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Normal equation of curve y = x ^ 3-3x at point (2,2) If the problem, solve!
A:
Derivation y '= 3x ^ 2-3
When x = 2, y '= 9
So the normal equation is y = - 1 / 9x + K. because over (2,2), k = 20 / 9
So the normal equation is: y = - X / 9 + 20 / 9, that is, 9y + x = 20
Solving the tangent plane equation of rotating paraboloid z = x ^ 2 + y ^ 2 at point (1,2,5)
Let f (x, y, z) = x ^ 2 + y ^ 2-z
Then f ` x (1,2,5) = 2x (1,2,5) = 2
f`y｜（1,2,5）=2y｜（1,2,5）=4
f`z｜（1,2,5）=-1｜（1,2,5）=-1
So the normal vector of this point is (2,4, - 1)
The tangent plane is 2 (x-1) + 4 (Y-2) - (Z-5) = 0
If a ∈ (π, 3 π / 2), then the inequality y
The flat area below, excluding the line itself
Vector a (COSA, Sina) vector b = (COSA, - COSA) y = f (x) = AB, (1) finding the monotone decreasing interval of F (x) (2) f (x)
Axis and center of symmetry
1、
f(x)=cos²a-sinacosa
=1/2*(1+cos2a)-1/2*sin2a
=-(√2/2)sin(2a-π/4)+1/2
The coefficient is less than 0
So if f (x) decreases, sin increases
The increasing interval of SiNx is (2k - π / 2,2k π + π / 2)
2k-π/2
It is known that a belongs to R. for the X inequality (1 + Sina + COSA) x ^ 2 - (1 + 2sina) x + Sina > 0, when XS belongs to [0,1], it is always true and the range of a is obtained
fast
How to prove sin2a = 2 * cosa * Sina?
An isosceles triangle ABC with waist length of 1 and vertex angle of 2a is constructed
Make the height ad on the bottom BC
Then ad = cosa, BD = Sina
So the area s (ABC) = 2S (ABD) = 2 * (1 / 2) * cosa * Sina = cosa * Sina
S (ABC) = (1 / 2) * 1 * 1 * sin2a = (1 / 2) * sin2a
So sin2a = 2 * cosa * Sina
The above proof is 0
Sin2a = 3 / 2 a {0.180} for Sina + cosa
SIN2A=2/3,0
Given Sina + cosa = 0.2, a ∈ (0 ~ 180), find the value of COTA
Senior one knowledge
sinA+cosA=1/5
sinA²+cosA²=1
A∈（0~180）
sinA=4/5
cosA=-3/5
cotA=-3/4
Does a satisfy Sina + cosa = 3 / 2
sina+cosa
=√2(√2/2*sina+√2/2cosa)
=√2(sinacosπ/4+cosasinπ/4)
=√2sin(a+π/4)≤√2
Find the tangent from point P (0.4) to the square of circle x + the square of y-4x-5 = 0
1. (x-2)^2+y^2=9
Let K be the slope
y=kx+4,kx-y+4=0
The distance from the center of the circle to the straight line = r = 3
(2k+4)^2/(1+k^2)=3^2=9
4k^2+16k+16=9+9k^2
5k^2-16k-16=0
(5k+4)(k-4)=0
k=-4/5,k=4
y=-4x/5+4;y=4x+4
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