# inverse direct variation

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• May 13th 2007, 05:42 PM
alwaysalillost
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?
• May 13th 2007, 06:59 PM
Jhevon
see http://www.mathhelpforum.com/math-he...ariations.html to see what "direct variation" and "inverse variation means"

Quote:

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

Quote:

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
• May 14th 2007, 05:45 AM
alwaysalillost
Thanks so much Jhevon! :)