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DeepakKumar Thakur
Unit 3
Q 7) find the second order derivatives of f(x,y) =y^2.e^x+y
Solution:
Here,
In order to find the second order derivatives of f(x,y)=y^2.e^x+y
We must need to find out the first order partial derivatives of f(x,y)=y^2.e^x+y first.
Thus ,we have
fx(x,y)=df/dx=d(y^2.e^x+y )/dx
fx(x,y)=y^2.e^x
fy(x,y)=df/dy=d(y^2.e^x+y )/dy
fy(x,y)=2y.e^x+1
So,that
fxx(x,y)=d^2f/dx^2=d/dx(df/dx)=d(y^2.e^x)/dx =y^2.e^x
fxy(x,y)=d^2f/dy.dx=d/dy(df/dx)=d(y^2.e^x)/dy = 2y.e^x
fyx(x,y)=d^2f/dx.dy=d/dx(df/dy)=d(2y.e^x+1)
/dx = 2y.e^x
fyy(x,y)=d^2f/dy^2=d/dy(df/dy)=d(2y.e^x+1)/dy =2e^x
Right answer
Mam,can we use “•” this for multiplication in place of “x” as it may produce confusion.
u can use*for multiplication
DeepakKumar Thakur
Unit 3
Q.8) Find the directional derivative of f(x,y,z)=x^2 × y−y × z^3+z at the point (1, 2, 0) in the direction of the vector a=2i+j−2k.
Solution:
Here, f(x,y,z)=x^2×y−y×z^3+z and (x0,y0,z0)=(1,-2,0)
Now,the partial derivatives of are
fx(x,y,z)=2xy
fy(x,y,z)=x^2-z^3
fx(x,y,z)=-3yz^2+1
At point (x0,y0,z0)=(1,-2,0),
fx(1,-2,0)=-4
fy(1,-2,0)=1
fz(1,-2,0)=1
Here,since a is not a unit vector, we normalize it,thus getting
u=a/||a||=1(2i+j-2k)/√9=2/3i+1/3j-2/3k
Lastly, Duf(x0,y0,z0)=fx(x0,y0,z0)u1+fy(x0,y0,z0)+fz(x0,y0,z0)
Therefore,
Duf(1,-2,0)=(-4)(2/3)+1/3-2/3=-3
Thus, the directional derivative is
Duf(1,-2,0)=-3.
Right answer
Rishabh Sharma . F111 FYCS
Q.2]Solve dy/dx=1-y; y(0)=0 , find y(0.1) and y(0.3) using Eulars method. Taking h=0.1.
Ans.]
dy/dx=1-y=x^0-y
y(0)=0
h=0.1
x0=0, x1=0.1, x2=0.2, x3=0.3
therefor f(x,y)=1-y, x0=0, y0=0
the approximations are:-
y1= y(0.1)=y0+f(x0,y0)delta(x)
=0+0+f(1-0)(0.1)
=0.1
y2= y(0.2)=y1+f(x1,y1)delta(x)
=0.1+f(1-0)(0.1)
=0.2
y3=y(0.3)=y2+f(x2,y2)delta(x)
=0.2+f(1-0)(0.1)
=0.3
value of y2 and y3 is wrong
rishabh sharama f111 fycs
unit 2
Q.3]solve the differential equation dy/dx=-xy
Ans.] dy/dx=-xy
dy/dx+xy=0
So given differential equation is in the form of
dy/dx+p(x)y =q(x)
p(X)=x and q(x)=0
step 1
put y=uv
dy/dx = u d/dx v + v d/dx u
u dv/dx + v du/dx + x uv = 0
step 2
u dv/dx +v (dv/dx + xu)=0
step 3
put the v term equal to zero then
du/dx + xu = 0
du/dx = -xu
du/u = -xdx
put integral sign
int du/u = -int xdx
ln u = -x^2/2 + C
ln u = -x^2/2-ln C …………….. C= -ln(k)
ln u + ln k = -x^2/2
log uk = -x^2/2
log uk = -x^2/2
uk = e^-x^2/2
u = e ^-x^2/2 / k
step 4
substitute u in back in the
e ^-x^2/2 / k dv/dx = 0
step 5
dv/du= e ^-x^2/2
Method is wrong. You can use separation of variable method
rishabh sharma f111 fycs
unit 2
Q.4]Find the approximate value of int^2^^1 1/x^2 dx using simpsons rule formula with
n=10
Ans.]
Delta x = b-a/n = 2-1/10 = 1/10 = 0.1
Calculate the val of the function as
x 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0
y=1/x^2 1 0.826 0.699 0.591 0.510 0.44 0.39 0.396 0.308 0.377 0.25
y0 y1 y2 y3 y4 y5 y6 y67 y8 y9
int ^b ^^a f(x) dx= delta x /3 [y0 + 4(y1+y3+y5+y7+y9)+2(y2+y4+y6+y8)+y10]
Substituting the values we get :-
= 0.1/3 [1 + 4(2.48)]
= 1.4974/3
= 0.499
Right answer
rishabh sharma f111 fycs
unit 2
Q.5] Find the area of region that is enclosed between the curve y = x^2 and y = x + 6
Ans.]
y= x^2
y= x+6
therefore x=y-6
y2-13y-36=0
y2-9y-4y+36=0
(y-4)(y-9)=0
y=4 or y=9
A =int^9 ^^4 [y-6 -(13y – 36)]dy
=int^9 ^^4 [y-6 -13y + 36] dy
=int^9 ^^4 [30 – 12y] dy
=[30y – 6y]^9 ^^4
=[270 – 54] – [24]
=192
Wrong answer
DeepakKumar Thakur
Unit 3
Q9) Find the gradient vector of f(x, y) if f(x, y) =x^3+2xy^2. Evaluate it at (-3, -4).
Solution:
Here,f(x,y)=x^3+2xy^2 and the point is(-3,-4).
Now, inorder to find the partial derivatives first,
We have,
fx(x,y)=df/dx=d(x^3+2xy^2)/dx
fx(x,y)=3x^2+2y^2
fy(x,y)=df/dy=d(x^3+2xy^2)/dy
fy(x,y)=4xy
Now,for a function of x and y,the gradient of f is defined by
∇f(x,y)=fx(x,y)i+fy(x,y)j
Therefore,
∇f(x,y)=(3x^2+2y^2)i+(4xy)j
Now,the gradient of f at (-3,-4) is
∇f(-3,-4)=[3(-3)^2+2(-4)^2]i+[4(-3)(-4)]j
∇f(-3,-4)=59i+48j
Therefore,the gradient of f at (-3,-4) is
∇f(-3,-4)=59i+48j .
Right answer
Saeel Pandurang Sawant FYCS 107
Unit 1
Q1] Show that the function f(x) = x^3-9x^2+30x+7 is always increasing
Solution:
f(x)= x^3-9x^2+30x+7
Therefore,
f'(x)=3x^2-18x+30
Therefore,
f'(x)>0
Therefore,
f'(x)=3x^2-18x+30>0
f'(x)=3(x^2-6x+10)>0
f'(x)=(x^2-6x+10)>0
f'(x)=(x^2-6x+9+)>0
f'(x)=[(3+x)^2+1]>0
Since, The square of real number is always a non-negative,
Therefore,
(3+x)^2>0
Therefore,
f(x) is always a increasing function
Half of the answer is wrong. Refer answer key.
Saeel Pandurang Sawant FYCS 107
Unit 1
Q2]Find the absolute maximum and minimum values of f(x)=(x-2)^2 in [1,4]
Solution:
f(x)=(x-2)^2
f(X)=x^2-4x+4
f'(x)=2x-4
if x=0 then f'(x)=0
Therefore,
f'(x)=0>>>2x-4=0
2(x-2)=0
Therefore,
x-2=0
Therefore,
x=2
at x=1, f(1)=(1)^2-4(1)+4
=1 -4 +4…………..(-4 and +4 gets cancel)
=1
at x=4, f(4)=(4)^2-4(4)+4
=16 – 16 + 4……………..(+16 and -16 also gets cancel)
=4
So we have a stationary point 0
Therefore,
absolute maximum at 4 at x =4 and,
absolute minimum at 1 at x =1
Answer is incomplete. Refer answer key.
Saeel Pandurang Sawant FYCS 107
Q3]Using Newton’s method find the approximate root for the equation f(x)=x-cos x
Formula: xn+1= x n+ f(x n / f'(x n)
Solution:
f(x)=x-cos x
Therefore,
f'(x)= 1+ sin x
Taking range x=0
Therefore,
x1 = 0 + x – cos x / 1 + sin x
x1 = 0 + 0 – cos 0 / 1 + sin 0
x1 = -1/1
x1 = -1
Therefore,
x = 1
x2 = x1 + f(x1) / f'(x1)
x2 = -1 + (-1)
x2 = -2
Therefore,
The approximate root for the equation f(x) = x – cos x is -1 and -2.
Wrong answer. Refer answer key.
Saeel Pandurang Sawant FYCS 107
Unit 1
Q4]Find the relative extrema of f(x) = 3x^5 – 5x^3
Solution:
f(x) = 3x^5 – 5x^3
Therefore,
f'(x) = 15x^4 – 15x^2
f'(x) = 15×2 ( x^2 – 1)
f'(x) = 15x^2 ( x – 1) (x + 1)
critical points are 0, -1, +1,
Finding the derivative of f'(x) from,
f'(x) = 60x^3 – 30x
f'(x) = 30x (2x^2 – 1)
(2x^2-1) f'(x)
x = – 1 -30 –
x = 0 0 0
x = 1 30 +
Therefore,
There is no relative maxima and minima become no sign change.
Wrong answer
DeepakKumar Thakur F115 FYCS
Unit 3
Q10) Locate all relative extrema and saddle points of f(x, y)=x^3+2y^3-3x^2-24y+16.
Solution:
Here,f(x, y)=x^3+2y^3
-3x^2
-24y+16
Now, the partial derivatives are
fx(x,y)=df/dx=d(x^3+2y^3
-3x^2
-24y+16)/dx
fx(x,y)=3x^2-6x
fy(x,y)=df/dy=d(x^3+2y^3
-3x^2
-24y+16)/dy
fy(x,y)=6y^2-24
Since fx(x,y)=3x^2-6x and fy(x,y)=6y^2-24, the critical points of f satisfy the equations;
3x^2-6x=0 and 6y^2-24=0
By solving 3x^2-6x=0 we get,
x=0 and x=2
By solving 6y^2-24=0 we get,
y=2 and y=-2
Therefore, (0,2) , (0,-2) , (2,2) , (2,-2) are the critical points of f.
Now,the second order partial derivatives are
fxx(x,y)=d^2 f/dx^2=(d/dx)(df/dx)=d(3x^2-6x)/dx =6x-6
fyy(x,y)=d^2 f/dy^2=(d/dy)(df/dy)=d(6y^2-24)/dy =12y
fxy(x,y)=d^2 f/dy.dx=(d/dy)(df/dx)=d(3x^2-6x)/dy =0
At the point (0,2) , we have
D=fxx(0,2)fyy(0,2)-fxy^2(0,2)=(-6)(24)-0
=-144 < 0
Here,since D 0 and fxx(0,-2)=-6 0 and fxx(0,-2) 0 and fxx(2,2)=6 > 0
Here,since D>0 and fxx(0,-2)>0
Therefore,f has a relative minimum at (2,2)
At the point (2,-2) , we have
D=fxx(2,-2)fyy(2,-2)-fxy^2(2,-2)=(6(-24)-0
=-144 < 0
Here,since D<0
Therefore,f has a saddle point at(2,-2)
Thus,the function f has relative maximum at the point (0,-2) and relative minimum at the point (2,2) as well as f has saddle points at the points(0,2) and (2,-2) respectively .
Right answer
DeepakKumar Thakur F115 FYCS
Unit 3
Locate all relative extrema and saddle points of f(x, y)=x^3+2y^3-3x^2-24y+16.
Solution:
Here,f(x, y)=x^3+2y^3
-3x^2
-24y+16
Now, the partial derivatives are
fx(x,y)=df/dx=d(x^3+2y^3
-3x^2
-24y+16)/dx
fx(x,y)=3x^2-6x
fy(x,y)=df/dy=d(x^3+2y^3
-3x^2
-24y+16)/dy
fy(x,y)=6y^2-24
Since fx(x,y)=3x^2-6x and fy(x,y)=6y^2-24, the critical points of f satisfy the equations;
3x^2-6x=0 and 6y^2-24=0
By solving 3x^2-6x=0 we get,
x=0 and x=2
By solving 6y^2-24=0 we get,
y=2 and y=-2
Therefore, (0,2) , (0,-2) , (2,2) , (2,-2) are the critical points of f.
Now,the second order partial derivatives are
fxx(x,y)=d^2 f/dx^2=(d/dx)(df/dx)=d(3x^2-6x)/dx =6x-6
fyy(x,y)=d^2 f/dy^2=(d/dy)(df/dy)=d(6y^2-24)/dy =12y
fxy(x,y)=d^2 f/dy.dx=(d/dy)(df/dx)=d(3x^2-6x)/dy =0
At the point (0,2) , we have
D=fxx(0,2)fyy(0,2)-fxy^2(0,2)=(-6)(24)-0
=-144 < 0
Here,since D 0 and fxx(0,-2)=-6 0 and fxx(0,-2) 0 and fxx(2,2)=6 > 0
Here,since D>0 and fxx(0,-2)>0
Therefore,f has a relative minimum at (2,2)
At the point (2,-2) , we have
D=fxx(2,-2)fyy(2,-2)-fxy^2(2,-2)=(6(-24)-0
=-144 < 0
Here,since D<0
Therefore,f has a saddle point at(2,-2)
Thus,the function f has relative maximum at the point (0,-2) and relative minimum at the point (2,2) as well as f has saddle points at the points(0,2) and (2,-2) respectively .
DeepakKumar Thakur
Unit 3
Locate all relative extrema and saddle points of f(x, y)=x^3+2y^3-3x^2-24y+16.
Solution:
Here,f(x, y)=x^3+2y^3
-3x^2
-24y+16
Now, the partial derivatives are
fx(x,y)=df/dx=d(x^3+2y^3
-3x^2
-24y+16)/dx
fx(x,y)=3x^2-6x
fy(x,y)=df/dy=d(x^3+2y^3
-3x^2
-24y+16)/dy
fy(x,y)=6y^2-24
Since fx(x,y)=3x^2-6x and fy(x,y)=6y^2-24, the critical points of f satisfy the equations;
3x^2-6x=0 and 6y^2-24=0
By solving 3x^2-6x=0 we get,
x=0 and x=2
By solving 6y^2-24=0 we get,
y=2 and y=-2
Therefore, (0,2) , (0,-2) , (2,2) , (2,-2) are the critical points of f.
Now,the second order partial derivatives are
fxx(x,y)=d^2 f/dx^2=(d/dx)(df/dx)=d(3x^2-6x)/dx =6x-6
fyy(x,y)=d^2 f/dy^2=(d/dy)(df/dy)=d(6y^2-24)/dy =12y
fxy(x,y)=d^2 f/dy.dx=(d/dy)(df/dx)=d(3x^2-6x)/dy =0
At the point (0,2) , we have
D=fxx(0,2)fyy(0,2)-fxy^2(0,2)=(-6)(24)-0
=-144 < 0
Here,since D 0 and fxx(0,-2)=-6 0 and fxx(0,-2) 0 and fxx(2,2)=6 > 0
Here,since D>0 and fxx(0,-2)>0
Therefore,f has a relative minimum at (2,2)
At the point (2,-2) , we have
D=fxx(2,-2)fyy(2,-2)-fxy^2(2,-2)=(6(-24)-0
=-144 < 0
Here,since D<0
Therefore,f has a saddle point at(2,-2)
Thus,the function f has relative maximum at the point (0,-2) and relative minimum at the point (2,2) as well as f has saddle points at the points(0,2) and (2,-2) respectively .
Right answer
Ravishankar singh F112
Find the area of the region bounded above by g=x+6 bounded below by y=x^2 and bounded on the sides by the lines x=0 and x=2
Formula : A =a^s^b [f(x)-g(x)]dx
X^2-x-6
X^2+2x-3x-6
X(x+2)-3(x+2)
(X-3) (x-2)
X=3 x=2
A=a^s^b[f(×)-g(x)]
=0^s^2[x+6-x^2]dx
0^s^2 x ^2
Wrong answer
DeepakKumar Thakur
Unit 3
Locate all relative extrema and saddle points of f(x, y)=x^3+2y^3-3x^2-24y+16.
Solution:
Here,f(x, y)=x^3+2y^3
-3x^2
-24y+16
Now, the partial derivatives are
fx(x,y)=df/dx=d(x^3+2y^3
-3x^2
-24y+16)/dx
fx(x,y)=3x^2-6x
fy(x,y)=df/dy=d(x^3+2y^3
-3x^2
-24y+16)/dy
fy(x,y)=6y^2-24
Since fx(x,y)=3x^2-6x and fy(x,y)=6y^2-24, the critical points of f satisfy the equations;
3x^2-6x=0 and 6y^2-24=0
By solving 3x^2-6x=0 we get,
x=0 and x=2
By solving 6y^2-24=0 we get,
y=2 and y=-2
Therefore, (0,2) , (0,-2) , (2,2) , (2,-2) are the critical points of f.
Now,the second order partial derivatives are
fxx(x,y)=d^2 f/dx^2=(d/dx)(df/dx)=d(3x^2-6x)/dx =6x-6
fyy(x,y)=d^2 f/dy^2=(d/dy)(df/dy)=d(6y^2-24)/dy =12y
fxy(x,y)=d^2 f/dy.dx=(d/dy)(df/dx)=d(3x^2-6x)/dy =0
At the point (0,2) , we have
D=fxx(0,2)fyy(0,2)-fxy^2(0,2)=(-6)(24)-0
=-144 is lessthan 0
Here,since D is less than 0
Therefore,f has a saddle point at(0,2)
At the point (0,-2) , we have
D=fxx(0,-2)fyy(0,-2)-fxy^2(0,-2)=(-6)(-24)-0
=144 is greater than0 and fxx(0,-2)=-6 is less than 0
Here,since Dis greater than 0 and fxx(0,-2) is less than 0
Therefore,f has a relative maximum at(0,-2)
At the point (2,2) , we have
D=fxx(2,2)fyy(2,2)-fxy^2(2,2)=(6)(24)-0
=144 > 0 and fxx(2,2)=6 > 0
Here,since D is greater than0 and fxx(0,-2) is less than 0
Therefore,f has a relative minimum at (2,2)
At the point (2,-2) , we have
D=fxx(2,-2)fyy(2,-2)-fxy^2(2,-2)=(6(-24)-0
=-144 < 0
Here,since D is less than 0
Therefore,f has a saddle point at(2,-2)
Thus,the function f has relative maximum at the point (0,-2) and relative minimum at the point (2,2) as well as f has saddle points at the points(0,2) and (2,-2) respectively .
DeepakKumar Thakur
Unit 3
Locate all relative extrema and saddle points of f(x, y)=x^3+2y^3-3x^2-24y+16.
Solution:
Here,f(x, y)=x^3+2y^3
-3x^2
-24y+16
Now, the partial derivatives are
fx(x,y)=df/dx=d(x^3+2y^3
-3x^2
-24y+16)/dx
fx(x,y)=3x^2-6x
fy(x,y)=df/dy=d(x^3+2y^3
-3x^2
-24y+16)/dy
fy(x,y)=6y^2-24
Since fx(x,y)=3x^2-6x and fy(x,y)=6y^2-24, the critical points of f satisfy the equations;
3x^2-6x=0 and 6y^2-24=0
By solving 3x^2-6x=0 we get,
x=0 and x=2
By solving 6y^2-24=0 we get,
y=2 and y=-2
Therefore, (0,2) , (0,-2) , (2,2) , (2,-2) are the critical points of f.
Now,the second order partial derivatives are
fxx(x,y)=d^2 f/dx^2=(d/dx)(df/dx)=d(3x^2-6x)/dx =6x-6
fyy(x,y)=d^2 f/dy^2=(d/dy)(df/dy)=d(6y^2-24)/dy =12y
fxy(x,y)=d^2 f/dy.dx=(d/dy)(df/dx)=d(3x^2-6x)/dy =0
At the point (0,2) , we have
D=fxx(0,2)fyy(0,2)-fxy^2(0,2)=(-6)(24)-0
=-144 is less than 0
Here,since D is less than 0
Therefore,f has a saddle point at(0,2)
At the point (0,-2) , we have
D=fxx(0,-2)fyy(0,-2)-fxy^2(0,-2)=(-6)(-24)-0
=144 is greater than 0 and fxx(0,-2)=-6 is less than 0
Here,since D is greater than0 and fxx(0,-2) is less than 0
Therefore,f has a relative maximum at(0,-2)
At the point (2,2) , we have
D=fxx(2,2)fyy(2,2)-fxy^2(2,2)=(6)(24)-0
=144 > 0 and fxx(2,2)=6 > 0
Here,since D is greater than 0 and fxx(2,2) is greater than 0
Therefore,f has a relative minimum at (2,2)
At the point (2,-2) , we have
D=fxx(2,-2)fyy(2,-2)-fxy^2(2,-2)=(6(-24)-0
=-144 is less than 0
Here,since D is less than 0
Therefore,f has a saddle point at(2,-2)
Thus,the function f has relative maximum at the point (0,-2) and relative minimum at the point (2,2) as well as f has saddle points at the points(0,2) and (2,-2) respectively .
Right answer
Find the area of the region bounded above by g=x +6 bounded below by y=x^2 and x=2
Given: y=x+6÷y=x^2
X=0 and x=2
Formula : A=a^s^b[f(x)-g(x)]dx
X^2-x-6
X^2+2x-3x-6
X(x-2)-3(x+2)
(X-3) (x+2)
X=2, x=2
=A=a^s^b[f(x)-g(x)]dx
=a^s^b[x+b-×^2]dx
=a^s^b 2÷x^2 +sbx-lim sin 3÷x^3 dx
=(2÷4 + 12 – 3÷8)-(0)
=(14 – 8÷3)
=34÷3
=11.33
The area of region is 11.33
There also has one diagram
Right answer
Ravishankar singh F112 unit2
6) y=x+6 , x=0, x=2
A=b(lim)a [(x+6)-x^2]dx
=2(lim)a (2/x^2+6x-3/x^2)
2
=[2/x^2+6x-3/x^2] =[2/4+12-3/8]
0
=[2/0+0-3/0]
A=14-3/8=42-3/8=3/34
7) using eulers method dx/dy =y-x,(x)=2
(0.2)
given f(x,y)=y-x
y(0)=2 x(0)=0
x(0)=0+4 x(2)=0.1+0.1 steq sin is o.1
=0+0.1=0.1 =0.2
x =0, 0.1 ,0.2, 0.3, 0.4, 0.5, 0.6 ,0.7, 0.8 ,0.9, 1
let x=0
y(1)=y0+0.1(x0,y0) f(x.y)=2-0=2
=2+0.2=2.2
=n=1
y(2)=y1+0.1(x1,y1) f(x,y)=2.2-0.1=2.1
=2.2+0.1(2.1)=2.41
n=2
y3=2.41+0.1(2.41-0.2)=2.631
8) evalute lim 1
(9cos^2x+4sin^2x)dx
lim 1
(9cos^2x+4sin^2x)dx
lim 1 + lim 1 dx
9cos^2x 4sin2x
lim 1 lim 1
9cos^2x-sin^2x + 4(2sinxcosx)
Wrong answers
Kishore thakare Fycs F114 Unit3
Q3)fx(1,3) and fy(1,3) for the function f(x,y)=2x^3*y^3+2y+4x.
Solution
fx(x,y)=6x^2*y^2+4
fx(1,3)=6(1)^2*(3)^2+4
fx(1,3)=54+4
fx(1,3)=58
fy(x,y)=4x^3*y+2
fy(1,3)=4(1)^3*(3)+2
fy(1,3)=12+2
fy(1,3)=14
Wrong answer
Kishore thakare Fycs F114 Unit3
Q4) evaluate lim y.log(x^2+y^2). By
(x,y)->(0,0)
converting to polar coordinates.
Solution
lim y.log(x^2+y^2)
(x,y)->(0,0)
0.log(0^2+0^2)
0.log0
Wrong answer
Kishore thakare Fycs F114 Unit3
Q6) find the direction derivative of f(x,y)=e^xy at (2,0) in the direction of unit vector that’s make an angle of π/3 with the positive x-axis
Solution
fx(x,y)=e^xy*y
fy(x,y)=e^xy*x
fx(2,0)=e^(2)(0)*0
=0
fy(2,0)=e^(2)(0)*2
=2
fx(x0,y0)=fx(x0,y0)*cos∅+fy(x0,y0)*sin∅
=0*cos(π/3)+2*sin(π/3)
=2*√3/2
=√3
Kishore thakare Fycs F114 Unit3
Q6) find the direction derivative of f(x,y)=e^xy at (2,0) in the direction of unit vector that’s make an angle of π/3 with the positive x-axis
Solution
fx(x,y)=e^xy*y
fy(x,y)=e^xy*x
fx(2,0)=e^(2)(0)*0
=0
fy(2,0)=e^(2)(0)*2
=2
du(x0,y0)=fx(x0,y0)*cos∅+fy(x0,y0)*sin∅
=0*cos(π/3)+2*sin(π/3)
=2*√3/2
=√3
Wrong answer
kishore thakare Fycs F114 unit 3
Q5) evaluate lim √x^2+y^2*log(x^2+y^2) by
_________ (x,y)->(0,0)
converting to polar coordinates.
Solution
lim √x^2+y^2*log(x^2+y^2)
(x,y)->(0,0)
lim √(0)^2+(0)^2*log((0)^2+(0)^2)
(x,y)->(0,0)
lim √0+log0
(x,y)->(0,0)
lim √log0
(x,y)->(0,0)
Wrong answer
Fardeen shaikh 109
Unit 1
Q.5) Discuss the continuity of the function f(x) = √4-x^2
Ans:
f(x) = √4-x^2
f(x) = lim √4-x^2
x ~ ∞
= √lim 4 – lim x^2
x ~ ∞ x ~ ∞
= √4 – 1
= √3
The function f(x)= √4-x^2 is continuous at lim √x
x ~ ∞
Wrong answer
Fardeen shaikh 109
Unit 1
Q.6) Divide 100 into two parts such as the sum of their squares is minimum
Ans :
Let x be one part of 100
therefore, x = 100 – x
therefore, According to the condition
sum of squares = x^2 + (100 – x^2)
= x^2 + 1000 – 200x + x^2
y = 2x^2 – 200x + 10000
= 2(x^2 – 100x + 5000)
dy/dx = 2(4x – 100)
2(4x – 100) = 0
8x – 200 = 0
x = 200/8
x = 25
100 – x = 75
The two parts where sum of squares is minimum are 25 and 75.
Wrong answer
Fardeen shaikh 109
Unit 1
Q.7) Show that |x| is continuous everywhere
Ans:
f(x) = |x|
lim |x| = lim x = 0
x~∞ x~0
f(x) = |x| is not differentiable
f ‘ (0) = lim f (0 + h) f (0)/h
x~∞
lim = u/lamda symbol integral sign 1 if h > 0
h~0 – 1 if h < 0
So the limit from right is 1 while from left is – 1 the two-sided limit doesn't exist
Hence |x| is continuous.
Wrong answer