Well, according to Mr. Taylor:Originally Posted by bobby77
Thus f(2) = 98/3 miles.
-Dan
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%.
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:Originally Posted by bobby77
and
and
So the first order Taylor expansion about x = 0 reads:
(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