参数方程求导习题

例1

已知 { x = t e t y = 2 t + t 2 \left\{\begin{matrix}x=te^t\\y=2t+t^2\end{matrix}\right. {x=tety=2t+t2,求 d y d x \dfrac{dy}{dx} dxdy

解:
d x d t = ( t + 1 ) e t \qquad\dfrac{dx}{dt}=(t+1)e^t dtdx=(t+1)et

d y d t = 2 + 2 t \qquad\dfrac{dy}{dt}=2+2t dtdy=2+2t

d y d x = d y d t d x d t = 2 + 2 t ( t + 1 ) e t = 2 e t \qquad\dfrac{dy}{dx}=\dfrac{\quad\frac{dy}{dt}\quad}{\frac{dx}{dt}}=\dfrac{2+2t}{(t+1)e^t}=\dfrac{2}{e^t} dxdy=dtdxdtdy=(t+1)et2+2t=et2


例2

已知 { x = 3 t 2 + 2 t e y sin ⁡ t − y + 1 = 0 \left\{\begin{matrix}x=3t^2+2t\\e^y\sin t-y+1=0\end{matrix}\right. {x=3t2+2teysinty+1=0,求 d y d x \dfrac{dy}{dx} dxdy

解:
d x d t = 6 t + 2 \qquad\dfrac{dx}{dt}=6t+2 dtdx=6t+2

\qquad e y sin ⁡ t − y + 1 = 0 e^y\sin t-y+1=0 eysinty+1=0两边同时对 t t t求导得

e y y ′ sin ⁡ t + e y cos ⁡ t − y ′ = 0 \qquad e^yy'\sin t+e^y\cos t-y'=0 eyysint+eycosty=0

\qquad 移项得 ( e y sin ⁡ t − 1 ) y ′ = − e y cos ⁡ t (e^y\sin t-1)y'=-e^y\cos t (eysint1)y=eycost

d y d t = y ′ = e y cos ⁡ t 1 − e y sin ⁡ t \qquad \dfrac{dy}{dt}=y'=\dfrac{e^y\cos t}{1-e^y\sin t} dtdy=y=1eysinteycost

d y d x = d y d t d x d t = e y cos ⁡ t ( 6 t + 2 ) ( 1 − e y sin ⁡ t ) \qquad\dfrac{dy}{dx}=\dfrac{\quad\frac{dy}{dt}\quad}{\frac{dx}{dt}}=\dfrac{e^y\cos t}{(6t+2)(1-e^y\sin t)} dxdy=dtdxdtdy=(6t+2)(1eysint)eycost


这题不仅考了参数方程求导,还考了隐函数求导。

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