Unit 4
Vector Differential Calculus
Q(1) A curve is given by the equation . Find the angle between tangents.
Ans-
Solution :
Where
Now
Q(2) if show that has a constant dim .
Ans-
Now
i.e
Both cant be true simultaneously
has constant direction.
Q(3) Calculate the curl for the following vector field. F⃗ =x3y2 i⃗ +x2y3z4 j⃗ +x2z2 k⃗ Ans- Solution: 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
Q(4) Find the directional derivative of Θ=x2y cos z at (1,2,π/2) in the direction of a = 2i+3j+2k. Ans- Solution : ∇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). =-
Q(5) In what direction from the point (2,1,-1) is the directional derivative of ϕ=x2yz3 maximum? What is its magnitude? Ans- 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
Q(6) 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) Ans- Sol. : (a) The field is conservative if cur͞͞͞͞͞͞F = 0. Now, cur͞͞͞F = ; 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 (c) now, work done = .d ͞r = 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
Q(7) If x2zi – 2y3z3j + xy2z2k find dvi and curl at (1,-1, 1) Ans- Solution : div = ∇.= = = 2xz – 6y2z3 + 2xy2z =(2-6+2) = -2 Curl = =i(2xyz2 + 6y3z2) – j(y2z2- x2) + k(0-0) =-8 at (1,-1,1)
Q(8). Find the angle between the normal to the surface xy = z2 at the points (1,4,2) and (-3,-3,3) Ans- Solution: 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= = .
Q(9). Find the directional derivative of at point A(1,-2,1) in the direction of AB where B is (2,6,-1). Ans- = - = (2-1)i + (6+2)j + (-1-1)k =I + 8j – 2k Hence, Directional Derivative at A is the direction of . =4(i-8j+4k). =
Q(10). Find the directional derivative of at (0,1) in direction making an angle of with positive x axis. Ans- Soln. Directional Derivative
=
=0i-j at (0,1) Unit vector making angle of with axis = = Hence, required directional derivative =(0i-j). [] =
Q(11). Prove that =0 where Ans- Soln. Let Then, =
=).[
= 0 Q(12). A vector field is given by =i + (Show that is irrotational and find its scalar potential. Ans- Soln. Curl = = =0i+0j+(2xy-2xy)k =0i+0j+0k Hence, is irrotational. If is the scalar potential, then Therefore, i + ( = Comparing the coefficients of i , j , k From above eq. common terms are:
Q(13). Prove that and hence find f if . Ans- Soln. We have = Here, =f(r) and f is a function of r and r is a function of (x,y,z) + j But 2r=2x ,
= Comparing with ; f(r)=+C = |
(2,6,-1).
= - = (2-1)i + (6+2)j + (-1-1)k
=I + 8j – 2k
Hence, Directional Derivative at A is the direction of .
=4(i-8j+4k). =
Q(14). Find the directional derivative of at (0,1) in direction making an angle of with positive x axis.
Ans-
Soln. Directional Derivative
=
=0i-j at (0,1)
Unit vector making angle of with axis
=
=
Hence, required directional derivative
=(0i-j). [] =
Q(15). Prove that =0 where
Ans-
Soln. Let
Then, =
=).[
= 0
Q(16). A vector field is given by =i + (Show that is irrotational and find its scalar potential.
Ans-
Soln. Curl =
=
=0i+0j+(2xy-2xy)k
=0i+0j+0k
Hence, is irrotational.
If is the scalar potential, then
Therefore, i + ( =
Comparing the coefficients of i , j , k
From above eq. common terms are:
Q(17) Prove that and hence find f if .
Ans-
Soln. We have =
Here, =f(r) and f is a function of r and r is a function of (x,y,z)
+ j
But
2r=2x
,
=
Comparing with ; f(r)=+C =.