A post is being driven into the ground by a mechanical hammer. How many blows does it take to drive the post at least 70cm?
The distance it is driven by the first blow is 8 cm. Subsequently, the distance it is driven by each blow is 9/10 of the distance it was driven by the previous blow.
(a) The post is to be driven a total distance of at least 70 cm into the ground. Find the smallest number of blows needed.
(b) Explain why the post can never be driven a total distance of more than 80 cm into the ground.
The distance the post is driven each time forms a geometric series
The formula for the sum of a geometric series is given by
`S_n = (a(1-r^n))/(1-r)`
We want to find the minimum integer value of`n`that gives `S_n > 70`
`implies ` `(8(1-r^n))/(1-r) > 70`
`implies` `8(1-r^n) > 70(1-r)`
`implies` `8r^n < 8 -70(1-r)`
`implies` `r^n < 1-(70/8)(1-r)`
`implies` `n > (log(1-(70/8)(1-r)))/logr` (remember log(r) is negative because r<1)
`implies` `n > log(1/8)/(log(9/10))`
`implies` `n > 19.7`
The minimum number of blows is therefore 20
The limiting sum of the series (the sum as `n ->oo`) is given by the formula
`S = a/(1-r)`
This is because `r^n -> 0` as `n -> oo` so `(1-r^n) -> 1`
`S = 8/(1/10) = 80`cm
a) minimum number of blows = 20
b) 80cm is the limiting sum of the geometric series