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The tangent plane equation at point P (1.1.1)
The square of surface y = 3x - the square of 2Z
F (x, y, z) = partial derivative at point P (1.1.1) of 3x ^ 2-y-2z ^ 2
ðF/ðx = 6x=6
ðF/ðy = -1,
ðF/ðz = -4z=-4
Normal vector of tangent plane n = {6, - 1, - 4}
Tangent plane equation 6 (x-1) - (Y-1) - 4 (Z-1) = 0
6x-y-4z = 1
Tangent plane equation of surface x ^ 2 + y ^ 2 + Z ^ 2 = 3 at (1,1,1)
x+y+z=3
Finding the tangent plane equation of surface 2 ^ (x / 2) + 2 (Y / 2) = 8 at point (2,2,1)
F(x,y,z) = 2^(x/2)+2^(y/2) - 8,
ðF/ðx = 2^(x/2) * ln2 * (1/2) = ln2,ðF/ðy = ln2,ðF/ðz = 0
The normal vector of tangent plane n = {LN2, ln2,0} / {1,1,0}
Tangent plane equation (X-2) + (Y-2) = 0, i.e. x + y = 4
If the height of a triangle is 10 square decimeters, then the height of a parallelogram is (_ ）Rice?
Lily Jiaying,
If the height of a triangle is 10 decimeters, then the height of a parallelogram is (10 △ 2 = 5 decimeters = 0.5 meters)
In the figure below, the area of the shadow triangle is 4 / 5 square decimeters, and the area of the blank part is () square decimeters
Parallelograms and triangles are of equal base and height
So the area of a parallelogram is twice that of a triangle
So the area of the blank part is equal to the area of the triangle, which is 4 / 5 square decimeters
A rectangular figure is divided into four unequal triangles, which are 20 square decimeters, 18 square decimeters and 10 square decimeters respectively. Find the area of the remaining triangles
1： 20 + 10-18 = 12; 18 + 20-10 = 28; 10 + 18-20 = 8
2： The area sum of three triangles with the base on the same side is equal to the area sum of another triangle
20+10+18=48
As shown in the figure, the upper bottom of trapezoidal aebd is 1.4 decimeters, the lower bottom is 3.2 decimeters, and the area of triangle Abe is 1.54 square decimeters
1.54 × 2 △ 1.4 = 2.2 (decimeter), trapezoid area: (1.4 + 3.2) × 2.2 △ 2, = 4.6 × 2.2 △ 2, = 5.06 (square decimeter), rectangle area: (5.06-1.54) × 2 = 7.04 (square decimeter); answer: rectangle ABCD area is 7.04 square decimeter
If the area of the triangle formed by the image of the linear function y = 2x + B and the two coordinate axes is 8, then B =?
X = 0, y = B, so the intersection of Y axis is (0, b)
Y = 0, x = - B / 2, so the intersection of X axis is (- B / 2,0)
The area of triangle is 1 / 2 * LBL * L-B / 2L = 8
b^2=8
B = 2 radical 2 or - 2 radical 2
The image of a function of degree is known to pass through points (3,5), (- 4, - 9). (1) find the intersection coordinates of the image and the coordinate axis and the area of the enclosed triangle
Let y = KX + B, slope k = (5 - (- 9)) / (3 - (- 4)) = 2
5 = 2 * 3 + B get b = - 1 get y = 2x-1 get intersection (1 / 2,0) and (0, - 1)
Area s = 1 / 2 * 1 / 2 * 1 = 1 / 4
The image of the first-order function y = - 2 / 3x + 2 intersects the x-axis and y-axis at points a and B respectively. Take the line AB as the edge to make the isosceles right triangle ABC, and the angle BAC = 90 degrees. The analytical formula of the two-point straight lines B and C is obtained
The length of a (3,0), B (0,2) AB = √ 13 ∵ BA ⊥ AC is perpendicular to the x-axis and intersects the x-axis with D △ ABO ≌ △ AOCD (corner) ob = ad = 2 CD = OA = 3. The coordinate of C point is (5,3). Let the analytic expression of BC line be y = KX + B
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