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QveST [7]
3 years ago
15

Consider F and C below. F(x, y, z) = yz i + xz j + (xy + 6z) k C is the line segment from (2, 0, −2) to (5, 4, 2) (a) Find a fun

ction f such that F = ∇f. f(x, y, z) = Correct: Your answer is correct. (b) Use part (a) to evaluate C ∇f · dr along the given curve C.
Mathematics
1 answer:
WITCHER [35]3 years ago
3 0

Answer:

Required solution (a) f(x,y,z)=xyz+3z^2+C (b) 40.

Step-by-step explanation:

Given,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k

(a) Let,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}

Then,

f_x=yz,f_y=xz,f_z=xy+6z

Integrating f_x we get,

f(x,y,z)=xyz+g(y,z)

Differentiate this with respect to y we get,

f_y=xz+g'(y,z)

compairinfg with f_y=xz of the given function we get,

g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)

Then,

f(x,y,z)=xyz+h(z)

Again differentiate with respect to z we get,

f_z=xy+h'(z)=xy+6z

on compairing we get,

h'(z)=6z\implies h(z)=3z^2+C   (By integrating h'(z))  where C is integration constant. Hence,

f(x,y,z)=xyz+3z^2+C

(b) Next, to find the itegration,

\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40

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3 years ago
Orbet used 24 square tiles to make a rectangle. The perimeter of the rectangle is 20 units. What would be the perimeters of the
Dmitriy789 [7]

Answer:

The perimeter of other rectangles are 50, 28, 22 units

The area of all possible rectangles are same and equals to 24 unit²

Step-by-step explanation:

Given - Orbet used 24 square tiles to make a rectangle. The perimeter of the rectangle is 20 units.

To find -  What would be the perimeters of the other rectangles Orbet could make using only these 24 tiles? How do the areas of the rectangles compare?

Proof -

We know that

Are of Rectangle = Length × Breadth

And Perimeter of Rectangle = 2 (Length + Breadth )

Now,

Given that,

Orbet used 24 square tiles

And perimeter of the rectangle made = 20

So,

Possible Length of rectangle = 6

Possible breadth of Rectangle = 4

Or Vice-versa.

Total number of Rectangles possible = 4

Possibilities are -  1 × 24, 2 × 12, 3 × 8, 4 × 6

Case I :

Length of rectangle = 1

Breadth of rectangle = 24

∴ Perimeter of rectangle = 2(1 + 24) = 2(25) = 50 units

Area of rectangle = 1 × 24 = 24 unit²

Case II :

Length of rectangle = 2

Breadth of rectangle = 12

∴ Perimeter of rectangle = 2(2 + 12) = 2(14) = 28 units

Area of rectangle = 2 × 12 = 24 unit²

Case III :

Length of rectangle = 3

Breadth of rectangle = 8

∴ Perimeter of rectangle = 2(3 + 8) = 2(11) = 22 units

Area of rectangle = 3 × 8 = 24 unit²

Case IV :

Length of rectangle = 4

Breadth of rectangle = 6

∴ Perimeter of rectangle = 2(4 + 6) = 2(10) = 20 units

Area of rectangle = 4 × 6 = 24 unit²

∴ we get

The perimeter of other rectangles are 50, 28, 22 units

The area of all possible rectangles are same and equals to 24 unit²

5 0
3 years ago
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