高専の数学解説

第1章 1.8

(1)

x=2tx = 2t

y=t2−1y = t^2 - 1

dxdt=2dydt=2t\frac{dx}{dt} = 2 \qquad \frac{dy}{dt} = 2t

dydx=dydtdxdt=2t2=t\frac{dy}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} = \frac{2t}{2} = t

(2)

x=pt2x = pt^2

y=2pty = 2pt

dxdt=2ptdydt=2p\frac{dx}{dt} = 2pt \qquad \frac{dy}{dt} = 2p

dydx=2p2pt=1t\frac{dy}{dx} = \frac{2p}{2pt} = \frac{1}{t}

(3)

x=11+t2x = \frac{1}{1+t^2}

y=t1+t2y = \frac{t}{1+t^2}

dxdt=−2t(1+t2)2dydt=(1+t2)−t⋅2t(1+t2)2=1−t2(1+t2)2\frac{dx}{dt} = \frac{-2t}{(1+t^2)^2} \qquad \frac{dy}{dt} = \frac{(1+t^2) - t \cdot 2t}{(1+t^2)^2} = \frac{1-t^2}{(1+t^2)^2}

dydx=1−t2(1+t2)2−2t(1+t2)2=1−t2−2t=t2−12t\frac{dy}{dx} = \frac{\dfrac{1-t^2}{(1+t^2)^2}}{\dfrac{-2t}{(1+t^2)^2}} = \frac{1-t^2}{-2t} = \frac{t^2-1}{2t}

(4)

x=a1cos⁡tx = a\dfrac{1}{\cos t}

y=btan⁡ty = b\tan t

dxdt=a⋅−(−sin⁡t)cos⁡2t=a⋅sin⁡tcos⁡2tdydt=b⋅1cos⁡2t\dfrac{dx}{dt} = a \cdot \dfrac{-(-\sin t)}{\cos^{2} t} = a \cdot \dfrac{\sin t}{\cos^{2} t} \qquad \dfrac{dy}{dt} = b \cdot \dfrac{1}{\cos^{2} t}

dydx=b⋅1cos⁡2ta⋅sin⁡tcos⁡2t=basin⁡t\dfrac{dy}{dx} = \dfrac{b \cdot \dfrac{1}{\cos^{2} t}}{a \cdot \dfrac{\sin t}{\cos^{2} t}} = \dfrac{b}{a\sin t}

解説: r31bn1z