g A pipe of negligible diameter is to be carried horizontally around a corner from a hallway 8 ft wide into a hallway 4 ft wide. What is the longest length that the pipe can be

The length of the longest pipe that can be carried horizontally is given by the minimum value of the function of the length of the pipe.

The longest length that the pipe can be is approximately 16.648 feet.

Reasons:

The length of the pipe, L

Width of the first hallway = 8 ft.

The width of the second hallway = 4 ft.

We have;

(L - a)·sin(θ) = 8

a·cos(θ) = 4

Therefore;

L - a = 8·csc(θ)

a = 4·sec(θ)

L(θ) = 8·csc(θ) + 4·sec(θ)

At the minimum length, we have;

L'(θ) = 4·sec(θ)·tan(θ) - 8·csc(θ)·cot(θ) = 0

4·sec(θ)·tan(θ) = 8·csc(θ)·cot(θ)

Therefore;

[tex]\displaystyle \frac{sec(\theta) \cdot tan(\theta)}{csc(\theta) \cdot cot(\theta)} = \frac{sin(\theta) \cdot tan(\theta) \cdot tan(\theta) }{cos(\theta) } =tan^3(\theta) = \frac{8}{4}[/tex]

[tex]\displaystyle tan(\theta) =\sqrt[3]{\frac{8}{4} } ={\sqrt[3]{2} }[/tex]

θ = arctan([tex]\sqrt[3]{2}[/tex]) ≈ 51.56°

a = 4×sec(arctan([tex]\sqrt[3]{2}[/tex]))

a ≈ 6.434

L = 8·csc(θ) + a = 8 × csc(arctan([tex]\sqrt[3]{2}[/tex])) + 4×sec(arctan([tex]\sqrt[3]{2}[/tex])) ≈ 16.648

The longest length that the pipe can be, L ≈ 16.648 ft.

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The longest length of the pipe is an illustration of trigonometry functions

The longest length of the pipe is 16.64 ft

The widths of the hallways are given as:

[tex]\mathbf{w_1 = 8}[/tex]

[tex]\mathbf{w_2 = 4}[/tex]

See attachment for the diagram that represents the scenario

From the diagram, we have:

[tex]\mathbf{sin(\theta) = \frac 4x}[/tex]

[tex]\mathbf{cos(\theta) = \frac 8y}[/tex]

Make x and y the subject

[tex]\mathbf{x = \frac 4{sin(\theta)}}[/tex]

[tex]\mathbf{y = \frac 8{cos(\theta)}}[/tex]

Where, the total length (L) of the pipe is:

[tex]\mathbf{L = x + y}[/tex]

So, we have:

[tex]\mathbf{L = \frac 4{sin(\theta)} + \frac 8{cos(\theta)}}[/tex]

Differentiate

[tex]\mathbf{L' = -\frac{4cos(\theta)}{sin^2(\theta)} + \frac{8sin(\theta)}{cos^2(\theta)}}[/tex]

Set to 0

[tex]\mathbf{-\frac{4cos(\theta)}{sin^2(\theta)} + \frac{8sin(\theta)}{cos^2(\theta)} = 0}[/tex]

Rewrite as:

[tex]\mathbf{\frac{4cos(\theta)}{sin^2(\theta)} = \frac{8sin(\theta)}{cos^2(\theta)}}[/tex]

Divide through by 4

[tex]\mathbf{\frac{cos(\theta)}{sin^2(\theta)} = \frac{2sin(\theta)}{cos^2(\theta)}}[/tex]

Cross multiply

[tex]\mathbf{2sin^3(\theta) = cos^3(\theta)}[/tex]

Divide both sides by [tex]\mathbf{2cos^3(\theta)}[/tex]

[tex]\mathbf{tan^3(\theta) = \frac 12}[/tex]

Take cube roots of both sides

[tex]\mathbf{tan(\theta) = \sqrt[3]{\frac 12}}[/tex]

[tex]\mathbf{tan(\theta) = 0.7937}[/tex]

Take arc tan of both sides

[tex]\mathbf{\theta = tan^{-1}(0.7937)}[/tex]

[tex]\mathbf{\theta = 38.4}[/tex]

Substitute [tex]\mathbf{\theta = 38.4}[/tex] in [tex]\mathbf{L = \frac 4{sin(\theta)} + \frac 8{cos(\theta)}}[/tex]

[tex]\mathbf{L = \frac 4{sin(38.4)} + \frac 8{cos(38.4)}}[/tex]

Evaluate

[tex]\mathbf{L = 6.43+ 10.21}}[/tex]

[tex]\mathbf{L = 16.64}}[/tex]

Hence, the longest length of the pipe is 16.64 ft

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SUBIECTUL I

1. 1p Să se determine primul termen al unei progresii aritmetice (an

)n≥1 știind că rația este

egală cu 5 și suma primilor zece termeni ai progresiei este egală cu 245.
2. 1p Să se determine punctele de intersecție ale graficului funcției
f: R → R, f (x) = x

2 − 7x + 12, cu axele de coordonate.
3. 1p Să se determine numărul real a pentru care vectorii
u⃗ = (a − 2)i + (3a − 1)j și v = 2i +4j sunt coliniari.
4. 1p Să se calculeze ctg x, știind că sin x =
√3
2
, x ∈ (0, 90°
).
5. 1p Să se determine soluțiile reale ale ecuației 49x+2 = 7
x
2+5
.

6. 1p Să se rezolve ecuația : Ax
2+Cx
2 = 0, x∈ N , x≥ 2.

SUBIECTUL II
1 Se consideră matricea A(a) = (

1 0 √a
1 0 1
a 1 1

), unde a ∈ (0, ∞).

a) 0.5p Arătați că det(A(4)) = 1.
b) 0.5p Demonstrați că det(A(a) ∙ A(1) − A(a + 1)) > 0, pentru orice a ∈ (0, ∞).

2. Pe mulțimea de numere reale se definește legea de compoziție asociativă
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b) 0.5pDemonstrați că x ∘ y = √3(x − √3)(y − √3) + √3, pentru orice numere reale x și y.
SUBIECTUL III

1 0.5p Se consideră funcția f: R → R, f(x) = e
2x
(x − 5).

Arătați că f
′
(x) = e
2x
(2x − 9), x ∈ R .
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x
√x
2+1
.

Arătați că ∫ f(x)√x
2 + 1
2
0

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