设z=z(t)由x+y-z=e^z,xe^x=tan t,y=cos t所确定,求dz/dt/t=0
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设z=z(t)由x+y-z=e^z,xe^x=tan t,y=cos t所确定,求dz/dt/t=0
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x,y,z都是t的函数,记作x(t),y(t),z(t)
(1)首先求出x(0),y(0),z(0)
t = 0,
x(0)e^x(0)=tan 0 = 0
e^x(0) > 0,则
x(0) = 0
y(0) = cos 0 = 1
x(0)+y(0)-z(0)=e^z(0)得到
e^z(0) + z(0) -1 = 0
为求z(0),即解方程 e^z + z -1 = 0
令f(z) = e^z + z -1
f'(z) = e^z - 1
易得 z = 0 是f(z) 的唯一极值点
而f(0) = 0,即z=0是f(z)=0的唯一解
所以z(0) = 0
(2)求解dz/dt
xe^x=tan t,两边对t求导
dx/dt*e^x + xe^x*dx/dt = 1/(cost)^2
dx/dt = 1/( (cost)^2 * (1+x)e^x )
y=cos t,两边对t求导
dy/dt = -sint
x+y-z=e^z,两边对t求导
dx/dt + dy/dt - dz/dt = e^z *dz/dt
dz/dt = (dx/dt + dy/dt)/(1+e^z)
= ( 1/( (cost)^2 * (1+x)e^x ) -sint ) / (1+e^z)
当t=0时,已解出 x(0) = 0,y(0) = 1,z(0) = 0,代入
t=0时,dz/dt = ( 1/( (cos0)^2 * (1+0)e^0 ) -sin0 ) / (1+e^0) = 1/2
(1)首先求出x(0),y(0),z(0)
t = 0,
x(0)e^x(0)=tan 0 = 0
e^x(0) > 0,则
x(0) = 0
y(0) = cos 0 = 1
x(0)+y(0)-z(0)=e^z(0)得到
e^z(0) + z(0) -1 = 0
为求z(0),即解方程 e^z + z -1 = 0
令f(z) = e^z + z -1
f'(z) = e^z - 1
易得 z = 0 是f(z) 的唯一极值点
而f(0) = 0,即z=0是f(z)=0的唯一解
所以z(0) = 0
(2)求解dz/dt
xe^x=tan t,两边对t求导
dx/dt*e^x + xe^x*dx/dt = 1/(cost)^2
dx/dt = 1/( (cost)^2 * (1+x)e^x )
y=cos t,两边对t求导
dy/dt = -sint
x+y-z=e^z,两边对t求导
dx/dt + dy/dt - dz/dt = e^z *dz/dt
dz/dt = (dx/dt + dy/dt)/(1+e^z)
= ( 1/( (cost)^2 * (1+x)e^x ) -sint ) / (1+e^z)
当t=0时,已解出 x(0) = 0,y(0) = 1,z(0) = 0,代入
t=0时,dz/dt = ( 1/( (cos0)^2 * (1+0)e^0 ) -sin0 ) / (1+e^0) = 1/2
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