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3 years ago

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A remote ranch has a cylindrical water storage tank. It has a vertical central axis, a diameter of 16 ft, the sides are 5 ft hig

h, and it is full to the brim. The depth of this water is 4 ft. How much work (in ft-lb) would be required to pump all of the water over the upper rim? (Use the fact that water weighs 62.5 lb/ft3. Round your answer to the nearest integer.)

1 answer:

Vikentia [17]3 years ago

5 0

Answer and Step-by-step explanation:

Given: diameter = 16ft height = 5 ft. Depth = 4ft

Work =?

As we know that work = force x distance

W= distance x density x volume

W= ʃ04 (5-y) 62.5x ∏ (d/2)2dy (area = ∏r2 , r= d/2)

= 62.5 ʃ04 (5-y) ∏ (4y) dy

=62.5∏ ʃ04 (5-y) 4ydy

=62.5∏ ʃ04 20y-4y2dy

=62.5∏ |20y2/2 – 4y3/3|04

=62.5∏ |74.66|

=14659ft.lbs (approximately)

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garri49 [273]

Answer:

<h3>METHOD I:</h3>

(by using the first principle of differentiation)

We have the <u>"Limit definition of Derivatives"</u>:

$\boxed{\mathsf{f'(x)= \lim_{h \to 0} \{\frac{f(x+h)-f(x)}{h} \} ....(i)}}$

<em>Here, f(x) = sec x, f(x+h) = sec (x+h)</em>

<em>Substituting these in eqn. (i)</em>

$\implies \mathsf{f'(x)= \lim_{h \to 0} \{\frac{sec(x+h)-sec(x)}{h} \} }$ <em />

<em>sec x can be written as 1/ cos(x)</em>

<em /> $\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{1}{h} \{\frac{1}{cos(x+h)} -\frac{1}{cos(x)} \} }$ <em />

<em>Taking LCM</em>

<em /> $\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{1}{h} \{\frac{cos(x)-cos(x+h)}{cos(x)cos(x+h)} \} }$ <em />

<em>By Cosines sum to product formula, i.e.,</em>

$\boxed{\mathsf{cos\:A-cos\:B=-2sin(\frac{A+B}{2} )sin(\frac{A-B}{2} )}}$

<em>=> cos(x) - cos(x+h) = -2sin{(x+x+h)/2}sin{(x-x-h)/2}</em>

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{2sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: \lim_{h \to 0} \frac{sin(\frac{h}{2} )}{h} }$

<em>I shifted a 2 from the first limit to the second limit, since the limits ar ein multiplication this transmission doesn't affect the result</em>

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: \lim_{h \to 0} \frac{2sin(\frac{h}{2} )}{h} }$

<em>2/ h can also be written as 1/(h/ 2)</em>

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: \lim_{h \to 0} \frac{1\times sin(\frac{h}{2} )}{\frac{h}{2} } }$

<em>We have limₓ→₀ (sin x) / x = 1. </em>

<em /> $\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: 1 }$ <em />

<em>h→0 means h/ 2→0</em>

<em>Substituting 0 for h and h/ 2</em>

<em /> $\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+0)}{cos(x+0)cos(x)} }$ <em />

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x)}{cos(x)cos(x)} }$

$\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x)}{cos(x)}\times \frac{1}{cos x} }$

<em>sin x/ cos x is tan x whereas 1/ cos (x) is sec (x)</em>

$\implies \mathsf{f'(x)= tan(x)\times sec(x) }$

Hence, we got

$\underline{\mathsf{\overline{\frac{d}{dx} (sec(x))=sec(x)tan(x)}}}$

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3>METHOD II:</h3>

(by using other standard derivatives)

$\boxed{ \mathsf{ \frac{d}{dx} ( \sec \: x) = \sec x \tan x }}$

sec x can also be written as (cos x)⁻¹

We have a standard derivative for variables in x raised to an exponent:

$\boxed{ \mathsf{ \frac{d}{dx}(x)^{n} = n(x)^{n - 1} }}$

Therefore,

$\mathsf{ \frac{d}{dx}( \cos x)^{ - 1} = - 1( \cos \: x) ^{( - 1 - 1} } \\ \implies \mathsf{\ - 1( \cos \: x) ^{- 2 }}$

Any base with negative exponent is equal to its reciprocal with same positive exponent

$\implies \: \mathsf{ - \frac{1}{ (\cos x) {}^{2} } }$

The process of differentiating doesn't just end here. It follows chain mechanism, I.e.,

<em>while calculating the derivative of a function that itself contains a function, the derivatives of all the inner functions are multiplied to that of the exterior to get to the final result</em>.

The inner function that remains is cos x whose derivative is -sin x.

$\implies \mathsf{ - \frac{1}{ (\cos x )^{2} } \times ( - \sin x) }$

cos²x can also be written as (cos x).(cos x)

$\implies \mathsf{ \frac{ \sin x }{ \cos x } \times ( \frac{1}{cos x} ) }$

<u>sin x/ cos x</u> is tan x, while <u>1/ cos x</u> is sec x

$\implies \mathsf{ \tan x \times \sec x }$

= sec x. tan x

<h3>Hence, Proved!</h3>

7 0

3 years ago

A farmer has some cows and horses. All the animals are either brown or black. The table shows how many of each

ludmilkaskok [199]

Answer:

P = 1/5 = 0.20 = (20%)

Step-by-step explanation:

there are 12 horses and 8 cows, in total there are 20 animals

of which 4 horses are brown

Then

$P=\frac{4}{20} =\frac{1}{5}$

Hope this helps

5 0

2 years ago

The coordinates of the following points represent the vertices of a

Setler [38]

I dont know i wish i could help ask me anytime with any thing else

7 0

3 years ago

Read 2 more answers

What’s the surface area in square units if right I’ll mark brainleiest

erik [133]

Answer: 349 square units

============================================

Work Shown:

L = 5.5 = length
W = 5 = width
H = 14 = height

SA = surface area of rectangular prism

SA = 2*(L*W + L*H + W*H)

SA = 2*(5.5*5 + 5.5*14 + 5*14)

SA = 2*(27.5 + 77 + 70)

SA = 2*(174.5)

SA = 349

The surface area is 349 square units.

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3 years ago

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I don't understand.......

devlian [24]

Answer:

Step-by-step explanation:

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8 0

3 years ago