抛体运动的射程 \( R \) 可以通过公式 \( R = \frac{v_0^2 \sin 2\theta}{g} \) 计算,其中 \( v_0 \) 代表初速度,\( \theta \) 代表抛射角度,\( g \) 代表重力加速度。相关知识点: 试题来源: 解析 \( R = \frac{v_0^2 \sin 2\theta}{g} \) ...
\rightarrow pic3:r^2=a^2sin2\theta 接下来关于双纽线相关数据计算,只考虑采用极坐标形式。 在计算整条曲线弧长、面积、旋转体积、旋转表面积之前优先考虑其对称性。 2.弧长(不用掌握,了解即可!) 记图2的弧长为L,其中 \displaystyle r'=-\frac{a\cdot sin2\theta}{\sqrt{cos2\theta}} \displaystyl...
यदि x=r sin theta cos phi, y=r sin theta sin phi तथा z=r cos theta हो, तो:
曲线r=2½sinθ与r²=cos2θ所围成图形面积为：pi/6+(1-√3)/2。解：本题利用了定积分的性质求解。因为r=√2sinθ表示圆，且圆心在点(√2/2,pi/2)处，半径为√2/2。r^2=cos2θ,表示双纽线。又有极角θ范围是[-pi,-3pi/4],[-pi/4,pi/4],[3pi/4,pi]再联立两...
r=3sin(2)θr=3sin(2)θ r=asin(nθ)r=asin(nθ) r=acos(nθ)r=acos(nθ) a≠0a≠0 nn >1>1 nn nn nn 2n2n r=3sin(2)θr=3sin(2)θ r=3sin(2)θr=3sin(2)θ ( ) | [ ] √ ≥
笛卡尔的心形公式中，r=a(1-sin (theta))中的a是一个常数。在百度百科上，没有详细说明a的具体意义，但可以看到a的值越大，心形线的大小也随之增大，实际上a控制着心形线的大小。进一步来看，2a等于凹陷点与突出点间线段的长度。当theta等于0时，r=a，这似乎是心形线弧长的起点，或者说是心形线...
rsin(θ)=3rsin(θ)=3 Sincesin(θ)=yrsin(θ)=yr, replacesin(θ)sin(θ)withyryr. ryr=3ryr=3 Multiplyboth sides byrr. Tap for more steps... both sides byr. r(ryr)=r⋅3 Simplifyr(ryr). rbyrby adding the. Mover. r⋅ryr=r⋅3 ...
Answer to: Find the area of the region that lies inside the curve r = 2 + sin theta and outside the curve r = 3 sin theta. By signing up, you'll...
Answer to: Find the area inside one loop of the lemniscate r^2 = 11sin(2theta). By signing up, you'll get thousands of step-by-step solutions to...
theta. has the nonlinear relationship with respect to time t, as expressed in the following equation:&thgr;=&ohgr;.sub. M t-K sin 2&ohgr;.sub.M... W Shimizu,J Toda,T Akama,... - US 被引量: 37发表: 1986年 Magnetize accretion and spin evolution in classical T Tauri stars 10 ...