参数方程求导
当我们知道dxdt,dydt\dfrac{dx}{dt},\dfrac{dy}{dt}dtdx,dtdy时,我们可以求出dydx\dfrac{dy}{dx}dxdy
dydx=dydtdxdt\dfrac{dy}{dx}=\dfrac{\quad \frac{dy}{dt}\quad}{\frac{dx}{dt}}dxdy=dtdxdtdy
然后我们也可以求d2ydx2\dfrac{d^2y}{dx^2}dx2d2y
d2ydx2=d(dydx)dt/dxdt\dfrac{d^2y}{dx^2}=\dfrac{d(\frac{dy}{dx})}{dt}/\dfrac{dx}{dt}dx2d2y=dtd(dxdy)/dtdx
例
{x=ln(1+t2)y=t−arctant\left\{\begin{matrix}x=\ln(1+t^2)\\y=t-\arctan t\end{matrix}\right.{x=ln(1+t2)y=t−arctant,求dydx,d2ydx2\dfrac{dy}{dx},\dfrac{d^2y}{dx^2}dxdy,dx2d2y
解:
dxdt=2t1+t2\qquad \dfrac{dx}{dt}=\dfrac{2t}{1+t^2}dtdx=1+t22t
dydt=1−11+t2=t21+t2\qquad\dfrac{dy}{dt}=1-\dfrac{1}{1+t^2}=\dfrac{t^2}{1+t^2}dtdy=1−1+t21=1+t2t2
dydx=dydtdxdt=t2\qquad \dfrac{dy}{dx}=\dfrac{\quad \frac{dy}{dt}\quad}{\frac{dx}{dt}}=\dfrac t2dxdy=dtdxdtdy=2t
d(dydx)dt=12\qquad \dfrac{d(\frac{dy}{dx})}{dt}=\dfrac 12dtd(dxdy)=21
d2ydx2=d(dydx)dt/dxdt=122t1+t2=1+t24t\qquad \dfrac{d^2y}{dx^2}=\dfrac{d(\frac{dy}{dx})}{dt}/\dfrac{dx}{dt}=\dfrac{\frac 12}{\frac{2t}{1+t^2}}=\dfrac{1+t^2}{4t}dx2d2y=dtd(dxdy)/dtdx=1+t22t21=4t1+t2