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2 years ago

The vector v⃗ in 2-space of length 3 pointing up at an angle of π/4 measured from the positive x-axis.

1 answer:

5 0

The vector v in 2-space of length 3 pointing up at an angle of π/4 measured from the positive x-axis is: (3/√2, 3√2) and The vector w in 3-space of length 1 lying in the yz-plane pointing upward at an angle of 2π/3 measured from the positive y-axis is: (0, -1/2, √3/2).

<h3>Vector</h3>

a. Vector (v)

Vector (v)=v (cos Ф, sin Ф)

V=1 while the counterclockwise angle that is measured from positive x=Ф=π/4

Hence:

Vector=3(cos π/4, sinπ/4)

Vector=(3/√2, 3√2)

b. Vector w:

Vector w=1(0, cos 2π/3, sin2π/3)

Vector w=(0, -1/2, √3/2)

Therefore the vector v in 2-space of length 3 pointing up at an angle of π/4 measured from the positive x-axis is: (3/√2, 3√2) and The vector ws: (0, -1/2, √3/2).

The complete question is:

Resolve the following vectors into components:

a. The vector v in 2-space of length 3 pointing up at an angle of π/4 measured from the positive x-axis.

(b) The vector w in 3-space of length 1 lying in the yz-plane pointing upward at an angle of 2π/3 measured from the positive y-axis.

Learn more about vector here:brainly.com/question/25705666

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A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot

ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = $(x\times x)$

$=x^2$

The volume of the box is = area of the base × height

$=x^2h$

Therefore,

$x^2h=272$

$\Rightarrow h=\frac{272}{x^2}$

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

$=20x^2$ cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

$=10x^2$ cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

$=4xh\times 1.5$ cents

$=6xh$ cents

Total cost = $(20x^2+10x^2+6xh)$

$=30x^2+6xh$

Let

C $=30x^2+6xh$

Putting the value of h

$C=30x^2+6x\times \frac{272}{x^2}$

$\Rightarrow C=30x^2+\frac{1632}{x}$

Differentiating with respect to x

$C'=60x-\frac{1632}{x^2}$

Again differentiating with respect to x

$C''=60+\frac{3264}{x^3}$

Now set C'=0

$60x-\frac{1632}{x^2}=0$

$\Rightarrow 60x=\frac{1632}{x^2}$

$\Rightarrow x^3=\frac{1632}{60}$

$\Rightarrow x\approx 3$

Now $C''|_{x=3}=60+\frac{3264}{3^3}>0$

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is $=\frac{272}{3^2}$

=30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

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Step-by-step explanation:

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Step-by-step explanation:

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