两道AP微积分题..1 the volume of a cone of radius r and height h is
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两道AP微积分题..
1 the volume of a cone of radius r and height h is given by V=1/3πr^2 if the radius increase at a constant rate of 1/2 centimeter per second,at what rate,in cubic centimeters per second,is the volume increasing when the height is 9 centimeters and the radius is 6 centimeters?
2 the volume of a cylindrical tin can with a top and a bottom is to be 16π cubic inches.If a minimum amount of tin is to be used to construct the can,what must be the height,in inches,of the can?
麻烦用微积分方法解啦...我是帮别人问所以麻烦详细一点 感激不尽!
1 the volume of a cone of radius r and height h is given by V=1/3πr^2 if the radius increase at a constant rate of 1/2 centimeter per second,at what rate,in cubic centimeters per second,is the volume increasing when the height is 9 centimeters and the radius is 6 centimeters?
2 the volume of a cylindrical tin can with a top and a bottom is to be 16π cubic inches.If a minimum amount of tin is to be used to construct the can,what must be the height,in inches,of the can?
麻烦用微积分方法解啦...我是帮别人问所以麻烦详细一点 感激不尽!
1) V = (1/3)πr^2 h
如果高度不变,只是半径增加,那么方程两边对时间求导得,
V' = (2/3)πrr' h = (2/3)π(6)(1/2)*9 = 18π cm^3/sec
2) A = 2π r^2 + 2π r h
限制条件:V = π r^2 h = 16π
=> h = 16/r^2 代入面积表达式,
A = 2π r^2 + 2π 16/r
对r求导,并令A' = 0,
2r - 16/r^2 = 0
r = 2 inches
h = 16/r^2 = 4 inches
美国大学理事部认证AP微积分AB和微积分BC教师
如果高度不变,只是半径增加,那么方程两边对时间求导得,
V' = (2/3)πrr' h = (2/3)π(6)(1/2)*9 = 18π cm^3/sec
2) A = 2π r^2 + 2π r h
限制条件:V = π r^2 h = 16π
=> h = 16/r^2 代入面积表达式,
A = 2π r^2 + 2π 16/r
对r求导,并令A' = 0,
2r - 16/r^2 = 0
r = 2 inches
h = 16/r^2 = 4 inches
美国大学理事部认证AP微积分AB和微积分BC教师
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