Thread: inverse direct variation

Solve. Leave in simplest form.

1. The distance an object falls is directly proportional to the square of the time it has been falling. After 6 seconds it has fallen 1296 feet. How long will it take to fall 2304 feet?

2. x varies directly as the square of s and inversely as t. How does x change when s is doubled? When both s and t are doubled?

see http://www.mathhelpforum.com/math-he...ariations.html to see what "direct variation" and "inverse variation means"

Originally Posted by alwaysalillost

Solve. Leave in simplest form.

1. The distance on object falls is directly proportional to the square of the time it has been falling. After 6 seconds it has fallen 1296 feet. How long will it take to fall 2304 feet?

let d be the distance the object falls
let t be the time it takes to fall

then we have,
d = k(t^2)

when t = 6, d = 1296
=> 1296 = k(6^2)
=> k = 1296/36 = 36

so our relationship is:
d = 36t^2

when d = 2304, we have:

2304 = 36t^2
=> t^2 = 2304/36 = 64
=> t = sqrt(64)
=> t = 8

so it takes 8 seconds for the object to fall 2304 feet

2. x varies directly as the square of s and inversely as t. How does x change when s is doubled? When both s and t are doubled?

we have:
x = k(s^2) and x = k/t

if s is doubled:
x = k((2s)^2) = 4k(s^2)

so we see when s is doubled, x increases by a factor of 4

when t is doubled:
x = k/(2t) = (1/2)(k/t)
when t is doubled, x is halved. so if both s and t are doubled, x will increase by a factor of 4 and decrease by a factor of 2 at the same time. that means it will change by a factor of 4(1/2) = 2, so x will double

Thanks so much Jhevon!