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Math Help - Weird problem

  1. #1
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    Post Weird problem

    Show that for small values of h, sqroot(4+h) is approx equal to 2 + (h/4)
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  2. #2
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    Quote Originally Posted by mightymax View Post
    Show that for small values of h, sqroot(4+h) is approx equal to 2 + (h/4)
    This is a linearzation problem.

    Consider the function,
    \sqrt{4+h}
    And its tangent line at point h=0.
    We know that a small change in the actual function \Delta h is approximately equal to a small change in the tangent line dh.
    That is,
    \Delta h\approx dh
    But the derivative is,
    (Hint: \sqrt{4+h}=(4+h)^{1/2})

    \frac{4+h}{2}=2+(h/2)
    (The reason why we look at the derivative is because the derivative is the tangent line. So a small change in the derivative is a small change in the function).
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  3. #3
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    Thanks for the help

    but I dont really understand how to answer the question. And do I look at the derivative of the sqroot(4+h) function? and if so what do i do next?
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  4. #4
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    Quote Originally Posted by mightymax View Post
    Show that for small values of h, sqroot(4+h) is approx equal to 2 + (h/4)
    Hello,

    I'll try to explain what TPH meant.

    You've got a function v(h)=\sqrt{4+h} which gives for any h the value of the square-root. You know that v(0) = 2.

    Now consider the tangent line at this point. For small amounts of h the tangent line and the curve of the function v are practivally the same. (see attachment. I've enlarged a small surrounding of (0, v(0)))

    To calculate the tangent line you need the slope of the line which is the same as the gradient of v for h = 0:
    v'(h)=\frac{1}{2}\cdot \frac{1}{\sqrt{4+h}}. Thus v'(0)=\frac{1}{4}

    Therefore the tangent line has the equation:
    t(h)=\frac{1}{4}\cdot h +2

    Because t(h) is the same as v(h) for small values of h, you can write (or calculate): \sqrt{4+h}\approx \frac{1}{4}\cdot h +2

    EB
    Attached Thumbnails Attached Thumbnails Weird problem-wrzl_tangkompl.gif  
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