Embed Size (px)
Citation preview
FSc-II / Ex 2.4 - 1
mathcity.org Merging man and maths
Exercise 2.4 (Solutions) Calculus and Analytic Geometry, MATHEMATICS 12
Available online @ http://www.mathcity.org, Version: 1.1.0
Question # 1 (i)
11
xyx
−=
+
Put 11
xux
−=
+
So y u= 12y u⇒ =
Now diff. u w.r.t. u
11
du d xdx dx x
− = +
( ) ( ) ( ) ( )
( )2
1 1 1 1
1
d dx x x xdx dx
x
+ − − − +=
+
( )( ) ( )( )( )2
1 1 1 11
x xx
+ − − −=
+
( )2
1 11
x xx
− − − +=
+
( )2
21
dudx x
−⇒ =
+
Now diff. y w.r.t. u
12dy d u
du du=
121
2u−
= 121 1
2 1xx
−− = +
121 1
2 1dy xdu x
+ ⇒ = −
Now by chain rule dy dy dudx du dx
= ⋅
( )2
121 1 2
2 1 1xx x
+ − = ⋅ − +
( )( ) ( )2
12
12
1 111
xxx
+ −= ⋅
+−
( ) ( )
1221
2
1
1 1x x −
−=
− +
( )
32
1
1 1x x
−=
− + Answer
Question # 1(ii)
y x x= +
Let u x x= + 12x x= +
y u⇒ =12u=
Diff. u w.r.t. x
12du d x x
dx dx = +
1211
2x
−= + 11
2 x= +
2 12
xx+
=
Now diff. y w.r.t. x
12dy d u
du du=
121
2u
−= 1
2
1
2u=
( )12
1
2 x x=
+
1
2
dydu x x
⇒ =+
Now by chain rule
dy dy dudx du dx
= ⋅
1 2 122
xxx x
+= ⋅
+
2 1
4
x
x x x
+=
⋅ + Answer
Question # 1 (iii) a xy xa x
+=
−
Put a xua x
+=
−
So y x u= ( )12x u=
Diff. w.r.t. x
( )12dy d x u
dx dx=
( ) ( )1 12 2d dx u u x
dx dx= +
( ) ( )1 12 21 (1)
2dux u udx
−= +
( ) ( )1 12 2
2dy x duu udx dx
−⇒ = + ……. (i)
Now diff. u w.r.t. x
FSc-II / Ex 2.4 - 2
du d a xdx dx a x
+ = −
( ) ( ) ( ) ( )
( )2
d da x a x a x a xdx dx
a x
− + − + −=
−
( )( ) ( )( )( )2
0 1 0 1a x a xa x
− + − + −=
−
( )( ) ( )( )( )2
1 1a x a xa x
− − + −=
−
( )2
a x a xa x
− + +=
−
( )22a
a x=
−
Using value of u and dudx
in eq. (i)
( )2
1 12 22
2x a x a a x
a x a xa xdydx
−+ ++
− −− =
( )( ) ( )
( )( )
2
1 12 2
1 12 2
a x a xaxa xa x a x
−
−
+ += ⋅ +
−− −
( ) ( )( )( )2
12
1 1 12 2 2
a xax
a x a x a x−
+= +
+ − −
( ) ( )( )( )
12
3 1122 2
a xax
a xa x a x
+= +
−+ −
( ) ( )31
2 2
( )( )ax a x a x
a x a x
+ + −=
+ −
( ) ( )
2 2
312 2
dy ax a xdx a x a x
+ −⇒ =
+ −
Question # 1 (iv) & (v) Do yourself as above
Question # 2 (i) 3 4 7 0x y+ + =
Diff. w.r.t. x .
( )3 4 7 (0)d dx ydx dx
+ + =
3(1) 4 0 0dydx
⇒ + + =
4 3dydx
⇒ = − 34
dydx
⇒ = −
Question # 2 (ii) 2 2xy y+ = Differentiating w.r.t. x
( ) ( )2 2d dxy ydx dx
+ =
( ) 2 0d dxy ydx dx
⇒ + =
2 0dy dx dyx y ydx dx dx
⇒ + + =
( )2 (1) 0dyx y ydx
⇒ + + =
( )2 dyx y ydx
⇒ + = −
( )2 dyx y ydx
⇒ + = −
2
dy ydx x y
−⇒ =
+
Question # 2 (iii) Do yourself
Question # 2(iv) 2 24 2 2 2 0x hxy by gx fy c+ + + + + = Differentiating w.r.t. x
( )2 24 2 2 2 (0)d d
x hxy by gx fy cdx dx
+ + + + + =
2 24 ( ) 2 ( ) ( )d d dx h xy b ydx dx dx
⇒ + +
2 ( ) 2 ( ) ( ) 0d d dg x f y cdx dx dx
+ + + =
4(2 ) 2 (1) 2dy dyx h x y b ydx dx
⇒ + + + ⋅
2 (1) 2 0 0dyg fdx
+ + + =
8 2 2 2dy dyx hx hy bydx dx
⇒ + + +
2 2 0dyg fdx
+ + =
( ) ( )2 2 4 0dy
hx by f x hy gdx
⇒ + + + + + + =
( ) ( )2 2 4dyhx by f x hy gdx
⇒ + + =− + + +
( ) ( )4dyhx by f x hy gdx
⇒ + + = − + + +
4dy x hy gdx hx by f
+ + +⇒ = −
+ +
FSc-II / Ex 2.4 - 3 Question # 2 (v)
1 1 0x y y x+ + + = ( ) ( )1 12 21 1 0x y y x⇒ + + + =
Differentiating w.r.t. x
( ) ( )1 12 21 1 (0)
d d dx y y x
dx dx dx⇒ + + + =
( ) ( )1 12 21 1d dxx y y
dx dx⇒ + + + ( ) ( )
1 12 21 1 0d dyy x x
dx dx+ + + + =
( ) ( )1 12 21 1 1 (1)
2dyx y ydx
−⇒ ⋅ + + + ( ) ( )1 12 21 1 (1) 1 0
2dyy x xdx
−+ ⋅ + + + =
( )( )
12
12
12 1
x dy ydxy
⇒ + ++
( )
( )12
12
1 02 1
y dyxdxx
+ + + =+
( )( )
12
12
12 1
x dyxdxy
⇒ + + +
( )( )
12
12
12 1
yyx
= − + + +
( ) ( )( )
1 12 2
12
2 1 1
2 1
x x y dydxy
+ + + ⇒ +
( ) ( )( )
1 12 2
12
2 1 1
2 1
x y y
x
+ + + = − +
2 (1 )(1 )2 1
x x y dydxy
+ + +⇒
+
2 (1 )(1 )2 1x y y
x
+ + += −
+
2 (1 )(1 ) 2 12 1 2 (1 )(1 )x y y ydy
dx x x x y+ + + +
⇒ = − ⋅+ + + +
( )( )
1 2 (1 )(1 )
1 2 (1 )(1 )
y x y ydydx x x x y
+ + + +⇒ = −
+ + + + Answer
Question # 2 (vi)
( )2 21 4y x x x− = +
Differentiating w.r.t x
( ) ( )2 2121 4d dy x x x
dx dx− = +
( ) ( ) ( ) ( )2 2 2 21 12 21 1 4 4d dy d dxy x x x x x
dx dx dx dx⇒ − + − = + + +
( ) ( ) ( ) ( )2 2 21 12 212 1 4 (2 ) 4 (1)
2dyy x x x x x xdx
−⇒ + − = + + +
( )( )
( )2
2 2
2
12
12
2 1 44
dy xxy x xdx x
⇒ + − = + ++
( )( )
( )2
2 2
2
12
12
1 4 24
dy xx x xydx x
⇒ − = + + −+
( )( )
( )2
2 2
2
12
12
1 4 24
dy xx x xydx x
⇒ − = + + −+
( ) ( )( )
2 2 22
212
4 2 41
4
x x xy xdyxdx x
+ + − +⇒ − =
+
( )( )2 2
2 2
2 4 2 4
1 4
x xy xdydx x x
+ − +⇒ =
− +
FSc-II / Ex 2.4 - 4
Question # 3 (i)
Since 1x θθ
= +
1x θ θ −⇒ = + Differentiating x w.r.t. θ
( )1dx dd d
θ θθ θ
−= +
21 θ −= − 211
θ= −
2
21θ
θ−
=
2
2 1ddxθ θ
θ⇒ =
−
Now 1y θ= + Diff. w.r.t. θ
( )1dy dd d
θθ θ
= + 1dydθ
⇒ =
Now by chain rule dy dy ddx d dx
θθ
= ⋅
dy dd dx
θθ
= ⋅ 2
211
θθ
= ⋅−
2
2 1dydx
θθ
⇒ =−
Question # 3 (ii)
Since ( )2
2
11
a tx
t−
=+
*Correction
Diff. w.r.t. t
2
211
dx d tadt dt t
−= +
( ) ( ) ( ) ( )
( )
2 2 2 2
22
1 1 1 1
1
d dt t t tdt dta
t
+ − − − +=
+
( )( ) ( )( )
( )
2 2
22
1 2 1 2
1
t t t ta
t
+ − − −=
+
( )
3 3
22
2 2 2 2
1
t t t tat
− − − +=
+
( )22
4
1
dx atdt t
−⇒ =
+
( )2214tdt
dx at+
⇒ =−
Now 22
1btyt
=+
y
Diff. w.r.t. t
22
1dy d btdt dt t
= +
( ) ( )
( )
2 2
22
1 2 2 1
1
d dt bt bt tdt dt
t
+ − +=
+
( ) ( )( )
2
22
1 2 (1) 2 2
1
t b bt t
t
+ −=
+
( )
2 2
22
2 2 4
1
b bt bt
t
+ −=
+ ( )2
22
2 2
1
b bt
t
−=
+
( )( )
2
22
2 1
1
b t
t
−=
+
Now by chain rule
dy dy dtdx dt dx
= ⋅
( )
( )( )22 2
22
2 1 141
b t tatt
− += ⋅
−+
( )212
b tdydx at
−⇒ = −
Question # 4
Since 2
211
txt
−=
+
Differentiating w.r.t. t , we get (solve yourself as above)
24
1dx tdt t
−=
+
214
dt tdx t
+⇒ =
−
Now 22
1btyt
=+
Differentiating w.r.t. t , we get (solve yourself as above)
( )
( )
2
22
2 1
1
tdydt t
−=
+
Now by chain rule dy dy dtdx dt dx
= ⋅
( )
( )( )22 2
22
2 1 141
t ttt
− += ⋅
−+
21
2dy tdx t
−⇒ = −
Multiplying both sides by y
21
2dy ty ydx t
−⇒ = − ⋅
2
22 1
1 2t tt t
−= − ⋅
+
FSc-II / Ex 2.4 - 5
2
211
dy tydx t
−⇒ = −
+
dyy xdx
⇒ = − 2
211
txt
−=
+∵
0dyy xdx
⇒ + = Proved.
Question # 5(i)
Suppose 22
1y xx
= − and 4u x=
Diff. y w.r.t x 2
21dy d x
dx dx x = −
( )2 2d x xdx
−= − 32 2x x−= +
312 xx
= +
4
312dy x
dx x +
⇒ =
Now diff. u w.r.t x
( )4du d xdx dx
=
34du xdx
⇒ =
Now by chain rule dy dy dxdu dx du
= ⋅
1dydudxdx
= ⋅
4
3 31 12
4dy xdu x x
+⇒ = ⋅
4
61
2dy xdu x
+⇒ =
Question # 4(ii)
Let ( )21n
y x= + and 2u x= Differentiation y w.r.t x
( )21ndy d x
dx dx= +
( ) ( )12 21 1n dn x x
dx−
= + +
( ) ( )121 2
nn x x
−= +
( ) 122 1n
nx x−
= + Now differentiating u w.r.t x
2du d xdx dx
=
2x= 12
dxdu x
⇒ =
Now by chain rule
dy dy dxdu dx du
= ⋅
( ) 12 12 12
ndy nx xdu x
−⇒ = + ⋅
( ) 121ndy n x
du−
⇒ = +
Question # 5 (iii)
Let 2
211
xyx
+=
− and 1
1xux
−=
+
Diff. y w.r.t x
2
211
dy d xdx dx x
+= −
= Solve yourself ( )2
41
xx−
=−
Now diff. u w.r.t x
11
du d xdx dx x
− = +
= Solve yourself ( )2
21x
=+
( )212
xdudx
+⇒ =
Now by chain rule
dy dy dxdu dx du
= ⋅
( )
( )2
2
1421
xxx
+−= ⋅
−
( ) ( ) ( )22 1
1 1x x
x x−
= ⋅ +− +
( )( )2 1
1x xdy
dx x− +
⇒ =−
Question # 5(iv)
Let ax bycx d
+=
+ and
2
2ax buax d
+=
+
Diff. y w.r.t. x dy d ax bdx dx cx d
+ = +
FSc-II / Ex 2.4 - 6
( ) ( ) ( ) ( )
( )2
d dcx d ax b ax b cx ddx dx
cx d
+ + − + +=
+
( )( ) ( )( )( )2
cx d a ax b ccx d
+ − +=
+
( )2
acx ad acx bccx d
+ − −=
+
( )2
dy ad bcdx cx d
−⇒ =
+
Now diff. u w.r.t x 2
2du d ax bdx dx ax d
+= +
( )2 2 2 2
22
( ) ( ) ( ) ( )d ddx dxax d ax b ax b ax d
ax d
+ + − + +=
+
( )( ) ( )( )
( )
2 2
22
2 2ax d ax ax b ax
ax d
+ − +=
+
( )
( )
2 2
22
2ax ax d ax b
ax d
+ − −=
+
( )( )22
2ax d b
ax d
−=
+
( )( )
22
2ax ddx
du ax d b+
⇒ =−
Now by chain rule
dy dy dxdu dx du
= ⋅
( )( )
( )
22
2 2ax dad bcax d bcx d
+−= ⋅
−+
( )( )
( ) ( )
22
22
ad bc ax ddydx ax cx d d b
− +⇒ =
+ −
Question # 5 (v)
Let 2
211
xyx
+=
− and 3u x=
Diff. y w.r.t x
2
211
dy d xdx dx x
+= −
= Solve yourself
( )22
4
1
x
x
−=
−
Now diff. u w.r.t x
3du d xdx dx
=
23x=
21
3dxdu x
⇒ =
Now by chain rule dy dy dxdu dx du
= ⋅
( )2 22
4 131
xxx
−= ⋅
−
( )22
4
3 1
dydx x x
−⇒ =
−
² ---- The End --- ²
Note: If you find any mistake in these notes, please inform to correct it.
