大一高数,求具体过程?
1个回答
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解:
x=3t²+2t+3方程两边对t求导
dx/dt
= 6t+2
(e^y)sint-y+1=0方程两边对t求导
dy/dt
=(e^y) * cost / (1 - e^y * sint) = e^y * cost / (2 - y)
所以根据参数方程的求导公式
dy/dx
= (dy/dt) / (dx/dt)
= e^y * cost / [(6t+2)(2-y)]
采用用对数求导法化简上式:
对上式先求对数
ln(dy/dx)
= y + lncost - ln(6t+2) - ln(2-y)
然后对t求导:
d[ln(dy/dx)]/dt
=d(dy/dx)/dt / (dy/dx)
= dy/dt - tant - 6/(6t+2) + (dy/dt)/(2-y)
代入数据t=0
e^ysint-y+1=0可得y=1
dx/dt = 6t+2 = 2
dy/dt=e^y * cost / (2 - y) = e
dy/dx = e^y * cost / [(6t+2)(2-y)]=e/2
d(dy/dx)/dt
= (dy/dx)[dy/dt - tant - 6/(6t+2) + (dy/dt)/(2-y)] = e(2e-3)/2
所以:
d²y/dx²
=d(dy/dx)/dt / (dx/dt)
= e(2e-3)/4
x=3t²+2t+3方程两边对t求导
dx/dt
= 6t+2
(e^y)sint-y+1=0方程两边对t求导
dy/dt
=(e^y) * cost / (1 - e^y * sint) = e^y * cost / (2 - y)
所以根据参数方程的求导公式
dy/dx
= (dy/dt) / (dx/dt)
= e^y * cost / [(6t+2)(2-y)]
采用用对数求导法化简上式:
对上式先求对数
ln(dy/dx)
= y + lncost - ln(6t+2) - ln(2-y)
然后对t求导:
d[ln(dy/dx)]/dt
=d(dy/dx)/dt / (dy/dx)
= dy/dt - tant - 6/(6t+2) + (dy/dt)/(2-y)
代入数据t=0
e^ysint-y+1=0可得y=1
dx/dt = 6t+2 = 2
dy/dt=e^y * cost / (2 - y) = e
dy/dx = e^y * cost / [(6t+2)(2-y)]=e/2
d(dy/dx)/dt
= (dy/dx)[dy/dt - tant - 6/(6t+2) + (dy/dt)/(2-y)] = e(2e-3)/2
所以:
d²y/dx²
=d(dy/dx)/dt / (dx/dt)
= e(2e-3)/4
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