeling all relevant information.

    Than, set up a system of equations, making sure to include the constraint of the problem as one of the equations. Given and , a point is a critical point of if is not differentiable at or . From this information, calculate the critical points of the function of the system of equations, first expressing the constraint in terms of one variable and than substituting that into the function. After that, you must simply express the problem algebraically and use your function to solve for an open sentence.

    Consider the following example problem:

    find the rectangle with the least perimeter that has an area equal to 1500 square units.

    First of all, sketch a diagram with appropriate labels. Than, recall that the perimeter of a rectangle is , where P is perimeter, l is length, and w is width. In this problem, our constrant is that , by solving for w, our constraint is expressed as . We can substitute the previous expression for our equation, which is now , which is equal to . Now we can differentiate, and . Let's express the previous equation as over a common denominator: . Find our critical points now, that is solve for where , which turns out to be . Let's solve w now. , which is obviously . This means that a rectangle with the least perimeter that has an area equal to 1500 square units is a square. And in fact, using a method similar to the one that I just used, we can show that a square always has the lowest perimeter for its area as well with a comparison in relation to rectangles (and perhaps any other one-dimensional manifold).

    As you can see, the main point is that using a set algorithm and understanding how it works makes calculus problems, including those involving local and global maxima and minima rather easy.
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