Calculus III: Partial Derivatives and Surface Normals

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Question 1
[ x , y ]=meshgrid (0 :2∗pi , 0:2∗pi) ;
A=((2+cos ( x)).∗cos( y));
B=((2+cos( x )).∗sin ( y ));
C=( sin( x));
W =( √( A .2+ B .2)−2).2 +C .2
surf ( x , y , W )
xlabel(x) ;
ylabel (y );
zlabel (W) ;
Figure 1: Code Execution on Matlab

Figure 2: Surface Plot for Given Cartesian Equation
Question 2
f ( x , y , z ) =( √ x2 + y2−2)2
+z2=1
∇ f ( x , y , z ) = [ f x ( x , y , z ) , f y ( x , y , z ) , f z ( x , y , z ) ]
∇ f ( x , y , z ) = ∂ f
∂ x ^i + ∂ f
∂ y ^j+ ∂ f
∂ z ^k
So here,
∂ f
∂ x = ∂
∂ x ( √ x2+ y2−2)2
+ z2−1 ¿
∂ f
∂ x =2 ( √ x2+ y2−2 )[ ( x2+ y2 )
−1
2 × 2 x ]
∂ f
∂ x = ( 2 √ x2+ y2−4 ) [ 2 x
√ x2 + y2 ]
∂ f
∂ x = 4 x √ x2 + y2−8 x
√ x2 + y2

∂ f
∂ x =4 x− 8 x
√x2 + y2
∂ f
∂ x =4 x [ 1− 2 x
√ x2 + y2 ]
Similarly,
∂ f
∂ y = ∂
∂ y ( √ x2 + y2−2)2
+ z2−1¿
∂ f
∂ y =2 ( √ x2 + y2−2 ) [ ( x2 + y2 )
−1
2 ×2 y ]
∂ f
∂ y = (2 √x2 + y2−4 ) [ 2 y
√ x2 + y2 ]
∂ f
∂ y = 4 y √x2 + y2−8 y
√ x2+ y2
∂ f
∂ y =4 y − 8 y
√ x2+ y2
∂ f
∂ y =4 y [1− 2 y
√ x2+ y2 ]
Now,
∂ f
∂ z = ∂
∂ z ( √ x2+ y2−2)2
+ z2−1 ¿
^n=(⃗
n
|n| )
|n|= √ x2+ y2+ z2
|n|= √(1− 2
√x2 + y2 )
2
+(1− 2
√ x2+ y2 )
2
+12

¿ 12 +( 2
√ x2 + y2 )
2
−2 ( 2
√ x2+ y2 ) +1+ ( 2
√ x2 + y2 )
2
+2 ( 2
√ x2 + y2 ) +1
Simplifying above equation, we get;
|n|=3+ 8
√ x2 + y2 ¿2 ¿
^U =(1− 2
√ x2+ y2 +1− 2
√ x2 + y2 + 1)/¿
Above vector is the outward unit normal to the surface. Hence proved
Question 3
Plane S=x+ y+z =0
Considering 3 coordinates as equal to 1(assumption: tangent plane is parallel to the plane S);
¿ 1− 2
√ x2 + y2 ,1− 2
√ x2+ y2 ,1>¿ ---------- (1)
¿ 1,1,1>¿ ----------- (2)
x (1− 2
√ x2 + y2 )=1
Z(1)=1
X = 1
1− 2
√ x2+ y2
y= 1
1− 2
√ x2+ y2
z=1
x= √ x2 + y2
√ x2 + y2−2

y= √x2 + y2
√ x2+ y2 −2
z=1
So the point on plane S, on which the tangent plane is parallel to the plane, is;
( √ x2 + y2
√ x2 + y2−2 , √ x2 + y2
√ x2 + y2−2 , 1)