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Mathematics-III (Differential Calculus)

Module 3

Vector calculus

Q1. Calculate the curl for the following vector field.

F⃗ =x3y2 i⃗ +x2y3z4 j⃗ +x2z2 k⃗

Sol: In order to calculate the curl, we need to recall the formula.

where P, Q, and R correspond to the components of a given vector field: F⃗ =Pi⃗ +Qj⃗ +Rk⃗

=((x2z2)−(x2y3z4) )i⃗ +((x3y2)−(xz2) )j⃗ +((x2y3z4)−(x3y2) )k⃗

=(0−4x2y3z3)i⃗ +(0−2xz2)j⃗ +(2xy3z4−2x3y)k⃗

Thus the curl is

=(−4x2y3z3)i⃗ +(−2xz2)j⃗ +(2xy3z4−2x3y)k⃗

Q2. Find the directional derivative of Θ=x2y cos z at (1,2,π/2) in the direction of a = 2i+3j+2k.

Sol : ∇ϕ = i + k

= 2xy cos zi+ x2 cos zj -x2y sin zk

At (1,2,π/2) ∇ϕ = 0i +0j-2k

Directional directive in the direction of 2i+3j+2k.

=(0i+0j-2k). =-

Q3. In what direction from the point (2,1,-1) is the directional derivative of ϕ=x2yz3 maximum? What is its magnitude?

Solution :∇ϕ= i + k

= -4i-4j+12k

Directional derivative is maximum in the direction of ∇Θ. Hence, directional derivative is maximum in the direction of -4i-4j+12k

Its magnitude = =4

Q4. Prove that ͞͞͞F = [y2 cos x +z3] i +(2y sin x – 4) j +(3xz2 + 2) k is a conservative field. Find (i) scalar potential for͞͞͞F (ii) the work done in moving an object in this field from (0, 1, -1) to (/ 2,-1, 2)

Sol. : (a) The field is conservative if cur͞͞͞͞͞͞F = 0.

Now, cur͞͞͞F = ̷̷ X / y / z

i j k

Y2COS X +Z3 2y sin x-4 3xz2 +2

; Cur = (0-0) – (3z2 – 3z2) j + (2y cos x- 2y cos x) k = 0

; F is conservative.

(b) Since F is conservative there exists a scalar potential ȸ such that

F = ȸ

(y2 cos x +z3) i + (2y sin x-4) j + (3xz2 + 2) k

= i + j + k

= y2 cos x + z3, = 2y sin x – 4, = 3xz2 + 2

Now, = dx + dy + dz

= (y2 cos x + z3) dx +(2y sin x – 4)dy + (3xz2 + 2)dz

= (y2 cos x dx + 2y sin x dy) +(z3dx +3xz2dz) +(- 4 dy) + (2 dz)

=d(y2 sin x + z3x – 4y -2z)

ȸ = y2 sin x +z3x – 4y -2z

= dx + (2y sin x – 4) dy + ( 3xz2 + 2) dz

= (y2 sin x + z3x – 4y + 2z) (as shown above)

= [ y2 sin x + z3x – 4y + 2z ]( /2, -1, 2)

= [ 1 +8 + 4 + 4 ] – { - 4 – 2} =4 + 15

Q5. If x2zi – 2y3z3j + xy2z2k find dvi and curl at (1,-1, 1)

Sol: div = ∇.= =

= 2xz – 6y2z3 + 2xy2z =(2-6+2) = -2

Curl =

=i(2xyz2 + 6y3z2) – j(y2z2- x2) + k(0-0)

=-8 at (1,-1,1)

Q6. Find the angle between the normal to the surface xy = z2 at the points (1,4,2) and (-3,-3,3)

Sol: let ϕ = xy-z2

∇ϕ= i =yi + xj -2zk =4i + j -4k

∇ϕ = 3i – 3j-6k

But these are the normal to the surface at given points. Angle between two vectors is given by (4i + j -4k).( 4i + j -4k)= |4i + j- 4k|.|-3i-3j -6K|cos θ.

If θ is the angle between then cos θ= = .

7. Evaluate where F= cos y.i-x siny j and C is the curve y= in the xy plae from (1,0) to (0,1)

Sol : The curve y= i.e x2+y2 =1. Is a circle with centre at the origin and radius unity.

= =-1

8. Evaluate where = (2xy +z2) I +x2j +3xz2 k along the curve x=t, y=t2, z= t3 from (0,0,0) to (1,1,1).

Sol: F x dr =

Put x=t, y=t2, z= t3

Dx=dt ,dy=2tdt, dz=3t2dt.

F x dr =

=(3t4-6t8) dt i – ( 6t5+3t8 -3t7) dt j +( 4t4+2t7-t2)dt k

=t4-6t3)dti –(6t5+3t8-3t7)dt j+(4t4 + 2t7 – t2)dt k

9. Prove that ͞͞͞F = [y2 cos x +z3] i +(2y sin x – 4) j +(3xz2 + 2) k is a conservative field. Find (i) scalar potential for͞͞͞F (ii) the work done in moving an object in this field from (0, 1, -1) to (/ 2,-1, 2)