form PDE by eliminating arbitrary function from lx+my+nz = f (x^2+y^2+z^2)
Answers
Answer:
Correct option is
B
(1+(y
1
)
2
)
3
=r
2
(y
2
)
2
(x−a)
2
+(y−b)
2
=r
2
By differentiating this equation, we get
2(x−a)+2(y−b)
dx
dy
=0
a=x+(y−b)
dx
dy
-Equation 1
Differentiating again, we get
0=1+(y−b)
dx
2
d
2
y
+(
dx
dy
)
2
−b=
dx
2
d
2
y
(−1−(
dx
dy
)
2
)
−y=
y
2
−1−y
1
2
−y-Equation 2
Now, substituting the values of a & b,
in (x−a)
2
+(y−b)
2
=r
2
, we get
(x−x−(y−b)
dx
dy
)
2
+(y−b)
2
=r
2
(y−b)
2
(
dx
dy
)
2
+(y−b)
2
=r
2
1+y
1
2
=
(y−b)
2
r
2
and y−b=
y
2
−1−y
1
2
1+y
1
2
=
(1+y
1
2
)
2
r
2
y
2
2
(1+y
1
2
)(1+y
1
2
)
2
=r
2
y
2
2
(1+y
1
2
)
3
=r
2
y
2
2
Step-by-step explanation: