標題:
f.4 maths variation~
發問:
it is given that z varies directly as x^(1/3) when y is constant and varies inverserly as y^2 when x is constant. show taht z^3 propotional to x^7 when x propotional to 1/y
Let z = k ?x / y2 , where k is a constant. then z3 = k3 x / y? ...... (*) when x propotional to 1/y , let x = k" (1/y) , where k" is a constant. i.e. y = k" / x , sub. into (*) : z3 = k3 x / (k" / x)? z3 = (k3 / k") x? , where (k3 / k") is a constant since k and k" is a constant. ∴ z3 propotional to x? when x propotional to 1/y. 2012-04-27 00:12:30 補充： The 2nd last line , (k3 / k") should be (k3 / k"^6) : z3 = (k3 / k"?) x? , where (k3 / k"?) is a constant since k and k" is a constant. Sorry for my mistakes. 2012-04-28 09:42:07 補充： z3 = (k3 / k") x? should be z3 = (k3 / k"^6) x?
其他解答:
f.4 maths variation~
發問:
it is given that z varies directly as x^(1/3) when y is constant and varies inverserly as y^2 when x is constant. show taht z^3 propotional to x^7 when x propotional to 1/y
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最佳解答:Let z = k ?x / y2 , where k is a constant. then z3 = k3 x / y? ...... (*) when x propotional to 1/y , let x = k" (1/y) , where k" is a constant. i.e. y = k" / x , sub. into (*) : z3 = k3 x / (k" / x)? z3 = (k3 / k") x? , where (k3 / k") is a constant since k and k" is a constant. ∴ z3 propotional to x? when x propotional to 1/y. 2012-04-27 00:12:30 補充： The 2nd last line , (k3 / k") should be (k3 / k"^6) : z3 = (k3 / k"?) x? , where (k3 / k"?) is a constant since k and k" is a constant. Sorry for my mistakes. 2012-04-28 09:42:07 補充： z3 = (k3 / k") x? should be z3 = (k3 / k"^6) x?
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