Question

The distance, in feet, a stone drops in tt seconds is given by d(t)=16t2d(t)=16t2. The depth of a hole is to be approximated by dropping a rock and listening for it to hit the bottom. Assume that time measurement is accurate to 2 / 10th of a second. Use a linear approximation to estimate an upper bound for the propagated error if the measured time is

Answer 1

Define the distance through the equation

$x=16t^2$

The distance against the time is given by

$\Delta x = 32 t \Delta t$

$\Delta t = \frac{2}{10}s$

When t=2s,

$\Delta x = 32(2)\frac{2}{10}$

$\Delta x = 12.8$

There is a error in 12.8feet.

b) When t=5sec

$\Delta x= 32t\Delta t$

$\Delta x= 32(5)*\frac{2}{10}$

$\Delta x = 32$

There is a error in 32 feet.