Question

急求高数题目一道

Answer 1

方法如下，
请作参考：

Answer 2

解如下图所示

Answer 3

答案: 反常积分计算∫(上限正无穷,下限0)dx/(√ (x*(x+1)^5))的值为() A.无穷 B.0 C.2/3 D.1 如果是u=1/√ (x*(x+1)^5)) ∫(上限正无穷,下限0)

Answer 4

假设直线ρcos(θ-π/6)=1上某一点(ρ,θ),那么这点到点(2,7π/6)的距离l应该满足:l^2 ... 很容易就能的出来,此时的l^2=8+1=9, 所以 l=3所以可以看出来,根本不用高数知识...

Answer 5

x=2t+3+arctant
x(0) = 3
x=2t+3+arctant
求导对应t
dx/dt=2+1/(1+t^2)
y=2-3t+ln(1+t^2)
y(0)=2
(x,y)|t=0 = (3,2)
y=2-3t+ln(1+t^2)
求导对应t
dy/dt=-3+2t/(1+t^2)
利用链式法则
dy/dx
=(dy/dt)/(dx/dt)
=[-3+2t/(1+t^2)]/[2+1/(1+t^2)]
dy/dx |t=0
=[-3+0]/[2+1]
=-1
切线方程 t=0
y-y(0) = dy/dx |t=0 . (x-x(0))
y-2 = (-1)(x-3)
x+y-5=0