求由x=asint,y=bcost确定的函数y=y(x)的导数dy/dx
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解决时间 2021-10-12 12:37
- 提问者网友：戎马万世
- 2021-10-12 02:55
要详细的过程，谢谢。
最佳答案
- 五星知识达人网友：酒者煙囻
- 2018-11-23 19:42
x=asin(t)
y=bcos(t)
---
x/a=sin(t)
y/b=cos(t)
---
(x/a)^2 + (y/b)^2 =[ sin(t) ]^2 + [ cos(t) ]^2 = 1
所确定的方程为：
x^2/a^2 + y^2/b^2 = 1
y =
(b*(a + x)^(1/2)*(a - x)^(1/2))/a
-(b*(a + x)^(1/2)*(a - x)^(1/2))/a
求导，得到：
dy/dx =
(b*(a - x)^(1/2))/(2*a*(a + x)^(1/2)) - (b*(a + x)^(1/2))/(2*a*(a - x)^(1/2))
(b*(a + x)^(1/2))/(2*a*(a - x)^(1/2)) - (b*(a - x)^(1/2))/(2*a*(a + x)^(1/2))
y=bcos(t)
---
x/a=sin(t)
y/b=cos(t)
---
(x/a)^2 + (y/b)^2 =[ sin(t) ]^2 + [ cos(t) ]^2 = 1
所确定的方程为：
x^2/a^2 + y^2/b^2 = 1
y =
(b*(a + x)^(1/2)*(a - x)^(1/2))/a
-(b*(a + x)^(1/2)*(a - x)^(1/2))/a
求导，得到：
dy/dx =
(b*(a - x)^(1/2))/(2*a*(a + x)^(1/2)) - (b*(a + x)^(1/2))/(2*a*(a - x)^(1/2))
(b*(a + x)^(1/2))/(2*a*(a - x)^(1/2)) - (b*(a - x)^(1/2))/(2*a*(a + x)^(1/2))
全部回答
- 1楼网友：渡鹤影
- 2019-01-22 13:39
你好！
x'(t)=acost ; y'(t)= -bsin(t);
dy/dx=y'(t)/x'(t)= -bsin(t); / acost = -(b/a)tant
打字不易，采纳哦！
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