第6章 6.6

(1)

$z = x^2 + 3y^2 \qquad x = \cos t \qquad y = 2\sin t$

$z_x = 2x \qquad z_y = -6y \qquad x' = -\sin t \qquad y' = 2\cos t$

$\frac{dz}{dt} = -2\cos t \cdot \sin t - 12\sin t \cdot 2\cos t = -26\sin t \cdot \cos t = -13\sin 2t$

(2)

$z = x^2 + y^2 \qquad x = e^t \cos t \qquad y = e^t \sin t$

$z_x = 2x \qquad z_y = 2y \qquad x' = e^t \cos t - e^t \sin t \qquad y' = e^t \sin t + e^t \cos t$

$\frac{dz}{dt} = 2(e^t \cos t)(e^t \cos t - e^t \sin t) + 2(e^t \sin t)(e^t \sin t + e^t \cos t)$

$= 2e^{2t}\{\cos^2 t - \cos t \cdot \sin t + \sin^2 t + \cos t \cdot \sin t\} = 2e^{2t}$

(3)

$z = \tan^{-1}\frac{y}{x} \qquad x = t - \sin t \qquad y = 1 - \cos t$

$z_x = \frac{-\dfrac{y}{x^2}}{1+\left(\dfrac{y}{x}\right)^2} = \frac{-y}{x^2+y^2} \qquad z_y = \frac{y}{x^2+y^2} \qquad x' = 1 + \cos t \qquad y' = \sin t$

$\frac{dz}{dt} = \frac{-(1-\cos t)(1+\cos t) + (t+\sin t)\cdot \sin t}{(t+\sin t)^2+(1-\cos t)^2} = \frac{t\sin t}{(t+\sin t)^2+(1-\cos t)^2}$