3. Correct the sentences and rewrite them in your notebook. A. i have many crayons B. i like the pink one Grade 1. English
Answers
Answer:
\large\underline{\sf{Solution-}}
Solution−
Given that
\rm \: x = a \: sint - b \: cost - - - (1)x=asint−bcost−−−(1)
and
\rm \: y = a \: cost + b \: sint - - - (2)y=acost+bsint−−−(2)
Now, Consider
\rm \: {x}^{2} + {y}^{2}x
2
+y
2
\rm \: = \: (asint - bcost)^{2} + (acost + bsint)^{2}=(asint−bcost)
2
+(acost+bsint)
2
\rm \: = {a}^{2} {sin}^{2}t + {b}^{2} {cos}^{2}t - 2absint \: cost \: + {a}^{2} {cos}^{2}t + {b}^{2} {sin}^{2}t + 2absint \: cost=a
2
sin
2
t+b
2
cos
2
t−2absintcost+a
2
cos
2
t+b
2
sin
2
t+2absintcost
\rm \: = {a}^{2} {sin}^{2}t + {b}^{2} {cos}^{2}t + {a}^{2} {cos}^{2}t + {b}^{2} {sin}^{2}t=a
2
sin
2
t+b
2
cos
2
t+a
2
cos
2
t+b
2
sin
2
t
\rm \: = {a}^{2} ({sin}^{2}t + {cos}^{2}t) + {b}^{2} ({cos}^{2}t + {sin}^{2}t)=a
2
(sin
2
t+cos
2
t)+b
2
(cos
2
t+sin
2
t)
\rm \: = \: {a}^{2} + {b}^{2}=a
2
+b
2
So, we get
\rm \: \: {x}^{2} + {y}^{2} = \: {a}^{2} + {b}^{2}x
2
+y
2
=a
2
+b
2
On differentiating both sides w. r. t. x, we get
\rm \: \: \dfrac{d}{dx}({x}^{2} + {y}^{2}) = \: \dfrac{d}{dx}( {a}^{2} + {b}^{2} )
dx
d
(x
2
+y
2
)=
dx
d
(a
2
+b
2
)
\rm \: 2x + 2y\dfrac{dy}{dx} = 02x+2y
dx
dy
=0
\rm \: x + y\dfrac{dy}{dx} = 0x+y
dx
dy
=0
\rm \: y\dfrac{dy}{dx} = - xy
dx
dy
=−x
\rm \: \dfrac{dy}{dx} = - \dfrac{x}{y}
dx
dy
=−
y
x
On differentiating both sides w. r. t. x, we get
\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = -\dfrac{d}{dx} \dfrac{x}{y}
dx
2
d
2
y
=−
dx
d
y
x
\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \bigg[\dfrac{y\dfrac{d}{dx}x - x\dfrac{d}{dx}y}{ {y}^{2} } \bigg]
dx
2
d
2
y
=−[
y
2
y
dx
d
x−x
dx
d
y
]
\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \bigg[\dfrac{y - x\dfrac{dy}{dx}}{ {y}^{2} } \bigg]
dx
2
d
2
y
=−[
y
2
y−x
dx
dy
]
can be further rewritten as using equation (1),
\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \bigg[\dfrac{y + x \times \dfrac{x}{y}}{ {y}^{2} } \bigg]
dx
2
d
2
y
=−[
y
2
y+x×
y
x
]
\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \bigg[\dfrac{y + \dfrac{ {x}^{2} }{y}}{ {y}^{2} } \bigg]
dx
2
d
2
y
=−[
y
2
y+
y
x
2
]
\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \bigg[\dfrac{ \dfrac{ {y}^{2} + {x}^{2} }{y}}{ {y}^{2} } \bigg]
dx
2
d
2
y
=−[
y
2
y
y
2
+x
2
]
\begin{gathered}\rm\implies \:\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \dfrac{ {x}^{2} + {y}^{2} }{ {y}^{3} } \\ \end{gathered}
⟹
dx
2
d
2
y
=−
y
3
x
2
+y
2
Hence, Proved
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
FORMULAE USED
\begin{gathered}\boxed{\tt{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }} \\ \end{gathered}
dx
d
x
n
=nx
n−1
\begin{gathered}\boxed{\tt{ \dfrac{d}{dx} \frac{u}{v} = \frac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} } \: }} \\ \end{gathered}
dx
d
v
u
=
v
2
v
dx
d
u−u
dx
d
v
\begin{gathered}\boxed{\tt{ \dfrac{d}{dx}x \: = \: 1 \: }} \\ \end{gathered}
dx
d
x=1
\begin{gathered}\boxed{\tt{ \dfrac{d}{dx}k \: = \: 0\: }} \\ \end{gathered}
dx
d
k=0
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
ADDITIONAL INFORMATION
\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}\end{gathered}
f(x)
k
sinx
cosx
tanx
cotx
secx
cosecx
x
logx
e
x
dx
d
f(x)
0
cosx
−sinx
sec
2
x
−cosec
2
x
secxtanx
−cosecxcotx
2
x
1
x
1
e
x
to
Answer:
it's already correct answer