已知x=t(1-cost),y=tcost,确定了y=f(x),求dy/dx和d^2y/dx^2,
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已知x=t(1-cost),y=tcost,确定了y=f(x),求dy/dx和d^2y/dx^2,
已知x=t(1-cost),y=tcost,确定了y=f(x),求dy/dx和d²y/dx²
y'=dy/dx=(dy/dt)/(dx/dt)=(cost-tsint)/(1-cost+tsint);
y''=d²y/dx²=dy'/dx=(dy'/dt)/(dx/dt)
={[(1-cost+tsint)(-sint-sint-tcost)-(cost-tsint)(sint+sint+tcost)]/(1-cost+tsint)²}/(1-cost+tsint)
=[(1-cost+tsint)(-2sint-tcost)-(cost-tsint)(2sint+tcost)]/(1-cost+tsint)³
=[-(1-cost+tsint)(2sint+tcost)-(cost-tsint)(2sint+tcost)]/(1-cost+tsint)³
=[(2sint+tcost)(1+cost-tsint-cost+tsint)]/(1-cost+tsint)³
=(2sint+tcost)/(1-cost+tsint)³
再问: 倒数第二行,第二个括号的1是不是应该为-1,因为倒数第三行的1前面有负号的~所以结果是不是也得加负号啊
再答: 你说的对!你很细心!我疏忽了!修正如下: 倒数第二行开始: =[(2sint+tcost)(-1+cost-tsint-cost+tsint)]/(1-cost+tsint)³ =-(2sint+tcost)/(1-cost+tsint)³
y'=dy/dx=(dy/dt)/(dx/dt)=(cost-tsint)/(1-cost+tsint);
y''=d²y/dx²=dy'/dx=(dy'/dt)/(dx/dt)
={[(1-cost+tsint)(-sint-sint-tcost)-(cost-tsint)(sint+sint+tcost)]/(1-cost+tsint)²}/(1-cost+tsint)
=[(1-cost+tsint)(-2sint-tcost)-(cost-tsint)(2sint+tcost)]/(1-cost+tsint)³
=[-(1-cost+tsint)(2sint+tcost)-(cost-tsint)(2sint+tcost)]/(1-cost+tsint)³
=[(2sint+tcost)(1+cost-tsint-cost+tsint)]/(1-cost+tsint)³
=(2sint+tcost)/(1-cost+tsint)³
再问: 倒数第二行,第二个括号的1是不是应该为-1,因为倒数第三行的1前面有负号的~所以结果是不是也得加负号啊
再答: 你说的对!你很细心!我疏忽了!修正如下: 倒数第二行开始: =[(2sint+tcost)(-1+cost-tsint-cost+tsint)]/(1-cost+tsint)³ =-(2sint+tcost)/(1-cost+tsint)³
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