Calculus - Application of Differentiation

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# h3cadiap2          →  rates of change

A 26 foot ladder is placed against a wall (Fig.1).   If the top of the ladder is sliding down the wall at 2 feet per second, at what rate is the bottom of the ladder moving away from the wall when the bottom of the ladder is 10 feet away from the wall?

(Problem credited by -- High School level)

Strategy the rule "Must have" of Mr.Zhang ® The ladder is a constant length, the bottom of the ladder will move away from the wall at the same rate which the top of the ladder is also moving down the wall. Let x be the distance of the bottom of the laddfer from the wall, and let Y be the distance of the top of the ladder on the wall (see Figure 1). Both x and y are changing with respect to time and can be thought of as functions of time ;that is , x=x(t) and y=y(t). x and y are related by ther Pythagorean relationship:                    x2 + y2 = 262   → (1)

The rate of of the bottom of the ladder moving away from the wall    →	dx dt
Also take diff both side from the equation(1) to find      →	dx dt

Solution

The ladder is a constant length, the bottom of the ladder will move away from the wall at the same rate that the top of the ladder is moving down the wall. At any moment in time, let x be the distance of the bottom of the laddfer from the wall, and let Y be the distance of the top of the ladder on the wall (see Figure 1). Both x and y are changing with respect to time and can be thought of as functions of time ; that is , x=x(t) and y=y(t). Furthermore, x and y are related by ther Pythagorean relationship:

x2 + y2 = 262    →  (2)

Take Diff both side;

dx dt

2x

+

dy dt

2y

=

0

→  (3)

Problem gives ;	dy dt	=-2 (y is decreasing at a constant rate of 2 feet per second),
problem is to find	dx dt	when x =10 feet

102 + y2	=	262
y	=	24 feet

Substitute

dy dt

= -2, x = 10, and y =24 into (2), then solve for

dx dt

2(10)

dx dt

+ 2(24)(-2) = 0

∴

dx dt

=

4.8 feet·sec-1

Ans.  4.8 feet·sec-1 is the rate of the bottom of the ladder moving away from the wall when the bottom of the ladder is 10 feet away from the wall

Fuel Injector

Calculus Applications of Differentiation

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# h3cadiap1

Water is poured into an inverted circular cone of base radius 5 cm and height 15 cm at the rate of 10 cm3•s-1  Calculate; a)the rate of increase of the height of water level. b)the rate of increase of the surface area of the water,when the water level is 4 cm high. (Leave answers in terms of π where applicable.)

(Problem credited by O-Level --Marshal Cavendish)

Strategy the rule "Must have" of Mr.Zhang ® Give the water level be h cm high after time t seconds, and the radius of the water surface r cm at this moment as the figure 1 which is a cross-section of the cone.

1. Similar triangle,find r in terms of h
2. Water volume       (V)   =		1 3	πr2h → (1)

3.  Take ;

dy

dx

   both side to get the rate of increase.

Solution
Take similar triangle ;
	r h	=	5 15
	r	=	h 3	→ (2)
From equation(1)	v	=	1 3	π r2h → (1)
Take r from equation(2) substitued to (1)	v	=	1 3	π	(	h 3	)2	h
⇒	v	=	π	h3 27
Take Diff	dV dh	=	π	h2 9
From	dV dt	=		dV dh	x	dh dt

dV

dt

=	10 cm3•s-1 and h = 4 , then ;

	10	=	π 42 9	x	dh dt
	10x9 16 π	=	dh dt
	45 8π	=	dh dt

Ans.   a) the rate of increase of the height when the height is 4 cm, is

45 8π

cm•s-1

b) The water surface is circular,therefore, the surface area

	A	=	πr2	at any instant
From equation(2),	r	=	h 3
Hence	A	=	πh2 9
Take Diff	dA dh	=	2πh 9
And	dA dt	=	dA dh	x	dh dt

Substitue h=4 and

dh dt

=

45 8π

,

we get

dA dt

=

2π(4) 9

x

45 8π

⇒

dA dt

=

5 cm2•s-1

Ans.   the surface rate of the water is increasing at the rate of 5 cm2•s-1 when the water level height is 4 cm.

Fuel Injector

Calculus Integration

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# h3macaar1

Calculate the area enclosed by the curve y = x2+1, the y-axis and the lines, y=2,y=3 for the portion in the first quadrant only.

(Problem credited by O-Level --Marshal Cavendish)

Strategy - the rule "Must have" of Mr.Zhang ® Find the area ;   → (1) To calculate area A, x must be expressed in terms of y for the intregation.

Solution	We have ;
y	=	x2 + 1
x2	=	y - 1
x	=	(y-1) 1/2

We bring the equation(1) to find the required area;
	=
	=
	=	2 3	(2)3/2   -	2 3	(1)3/2
	=	2 3	(2√2)   -	2 3
	=	2 3	(2√2 - 1)

.

Ans.    The area is

2 3

(2√2 - 1) units2

Fuel Injector

Calculus applied

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# h3macasp1

A force of 40 N. is required to hold a spring that has been stretched from its natural length of 10 cm to a length of 15 cm.    How much work is done in stretching the spring from 15 cm to 18 cm?

Strategy - the rule "Must have" of Mr.Zhang ®
Find the value of k ;	f(x)=kx	→ (1)
And later find work done (w) from ;		→ (2)

Solution

Find k(constant of spring) from equation(1)

40	=	0.05k
k	=	40 0.05
	=	800

Thus, f(x)=800x

And the work done in stretching the spring from 15 cm to 18 cm is

Find work done(w) from equation(2)

w	=
	=
	=	400[(0.18)2-(0.15)2]
	=	3.96 J

Ans.    3.96 J work is used for stretching the spring from 15 cm to 18 cm.

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