วันจันทร์ที่ 14 เมษายน พ.ศ. 2557

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

วันพฤหัสบดีที่ 10 เมษายน พ.ศ. 2557

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
HenceA  = 
πh2
9
Take Diff
dA
dh
 = 
2πh
9
And
dA
dt
 = 
dA
dh
 x
dh
dt
Substitue h=4 and
dh
dt
 = 
45
, we get  
dA
dt
=
2π(4)
9
x
45
⇒                          
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.

วันพุธที่ 9 เมษายน พ.ศ. 2557

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


วันอังคารที่ 8 เมษายน พ.ศ. 2557

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.