The difference between solving normal vector and tangent vector The meaning is different, it should be vertical, but why is it the same, all are partial derivatives. One (1, YX, ZX), and another (FX, FY, Fz)

The difference between solving normal vector and tangent vector The meaning is different, it should be vertical, but why is it the same, all are partial derivatives. One (1, YX, ZX), and another (FX, FY, Fz)

The difference is that tangent vector is for curve and normal vector is for surface
For example,
Curve X ^ 2 + y ^ 2 = a ^ 2
As a curve, the tangent vector of the upper point is s = (x'x, y'x, Z'x) = (1, - X / y, 0)
If this curve can be extended into a cylinder in space, as a cylinder, its normal vector is n = (x, y, 0)
It can be seen that the two are vertical

The difference between parallel vector and collinear vector in Higher Mathematics
When reviewing the space analytic geometry and vector algebra of Tongji edition of higher mathematics, I didn't understand the concepts of collinear and parallel
If two non-zero vectors have the same or opposite directions, they are said to be parallel. When the starting points of the two parallel vectors are placed at the same point, their end points and common starting points should be in a straight line. Therefore, two vectors are parallel, also known as two vectors collinear
There is a saying on the Internet that "a collinear vector is not necessarily a parallel vector, but a parallel vector must be a collinear vector."
Why are collinear vectors not necessarily parallel vectors
Can you give an example

"When the starting points of these two parallel vectors are placed at the same point, their end points and common starting points should be in a straight line. Therefore, two vectors are parallel, also called two vectors collinear."
As discussed in this passage, if two vectors are collinear, then they must be parallel vectors, so the proposition is wrong
If we have to get to the bottom of the problem, then the correct statement of the proposition should be "if two vectors are parallel, but they are not necessarily collinear", because for example, zero vector is parallel to any vector, but you can't say which vector it is collinear with

Find the tangent equation of curve y = x (ln-1) at point (E, 0)
Find the tangent equation of curve y = x (lnx-1) at point (E, 0)

Y = x (lnx-1) finding derivative is the slope of tangent. Y '= (lnx-1) + X * 1 / x = LNX
In (E, 0), the tangent slope is k = lne = 1
So y-0 = 1 * (x-e)
Y = x-e is tangent

The solution of (7x + 70): (4x + 70) = 8:5

Cross multiplication (7x + 70) × 5 = (4x + 70) × 8
Simplify 3x = 210
x＝70

It is known that the equation cos & # 178; X + 4sinx-a = 0 has a solution, and the value range of A
Do you want to consider Delta ≥ 0?

Don't consider the discriminant
That is, a = cos & # 178; X + 4sinx
=-sin²x+4sinx+1
=-(sinx-2)²+5
∵ sinx∈[-1,1]
∴ a∈[-4,4]

The product of a number and its reciprocal plus five-thirds of a is used to find a

X*（1/X）+a=5/3;
1+a=5/3;
a=2/3

Find the equation of the line parallel to the line 3x + 4y-12 = 0 and its distance is 7

Firstly, let the required equation be 3x + 4Y + C = 0.. then, by using the distance formula between parallel lines, we get d = | C + 12 | / √ 3 & # 178; + 4 & # 178; = 7.. that is to say, | C + 12 | = 35.. then c + 12 = 35 or C + 12 = - 35 then C = 23 or - 47. So far, the obtained equation is 3x + 4Y + 23 = 0 or 3x + 4y-47 = 0

Given the function f (x) = AX2 + 3x + 1x + 1 and this function has only one zero point in its domain of definition. (1) find the value set of real number A. (2) when a ∈ n *, let the sum of the first n terms of the sequence {an} be Sn, and Sn = n · f (n), find the general term formula of {an}. (3) under the condition of (2), if the sequence {an} is a finite sequence with fixed N terms, we extract a term (excluding the first one) from it The average value of the remaining items is 31. Find the number of items in this sequence and point out which item is removed

(1) The domain of a function is {x ∈ R | x ≠ - 1}. Because a function has and has only one zero point in its domain, when a = 0, the function has only one zero point, x = - 13 (1) when a ≠ 0, there is only one solution from AX2 + 3x + 1 = 0 x + 1 ≠ 0, which can be divided into two cases: (1) the quadratic equation AX2 + 3x + 1 = 0

The coordinates of the intersection of the parabola y = - 1 / 2 (x-3) &# + 2 and the Y axis are, and the coordinates of the intersection of the parabola y = - 1 / 2 (x-3) &# + 2 and the X axis are

When x = 0, y = - 5 / 2, so the intersection coordinates with y axis are (0, - 5 / 2);
When y = 0, that is: - (x-3) &# / 2 + 2 = 0
-(x-3)²+4=0
(x-3)²=4
x1=1,x2=5
So the coordinates of the intersection point with the X axis are (1,0) and (5,0)

It is known that the axis of symmetry of the parabola y = ax ^ 2 + BX-1 is x = - 1, and the highest point is on the straight line y = 2x + 4

A:
The symmetry axis of parabola y = ax ^ 2 + BX-1 x = - B / (2a) = - 1
So: B = 2A
The highest point is on the straight line y = 2x + 4, indicating that the opening of the parabola is downward, a