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Math Help - please hekp me to solve these

  1. #1
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    please hekp me to solve these

    1>Suppose that a plane is at location f(0) = 10 miles with velocity of f(0) = 10 miles/min, acceleration f(0) = 2 miles/min2 and f(0) = -1 miles/min3. Predict the location of the plane at time t = 2.


    2>The weight (force due to gravity) of an object of mass m and altitude x miles above the surface of the earth is w(x) = (mgR^2)/(R+x)^2, where R is the radius of the earth and g is the acceleration due to gravity. Show that w(x) is approximately mg(1-2x/R). Estimate how large x would need to reduce the weight by 10%.
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  2. #2
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    Quote Originally Posted by bobby77
    1>Suppose that a plane is at location f(0) = 10 miles with velocity of f(0) = 10 miles/min, acceleration f(0) = 2 miles/min2 and f(0) = -1 miles/min3. Predict the location of the plane at time t = 2.
    Well, according to Mr. Taylor:
    f(x)=f(0)+\frac{1}{1!}f'(0)t+\frac{1}{2!}f''(0)t^2  +\frac{1}{3!}f'''(0)t^3
    f(x)=10+10t+t^2-\frac{1}{6}t^3

    Thus f(2) = 98/3 miles.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by bobby77
    2>The weight (force due to gravity) of an object of mass m and altitude x miles above the surface of the earth is w(x) = (mgR^2)/(R+x)^2, where R is the radius of the earth and g is the acceleration due to gravity. Show that w(x) is approximately mg(1-2x/R). Estimate how large x would need to reduce the weight by 10%.
    We are dealing with the function w(x) where x is small. So consider the Taylor expansion for x near 0. Since x is small, we know that x >> x^2 and we need only consider the first two terms (the first term is a constant that doesn't depend on x) in the series:
    w(x)=\frac{mgR^2}{(R+x)^2} and w(0)=\frac{mgR^2}{R^2}=mg
    w'(x)=-\frac{2mgR^2}{(x+R)^3} and w'(0)=-\frac{2mgR^2}{R^3}=-\frac{2mg}{R}

    So the first order Taylor expansion about x = 0 reads:
    w(x) \approx w(0)+\frac{1}{1!}w'(0)x
    w(x) \approx mg - \frac{2mg}{R}x
    w(x) \approx mg \left ( 1 - \frac{2x}{R} \right )

    (This expansion is the reason that Intro Physics students are taught that the gravitational field of the Earth is constant. It is nearly constant for x small enough to neglect the linear term in the expansion above. That is to say for x small enough to neglect compared to the radius of the Earth: objects very near the Earth's surface.)

    -Dan
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