填充 10 橢圓
參數化 \( \begin{cases}
x & =\sqrt{2-t^{2}}\\
y & =t\\
z & =4-\sqrt{2-t^{2}}\end{cases} \), \( \begin{cases}
x' & =\frac{-t}{\sqrt{2-t^{2}}}\\
y' & =1\\
z' & =\frac{t}{\sqrt{2-t^{2}}}\end{cases} \), 令 \( t=1 \) 代入得，切線方向 \( (-1,1,1) \).
所以切線為 \( \frac{x-1}{-1}=\frac{y-1}{1}=\frac{z-3}{1} \).
計算1
(A) 請用三角形面積夾擠扇形
(B) 微分，二階微分，判斷增減和凹向
http://www2.wolframalpha.com/input/?i=Plot[Sin[x]/x,+{x,+-10,+10}]
計算2
(A) 還是微分，二階微分
\( x'=-3\cos^{2}t\sin t \), \( y'=3\sin^{2}t\cos t \).
\( \frac{dy}{dx}=-\frac{\sin t}{\cos t}=-\tan t \), \( \frac{d^{2}y}{dx^{2}}=\frac{\frac{dy}{dx}}{dt}\cdot\frac{1}{x'}=-\sec^{2}t\cdot\frac{1}{(-3)\cdot\cos^{2}t\sin t}=\frac{3}{\cos^{4}t\sin t} \).
http://www2.wolframalpha.com/input/?i=plot+x%3Dcos^3+t,+y%3Dsin^3+t
(B) 面積 \( =4\left|\frac{1}{2}\int_{0}^{\frac{\pi}{2}}yx'-xy'dt\right|=2\left|\int_{0}^{\frac{\pi}{2}}-3\cos^{2}t\sin^{4}t-3\cos^{4}t\sin^{2}tdt\right|=6\int_{0}^{\frac{t}{2}}\cos^{2}t\sin^{2}tdt=\frac{3}{8}\pi \).
(C) 弧長 \( =4\int_{0}^{\frac{\pi}{2}}\sqrt{x'^{2}+y'^{2}}dt=12\int_{0}^{\frac{\pi}{2}}\sin t\cos tdt=6 \).
