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Exercise 3
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org
Calculus and Analytic Geometry, MATHEMATICS 12
Merging man and maths
Available online @ http://www
...
org, Version: 1
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0
Question # 1(i)
y = x 2 − 1 …… (i)
x = 3 & δ x = 3
...
02
2
y + δ y = ( x + δ x ) −1
⇒ δ y = ( x + δ x ) − 1 − x2 + 1
2
= ( x + δ x ) − x2
Put x = 3 & δ x = 0
...
02 ) − (3) 2
⇒ 4 x 3 dx + 2 ydy = x ⋅ 2 ydy + y 2 dx
⇒ 2 ydy − 2 xydy = y 2 dx − 4 x3 dx
δ y = 0
...
02
dy = 2 ( 3)( 0
...
12
Question # 1(ii)
Do yourself as above
...
(i)
x = 4 & δ x = 4
...
41
1
y + δ y = ( x + δ x)2
1
⇒ δ y = ( x + δ x)2 − x2
Put x = 4 & δ x = 0
...
41) 2 − ( 4 ) 2
= 2
...
1
1
1
2x 2
dx
Put x = 4 & dx = δ x = 0
...
41
dy =
1 ( 0
...
1025
Question # 2(i)
xy + x = 4
Taking differential on both sides
d ( xy ) + dx = d ( 4 )
⇒ xdy + ydx + dx = 0
⇒ xdy + ( y + 1) dx = 0
⇒ xdy = − ( y + 1) dx
⇒
dy
y +1
= −
dx
x
dx
x
&
=−
dy
y +1
)
dy
y − 4x
=
dx
2 y (1 − x )
2
3
2 y (1 − x )
dx
=
dy
y 2 − 4 x3
Question # 2(iv)
xy − ln x = c
Taking differential
d ( xy ) − d ( ln x ) = d ( c )
1
⇒ xdy + ydx − dx = 0
x
1
⇒ xdy = dx − ydx
x
1
= − y dx
x
1 − xy
⇒ xdy =
dx
x
dy
1 − xy
⇒
=
dx
x2
dx
x2
&
=
dy
1 − xy
Question # 3(i)
Let
y = f ( x) = 4 x
where x = 16 and δ x = dx = 1
Taking differential of above
dy = d
( x)
4
1
= d ( x)4
1 1 −1
x 4 dx
4
1 −3
=
x 4 dx
4
1
=
dx
3
4x 4
Put x = 16 and dx = 1
1
dy =
(1)
3
4
4 (16 )
=
=
1
( )
4 2
3
4 4
=
1
= 0
...
1 - 2
Now f ( x + dx ) ≈ y + dy
= f ( x) + dy
⇒
⇒
4
4
16 + 1 ≈
17 ≈
4
(2 )
1
4 4
Q y = f ( x)
16 + 0
...
03125
Now dy = d ( sin x )
= cos x dx
Put x = 60o and dx = δ x = 0
...
01745 )
Now
Question # 3(ii)
Let y = f ( x) = ( x )
Where x = 8 & δ x = dx = 0
...
866 + 0
...
8747
1
dy = d ( x )3
1
=
1
( )
2
3 3
Question # 4
Let x be the length of side of cube where
x = 5 & δ x = 5
...
02
Assume V denotes the volume of the cube
...
2
1
dy =
0
...
2 )
=
3 2
1
( 0
...
2 )
(8
...
02
dV = 3 ( 5) ( 0
...
5
Hence increase in volume is 1
...
2
= 0
...
008725
= f ( x) + dy
⇒ sin ( 60 + 1) = sin 60o + 0
...
5 )( 0
...
03125 = 2
...
01667
2 + 0
...
01667
Question # 3(iii)
1
Let y = f ( x) = x 5
Where x = 32 & δ x = dx = −1
Try yourself as above
...
01745 rad
Where x = 30o & δ x = −1o = −
Now dy = d ( cos x )
= − sin x dx
Put x = 30o and dx = δ x = −0
...
01745 )
= − ( 0
...
01745) = 0
...
008725
⇒ cos 29o = 0
...
008725
= 0
...
01745 rad
Where x = 60o & δ x = 1o =
Question # 5
Let x denotes diameter of a disc
Where x = 44 cm & δ x = 44
...
4
x
Then radius =
2
Let A denotes the area of the disc
2
Then A = π ( radius )
x
π 2
= π =
x
2
4
Taking differential
π
dA = d x 2
4
π
π
=
⋅ 2 x ⋅ dx =
x dx
2
4
Put x = 44 and dx = δ x = 0
...
4 )
2
= ( 3
...
4 )
= 27
...
65 cm2
2
² ---- The End --- ²
Tuesday, 20 September 2005
By mathcity
...
com