首先求一阶导数: (dy)/(dx) = ((dy)/(dt))/((dx)/(dt)) = (bcos t)/(-asin t) = -b/acot t 然后求二阶导数: (d^2y)/(dx^2) = (d/(dt)(-b/acot t))/((dx)/(dt)) = (b/acsc^2 t)/(-asin t) = -b/(a^2)csc^3 t 因此,答案为 \circled{B} -b/(a...
Trigonometry 4sinθcosθ=2sinθ Linear equation y=3x+4 Arithmetic 699∗533 Matrix [2534][2−10135] Simultaneous equation {8x+2y=467x+3y=47 Differentiation dxd(x−5)(3x2−2) Integration ∫01xe−x2dx Limits x→−3limx2+2x−3x2−9 ...
Trigonometry 4sinθcosθ=2sinθ Linear equation y=3x+4 Arithmetic 699∗533 Matrix [2534][2−10135] Simultaneous equation {8x+2y=467x+3y=47 Differentiation dxd(x−5)(3x2−2) Integration ∫01xe−x2dx Limits x→−3limx2+2x−3x2−9 Back to top ...
4.任意角的三角函数任意角a的终边与单位圆交于点P(r,y)时,则 sinα=1516, cosα=\textcircled(16), tanα=\textcircled(17)(x≠q0) .三个三角函数的性质如下表:三角第一象第二象第第三象第四象定义域函数限符号限符号限符号限符号sin a18十十一--cos a19十一十tan a20+—+一 ...
The first thing to do is to understand that a circle is made can be created using two ways.1) a parametric equation x^2 + y^2 = R^2 where R is the radius and X and Y are the coordinates of each point.2) the sin (stands for sinus) and cos (stands for cosinus) function. x...
Listen,circleandwrie(1)7m∈NH*8H=(9√(35))/(98)=(2)1cosαcosα-sinα-α(n+20°)(3)cosα=2/(α+β)+α/(▱)+cosαcosβ=(α+cosα)/(α+β)(4)m=2cosx/(x+m)=1/2mg⋅(√2)/(36) 相关知识点: 试题来源:
The equation r=8sinθ+5cosθ represents a circle. Find the Cartesian equivalent of this equation. Polar Coordinates: To find the Cartesian equivalent of the circle above, we will use the relationships between the Cartesian and polar coordinates. Th...
Find the area inside the curve r = \sin \theta and outside of r = 1 + \cos \theta . Sketch the region for which you found the area. Find the area inside the circle r = 2 and outside of the polar curve r = 1 + \cos \theta. Compute the area under the graph ...
百度试题 结果1 题目Findondcircletheirfriends(1)(2sinα)/(cos)(2)∴(3)(2+2i)/(-2i)=i(4)(2-2g)/(3F-3)(5)(x+2)/(x+1) 相关知识点: 试题来源: 解析 (1)V(2)w(3)x(4)Y(5)Z 反馈 收藏
https://math.stackexchange.com/questions/1497061/why-do-i-see-sqrt-sin2-theta-sin-theta-in-integration-solutions This is because to be able to go back to the original variable, you have to use a bijective change of variable. Setting x−3=cosθ does not define...