Mathematics 2 BESS Assignment 1 Problems Solutions and Analysis

This document presents a detailed solution to a mathematics assignment, addressing problems from two exercises. Exercise 1 explores the application of Dini's theorem to an ellipse equation, finding derivatives, and determining the tangent line equation. Exercise 2 focuses on the natural domain of a function, limits, and derivatives, including directional derivatives and convexity. The solutions provide step-by-step explanations, including the use of the natural domain, limits, derivatives, and the concept of convexity. The assignment covers topics from calculus, including the study of implicit functions, the application of Dini's theorem and the analysis of the directional derivative. The solutions provided can be useful for students studying calculus and related subjects, offering a comprehensive understanding of the concepts involved.

APPLIED MATHS
Exercise 1.
Problem 1)
You can realize that f(x,0) =f(0,y)=0 as you approach the origin. This is an evidence that as you approach
the origin, from the two axes, both cases are realized.
Limx→0f(x,0)=0 ,and limy→0f(0,y)=0
However, it is wrong to conclude that the limit is equal to zero .As you move closer to (0,0) on the the
path of the line,y=mx(with m ∈ R),you get;
limx→0f(x,mx)= lim→0
x2
a2 + ( mx ) 2
b2 -1
As the result depends on m,the limit does not exist.Hence,we cannot use the Dini’s theorem at every
point of solution of f(x,y)=0.
Problem 2)
Consider the function f(x,y)= e6 x2
+ 2 y2−1
3 x2 + y2
Assume that t =3x2 +y2 ,and you will get ,
e6 x2
+2 y2−1
3 x2 + y2 = e2 t −1
t
But ,
When f(x,y) →(0,0),it implies that also t→0.
Hence,you will have;
Lim(x,y0→0f(x,y) = Limt→0
e2 t −1
t =2
Because the result does not depend on any factor m,the limit does exist.Thus ,we can use the Dini’s
theorem at every solution of f(x,y)=0

Problem 3)
Consider =-3x +10y=25 such that -3=b,10=b and 25 =cbe an ellipse equation.
Y= 3
10 + 5
2 since a2m2 +b2 =c2
This implies that m= 3
10 and c= 5
2
But 9
100a2 + b2 = 25
4 ,which implies that b2= 25
4 - 9
100a2
And this implies,
9b2 + 64
25 a2 =a2b2
→9( 25
4 - 9
100) + 64
25 a2 = a2( 25
4 - 9
100a2)→a2 -50a2 +625 =0
Hence a2 =25
This implies that ;
b 2 = 25
4 - 9
100*25 =4
Hence The equation of an ellipse x2
25 + y2
4 =1
xx 0
a2 + yy 0
b2 -1=0
Exercise 2
Problem 1.
Natural domain is the maximum set of values for which a function is defined.
F(x,y)=In(x+y2)
F(0,0)=Undefined
F(-1,1)=imposible

F(1,-1)=F(1,1)=0.6931
F(0,1)=f(1,0)=0
F(-2,-2)=0.6931
Since In(x+y2) is undefined for x=0, and also the given function In(x+y2) gives only the positive values,so
the domain of the function is (x,y)∈ (-∞,∞ ¿ U (−∞ , ∞ )
Problem 2
A is bounded between negative infinite to positive infinity. It is defined by some sets of X with real and
complex bounded values i.e R: ( x,y)∈ (-∞,∞ ¿ U (−∞ , ∞ )
Problem 3.
Lim (x,y)→0In(x+y2) =lim(x,y)→0( 1
x+ y2 ) is undefined .
You can see that f(x,y) =0 ,the derivative in (0,0) is such that
dy
dx (0,0) = dy
dx (0,0) =0
But f is differentiable at x0 if
F(x0+h) –f(x0)=∇f(x0) .h + 0(||h||) as h →0
Thus lim→0
f ( x 0+ h ) −f ( x 0 ) −∇ f ( x 0 ) . h
¿|h|∨¿ ¿
At x0=(0,0),you can say h=(h,k) since F(0,0) =0 and ∇ f ( 0,0 ) =0
Lim(h,k)→(0,0)f f ( h+ k ) −0
√ h2+ k2 =0, if A>0
Problem 4
Lim (x,y)→0In(x+y2) =lim(x,y)→0( 1
x+ y2 ) is undefined .
You can see that f(x,y) =0 ,the derivative in (0,0) is such that
dy
dx (0,0) = dy
dx (0,0) =0

But f is differentiable at x0 if
F(x0+h) –f(x0)=∇f(x0) .h + 0(||v||) as v →0
Thus lim→0
f ( x 0+ h ) −f ( x 0 ) −∇ f ( x 0 ) . h
¿|v|∨¿ ¿
At x0=(0,0),you can say h=(v,k) since F(0,0) =0 and ∇ f ( 0,0 ) =0
Lim(v,k)→(0,0)f f ( v + k ) −0
√ v +k 2 =0,
=lim(v,k)→0
1
x + v + 1
y2+ v
√1
√ ( x+ v )+( y2 +v)
=
1
0+1 + 1
0+1
√ ( 0+1 )+(0+ 1)
=
1
2
1
2
=1
Problem 5
The directional derivative is maximum in the direction of ∇f ,and the maximum value is ||∇ f ∨¿
Hence∇ f (x , y) =<fx,fy> =<Inx,Iny2>
=< 1
x , 1
y2 >
Therefore,
∇ f (1,1)=( 1
1, 1
1)=(1,1)
So that the direction is (1,1)
And the maximum value is √ 2
Problem 6
x = a + hx, y = b + hy, (x,y )∈ [1, 1].

Define f(x,y)=f(a+hx,b +hy)
F1(x,y)=fx
dx
dy +fy
dx
dy =hxf(x) +hy(y)
F11(x,y)= d f 1 ( x , y ) dy
dydx + df dy
dydx
= d
dx [hxf(x) +hyf(y)] hx + d
dy [hxf(x+hyf(y)]hy
=h2fxx+2hxhyf(x,y) +h2f(yy)
F(1,1) =f(1)(1-1) +f11(h)( 1−1
2 )
;
=f(0) +f1(1)(0) +f11(h)( 0
2 )
Problem 7
convex if for all a ∈ I, all b ∈ I, and all λ ∈ (1, 1) we have
f((1−λ)a + λb) ≤ (1 − λ)f(a) + λf(b).
At (1,1)
F(1-1)1 +1*1≤(1-1)f(1) +1(1)
F(0)1 +1 ≤0f(1) +1
1≤ 1
Hence the function is convex. It joins two x and y points such that no point lies above the graph.