1. ## 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%.

2. 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

3. 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 $w'(0)=-\frac{2mgR^2}{R^3}=-\frac{2mg}{R}$

$w(x) \approx w(0)+\frac{1}{1!}w'(0)x$
$w(x) \approx mg - \frac{2mg}{R}x$
$w(x) \approx mg \left ( 1 - \frac{2x}{R} \right )$