【解析】∵$$\left\{ \begin{matrix} x = r \cos \theta \\ y = r \sin \theta \end{matrix} \right.$$,$$ r = 1 - \cos \theta $$ ∴曲线方程转化为: $$\left\{ \begin{matrix} x = \cos \theta - \cos ^ { 2 } \theta \\ y = \sin \theta - \sin \theta \cos \...
我们班里没人知道r=1-cos(theta)是什麼意思 只看楼主 收藏 回复 Q22Rt 核心会员 7 ……我的数学才能废了……百无一用啊……小学生无敌了 zxjnttc 核心会员 7 你们班都是奇才!这是极坐标方程 Q529427402 正式会员 5 !!!这个,不知道很正常,只有有理想,有素质的年轻人才需要了解,楼主自己懂就好 ...
Answer to: Find the area of the region that lies inside both curves: r = 1 + cos(theta), r = 1 - cos(theta). By signing up, you'll get thousands of...
Find the arc length of the polar curve r = 1 - cos(theta) on the interval 0 less than or equal to theta less than or equal to pi. Find the arc length of a polar curve r = 1 - cos theta on the interval 0 less th...
解析 【解析】 $$ \frac { 5 } { 4 } $$π;提示$$ A = \int _ { 0 } ^ { \frac { \pi } { 3 } } ( 1 + \cos \theta ) ^ { 2 } d \theta + \int _ { \frac { \pi } { 3 } } ^ { \frac { \pi } { 2 } } 9 \cos ^ { 2 } \theta d $$ ...
【解析】原方程变形为$$ r = \frac { 1 } { 2 \sin ^ { 2 } \theta } , $$ 由$$ \tan \theta = \frac { y } { x } $$,得$$ \sin ^ { 2 } \theta = \frac { y ^ { 2 } } { x ^ { 2 } + y ^ { 2 } } , $$, 将$$ x ^ { 2 } + y ^ { ...
1.图像与表达式 1.1图像 图3 注:这里只展示θ在[0,2π]上的图像。 1.2表达式 极坐标: r=aθ,θ∈[0,+∞),a>0 参数方程: \displaystyle\left\{ \begin{array}{lc} x=a\theta\cdot cos\theta\\ y=a\theta\cdot sin\theta\\ \end{array} \right.(\theta\in[0,+\infty),a>0) 2.弧长...
微分式:\left\{ \begin{array}{} dx= (cos\theta )d\rho - \rho sin\theta d\theta\\ dy= (sin\theta )d \rho + \rho cos\theta d\theta \end{array} \right.;向量式:\left[ \begin{array}{ccc} dx \\ dy \end{array} \right]=d\rho \left[ \begin{array}...
$$x = f(\theta) \cos \theta \\ y = f(\theta) \sin \theta $$ Answer and Explanation:1 Given the polar equation of a curve, {eq}\displaystyle r = - \csc \theta {/eq} We need to graph the above curve on the coordinate plane. ...
【解析】 解$$ \int _ { 0 } ^ { \pi } x d x d y = \int _ { \frac { \pi } { 2 } } ^ { \frac { \pi } { 2 } } d \theta \int _ { 2 } ^ { 2 ( 1 + \cos \theta ) } r ^ { 2 } \cos \theta d r = \int _ { - \frac { \pi } {...