题目
(20)(本题满分11分)-|||-已知曲线 :y=dfrac (4)(9)(x)^2(xgeqslant 0), 点O(0,0),点A(0,1).设P是L上的动点,S是直线OA与直-|||-线AP及曲线L所围图形的面积,若P运动到点(3,4)时沿x轴正向的速度是4,求此时S关-|||-于时间t的变化率.
题目解答
答案
解：
解析
步骤 1：确定S的表达式
设P点坐标为(x,y)，则直线OA与直线AP及曲线L所围图形面积为
$S=\dfrac {1}{2}\times (1+\dfrac {4}{9}{x}^{2})x-{\int }_{0}^{x}\dfrac {4}{9}{t}^{2}dt$
步骤 2：计算积分
$S=\dfrac {1}{2}\times (1+\dfrac {4}{9}{x}^{2})x-{\int }_{0}^{x}\dfrac {4}{9}{t}^{2}dt$
$=\dfrac {1}{2}x+\dfrac {2}{9}{x}^{3}-\dfrac {4}{27}{x}^{3}$
$=\dfrac {1}{2}x+\dfrac {2}{27}{x}^{3}$
步骤 3：求导数
由题设知在点(3,4)处 $\dfrac {dx}{dt}=4$ ，又由 $S=\dfrac {1}{2}x+\dfrac {2}{27}{x}^{3}$ 知
$\dfrac {dS}{dt}=\dfrac {dS}{dx}\dfrac {dx}{dt}$
$=(\dfrac {1}{2}+\dfrac {2}{9}{x}^{2})\dfrac {dx}{dt}$
步骤 4：代入x=3
将x=3代入上式，得
$\dfrac {dS}{dt}=(\dfrac {1}{2}+\dfrac {2}{9}\times 9)\times 4$
$=(\dfrac {1}{2}+2)\times 4$
$=10$
设P点坐标为(x,y)，则直线OA与直线AP及曲线L所围图形面积为
$S=\dfrac {1}{2}\times (1+\dfrac {4}{9}{x}^{2})x-{\int }_{0}^{x}\dfrac {4}{9}{t}^{2}dt$
步骤 2：计算积分
$S=\dfrac {1}{2}\times (1+\dfrac {4}{9}{x}^{2})x-{\int }_{0}^{x}\dfrac {4}{9}{t}^{2}dt$
$=\dfrac {1}{2}x+\dfrac {2}{9}{x}^{3}-\dfrac {4}{27}{x}^{3}$
$=\dfrac {1}{2}x+\dfrac {2}{27}{x}^{3}$
步骤 3：求导数
由题设知在点(3,4)处 $\dfrac {dx}{dt}=4$ ，又由 $S=\dfrac {1}{2}x+\dfrac {2}{27}{x}^{3}$ 知
$\dfrac {dS}{dt}=\dfrac {dS}{dx}\dfrac {dx}{dt}$
$=(\dfrac {1}{2}+\dfrac {2}{9}{x}^{2})\dfrac {dx}{dt}$
步骤 4：代入x=3
将x=3代入上式，得
$\dfrac {dS}{dt}=(\dfrac {1}{2}+\dfrac {2}{9}\times 9)\times 4$
$=(\dfrac {1}{2}+2)\times 4$
$=10$