急!(F4 Maths) Inverse Variation

Willow

2010-10-29 5:39 am

1) If (x+y) varies inversely as (1/x + 1/y), show that xy varies directly as (x^2 + y^2).

2) It is given that p is inversely proportional to q, where both p and q are positive. When q=4,p=15. Find the range of possible values of q when p ≤ 5.

3)It is given that x varies inversely as y, and y varies inversely as z.
a) Show that x varies directly as z.
b) If y=2 when z=3; and x=4 when z=1/3, find an equation connecting x and y.

Thank you very much!! (:

更新1:

1) If (x+y) varies inversely as (1/x + 1/y), show that a) xy varies directly as (x^2 + y^2). b) (x+y)^2 varies directly as xy.

回答 (1)

知識就是力量

2010-10-29 6:03 am

✔ 最佳答案

1. x + y = k/(1/x + 1/y) , where k is non-zero constant.
x + y = k/[(x+y)/xy]
x + y = kxy/(x+y)
(x+y)^2 = kxy
x^2 + 2xy + y^2 = kxy
x^2 + y^2 = kxy - 2xy
x^2 + y^2 = (k-2)xy
(x^2+y^2)/(k-2) = xy
Since 1/(k-2) is a constant,
xy varies directly as (x^2+y^2)

2. p = k/q , where k is a non-zero constant.
Put q = 4 and p = 15,
15 = k/4
k = 60
Hence p = 60/q
When p ≤ 5,
60/q ≤ 5
60 ≤ 5q (where q is positive)
12 ≤ q
q ≥ 12

3.(a) x = a/y , where a is a non-zero constant ---(1)
y = b/z , where b is a non-zero constant ---(2)
Put (2) into (1),
x = a/(b/z)
x = az/b
Since a/b is a constant,
x varies directly as z.
(b) Put y = 2 and z = 3 into (2),
2 = b/3
b = 6
Put x = 4 and z = 1/3 into x = az/b,
4 = a(1/3)/b
4 = (a/3)/6
a = 72
Hence the equation is x = 72a/6, i.e. x = 12z

2010-10-28 22:04:59 補充：
Last line should be Hence the equation is x = 72z/6, i.e. x = 12z

2010-10-28 22:28:33 補充：
For 1(b)
x + y = k/(1/x + 1/y) , where k is non-zero constant.
x + y = k/[(x+y)/xy]
x + y = kxy/(x+y)
(x+y)^2 = kxy
Since k is a constant,
(x+y)^2 varies directly as xy

參考: Knowledge is power.

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