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Given that the side area of the cone is a fixed value of 8 π, the analytical function and definition domain of the generatrix length L with respect to the bottom radius R are obtained
Set the bottom radius as R and the bus length as L
Then the side area can be expressed as
[(2πr)/(2πl)]*πl ²= πrl=8π
RL = 8
l=8/r
In the right triangle composed of the bottom radius and the bus, the bus is an oblique edge and the bottom radius is a right angle edge
So π R ²
A hemisphere has a common bottom. If the volume of the cone is equal to that of the hemisphere, find the ratio of the height of the cone to the radius
Let the common radius be r, the common surface be s, and the height of the cone be h: the volume of the hemisphere be: 1 / 2 * 4 / 3 π ^ 3 --------------- one-half times four-thirds π R, and the volume of the cubic cone be: 1 / 3SH. Because s = π R ^ 2, the volume of the cone is converted to 1 / 3 * π R ^ 2 * h ------- the square of one-third π r times h, the known hemisphere
If the height of the cone is 4 and the radius of the bottom is 3, what is the volume of the ball? Wrong. Sorry, it's the side area of the cone
The side of the cone is triangular,
The triangle bottom is 6 in diameter and 4 in height
Area: 4 * 6 / 2 = 24 / 2 = 12
Derivation of trigonometric function Y = sin2x (Pie / 6-2x) derivation
Y = sin2x (PAI / 6-2x) = sin (PAI / 3 * x-4x ^ 2)
y'=cos(Pai/3*x-4x^2)*(Pai/3*x-4x^2)'
=cos(Pai/3*x-4x^2)*(Pai/3-8x)
The following questions are about derivative. These are some difficult problems I have come into contact with. Please write down the process, especially the trigonometric function a) y= In/e^x + 4 = b) Y = SiNx cosx cosx - SiNx formula is wrong c) Y = secx cosecx I'm the most confused d）y=Xarccosx e）y=tg lnx f）y= arctg^2X g) y=4 cos^4 X^4 h) y=(x^2-7X)^tg2X There's not much trouble, but I want to solve the following questions
（2）) y = sinx cosx
y'=(sinx cosx ) '=sinx'cosx+cosx'sinx=(cosx)^2-(sinx)^2=cos2x
c)y = secx cscx
y'=( secx cscx)'=secx'cscx+cscx'secx=(secxtanx)cscx+（-cscxcotx）secx=secx^2-cscx^2
Just a hint, in the form of y = a * B, y '= a'B + b'a. in other topics, and so on
Derivative formula with trigonometric function (SiN x) '= cos x
(cos x)'=-sin x
(tan x)'=(sec x)^2
(cot x)'=-(csc x)^2
(sec x)'=sec x*tan x
(csc x)'=-csc x*cot x
Derivation formula of trigonometric function
③ (sinx)' = cosx(cosx)' = - sinx(tanx)'=1/(cosx)^2=(secx)^2=1+(tanx)^2-(cotx)'=1/(sinx)^2=(cscx)^2=1+(cotx)^2(secx)'=tanx·secx(cscx)'=-cotx·cscx(arcsinx)'=1/(1-x^2)^1/2(arccosx)'=-1/(1-x^2)^1/2(arc...
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