【微分積分】3-2-3 媒介変数の導関数｜問題集

\(\displaystyle \frac{dx}{dt}=1\)
\(\displaystyle \frac{dy}{dt}=2t\)
よって、
\(\displaystyle \frac{dy}{dx}=\frac{2t}{1}=2t\)

\(\displaystyle \frac{dx}{dt}=\frac{1}{2}t^{-\frac{1}{2}}+t^{-2}\)
\(\displaystyle \frac{dy}{dt}=1-\frac{1}{2}t^{-\frac{3}{2}}\)
よって、
\(\displaystyle \frac{dy}{dx}=\frac{2t^2-t^{\frac{1}{2}}}{t^{\frac{3}{2}}+2}\)

\(\displaystyle \frac{dx}{dt}=\frac{3a(1+t^3)-3at・3t^2}{(1+t^3)^2}=\frac{3a(1-2t^3)}{(1+t^3)^2}\)
\(\displaystyle \frac{dy}{dt}=\frac{6at(1+t^3)-3at^2・3t^2}{(1+t^3)^2}=\frac{3at(2-t^3)}{(1+t^3)^2}\)
よって、
\(\displaystyle \frac{dy}{dx}=\frac{t(2-t^3)}{1-2t^3}\)

\(\displaystyle \frac{dx}{dt}=-a\sin t\)
\(\displaystyle \frac{dy}{dt}=a\cos t\)
よって、
\(\displaystyle \frac{dy}{dx}=-\frac{\cos t}{\sin t}=-\cot t\)

\(\displaystyle \frac{dx}{dt}=a(1-\cos t)\)
\(\displaystyle \frac{dy}{dt}=a\sin t\)
よって、
\(\displaystyle \frac{dy}{dx}=\frac{\sin t}{1-\cos t}\)

\(\displaystyle \frac{dx}{dt}=3a・\cos^2t・(-\sin t)=-3a\cos^2t\sin t\)
\(\displaystyle \frac{dy}{dt}=3b・\sin^2t・\cos t=3b\sin^2t\cos t\)
よって、
\(\displaystyle \frac{dy}{dx}=\frac{3b\sin^2t\cos t}{-3a\cos^2t\sin t}=-\frac{b}{a}\tan t\)

極座標を用いて表すと
\(x=\cos t,y=\sin t\)
\(\displaystyle \frac{dx}{dt}=-\sin t\)
\(\displaystyle \frac{dy}{dt}=\cos t\)
よって、
\(\displaystyle \frac{dy}{dx}=\frac{\cos t}{-\sin t}=-\frac{x}{y}\)

極座標を用いて表すと
\(x=a\cos t,y=b\sin t\)
\(\displaystyle \frac{dx}{dt}=-a\sin t\)
\(\displaystyle \frac{dy}{dt}=b\cos t\)
よって、
\(\displaystyle \frac{dy}{dx}=\frac{b\cos t}{-a\sin t}=-\frac{b}{a}\cot t\)