Vector Calculus Questions and Answers – Gradient of a Function and Conservative Field

This set of Basic Vector Calculus Questions and Answers focuses on “Gradient of a Function and Conservative Field”.

1. Del operator is also known as _________
a) Divergence operator
b) Gradient operator
c) Curl operator
d) Laplacian operator
View Answer

Answer: b
Explanation: This differential operator is not a vector itself but when it operates on a scalar function, for example, a vector ensues.

2. The gradient of a scalar field V is a vector that represents both magnitude and the direction of the maximum space rate of increase of V.
a) True
b) False
View Answer

Answer: a
Explanation: A gradient operates on a scalar only and gives a vector as a result. This vector has a magnitude and direction. The gradient is found by finding the speed that is by taking the partial differentiation.

3. The gradient is taken on a _________
a) tensor
b) vector
c) scalar
d) anything
View Answer

Answer: c
Explanation: Gradient is taken only on a scalar field. After taking gradient of a scalar field it becomes a vector. It is found by taking the partial differentiation.
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4. Find the gradient of a function V if V= xyz.
a) yz ax + xz ay + xy az
b) yz ax + xy ay + xz az
c) yx ax + yz ay + zx az
d) xyz ax + xy ay + yz az
View Answer

Answer: a
Explanation: V = xyz
Gradient of \(V = \frac{∂V}{∂x} a_x + \frac{∂V}{∂y}a_y + \frac{∂V}{∂z}a_z \)
= yz ax + xz ay + xy az

5. Find the gradient of V = x2 sin(y)cos(z).
a) 2x siny cos z ax + x2 cos(y)cos(z) ay – x2 sin(y)sin(z) az
b) 2x siny cos z ax + x2 cos(y)cos(z) ay + x2 sin(y)sin(z) az
c) 2x sinz cos y ax + x2 cos(y)cos(z) ay – x2 sin(y)sin(z) az
d) x siny cos z ax + x2 cos(y)cos(z) ay – x2 sin(y)sin(z) az
View Answer

Answer: a
Explanation: \(V = x^2 sin(y)cos(z) \)
Gradient of \(V = \frac{∂V}{∂x} a_x + \frac{∂V}{∂y} a_y + \frac{∂V}{∂z} a_z \)
\(= 2x siny cos z a_x + x^2 cos(y)cos(z) a_y – x^2 sin(y)sin(z) a_z \)

6. Find the gradient of the function W if W = ρzcos(ϕ) if W is in cylindrical coordinates.
a) zcos(ϕ)aρ – z sin(ϕ) aΦ + ρcos(ϕ) az
b) zcos(ϕ)aρ – sin(ϕ) aΦ + cos(ϕ) az
c) zcos(ϕ)aρ + z sin(ϕ) aΦ + ρcos(ϕ) az
d) zcos(ϕ)aρ + z sin(ϕ) aΦ + cos(ϕ) az
View Answer

Answer: a
Explanation: \(W = ρzcos(ϕ) \)
Gradient of \(W = \frac{1}{ρ} \frac{∂W}{∂ρ} a_ρ + \frac{1}{ρ} \frac{∂W}{∂ϕ}a_y + \frac{∂W}{∂z}a_z \)
\(= zcos(ϕ)a_ρ – z sin(ϕ) a_y + ρcos(ϕ) a_z \)

7. Find the gradient of A if A = ρ2 + z3 + cos(ϕ) + z and A is in cylindrical coordinates.
a) \(2ρz^3 \, a_ρ – \frac{1}{ϕ} sin(ϕ) \, aΦ + 3ρ^2 z^2 \, a_z \)
b) \(2ρz^3 \, a_ρ – \frac{1}{ρ} sin(ϕ) \, aΦ + 3ρ^2 z^2+1 \, a_z \)
c) \(2ρz^3 \, a_ρ – \frac{1}{ϕ} sin(ϕ) \, aΦ + 3ρ^2 z^2+1 \, a_z \)
d) \(2ρz^3 \, a_ρ – \frac{1}{ρ} sin(ϕ) \, aΦ + 3ρ^2 z^2 \, a_z \)
View Answer

Answer: b
Explanation: \(A = ρ^2 + z^3 + cos(ϕ) + z \)
Gradient of \(A = \frac{1}{ρ} \frac{∂A}{∂ρ} a_ρ + \frac{1}{ρ} \frac{∂A}{∂ϕ}a_y + \frac{∂A}{∂z}a_z \)
\(=2ρz^3 \, a_ρ – \frac{1}{ρ} sin(ϕ) \, aΦ + 3ρ^2 z^2 + 1 \, a_z \)
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8. Find gradient of B if B = ϕln(r) + r2 ϕ if B is in spherical coordinates.
a) \(\frac{ρ}{r}+ 2rθ \,a_r – r a_θ + \frac{lnr}{rsin(θ)} a_Φ \)
b) \(\frac{ρ}{r}+ 2rϕ \,a_r – r a_θ + \frac{lnr}{rsin(θ)} a_Φ \)
c) \(\frac{ρ}{r}+ 2rθ \,a_r – r^2 a_θ + \frac{lnr}{rsin(θ)} a_Φ \)
d) \( \frac{ρ}{r}+ 2rϕ \,a_r – r^2 a_θ + \frac{lnr}{rsin(θ)} a_Φ \)
View Answer

Answer: a
Explanation: \(B = ϕln(r)+r^2 ϕ \)
Gradient of \(B = \frac{∂B}{∂r} a_r + \frac{1}{r} \frac{∂B}{∂θ} a_θ + \frac{1}{rsin(θ)} \frac{∂B}{∂ϕ} a_Φ \)
\(= \frac{ρ}{r}+ 2rθ \,a_r – r a_θ + \frac{lnr}{rsin(θ)} a_Φ \)

9. Find gradient of B if B = rθϕ if X is in spherical coordinates.
a) \(θϕ \, a_r – ϕ \,a_θ + \frac{θ}{sin(θ)} a_Φ \)
b) \(rθϕ \, a_r – ϕ \,a_θ + r \frac{θ}{sin(θ)} a_Φ \)
c) \(θϕ \, a_r – ϕr \,a_θ + \frac{θ}{sin(θ)} a_Φ \)
d) \(θϕr \, a_r – ϕ \,a_θ + r\frac{θ}{sin(θ)} a_Φ \)
View Answer

Answer: a
Explanation: Gradient of \(B = \frac{∂B}{∂r} a_r + \frac{1}{r} \frac{∂B}{∂θ} a_θ + \frac{1}{rsin(θ)} \frac{∂B}{∂ϕ} a_Φ \)
\(B = rθϕ\)
Hence gradient of \(B = θϕ \, a_r -ϕ \, a_θ + \frac{θ}{sin(θ)} a_Φ \)
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10. If W = x2 y2 + xz, the directional derivative \( \frac{dW}{dl} \) in the direction 3 ax + 4 ay + 6 az at (1,2,0).
a) 5
b) 6
c) 7
d) 8
View Answer

Answer: b
Explanation: First find the gradient of W which is (2xy2+z) ax + 2yx2 ay + x az
At (1,2,0) the gradient of W is 8 ax + 4 ay + 1 az
\(\frac{dW}{dl} = \) (Gradient of W ) . al
\( = (8,4,1) . \frac{(3,4,6)}{\sqrt{(9+16+36)}} \)
\(=5.88897 = 6.\)

11. If W = xy + yz + z, find directional derivative of W at (1,-2,0) in the direction towards the point (3,6,9).
a) -0.6
b) -0.7
c) -0.8
d) -0.9
View Answer

Answer: c
Explanation: The gradient of W is = y ax + (x+z) ay + (y+1) az
At (1,-2,0) the gradient of the function W is -2 ax + ay – az
\(\frac{dW}{dl} = \)(Gradient of W ) . al
\(= \frac{(-2,1,-1).(3,6,9)}{11.22} \)
\(= -0.8.\)

12. Electric field E can be written as _________
a) -Gradient of V
b) -Laplacian of V
c) Gradient of V
d) Laplacian of V
View Answer

Answer: a
Explanation: Potential difference decreases in the direction of increase in Electric field. Hence Electric field is nothing but the negative of the gradient of potential difference.

13. Let F = (xy2) ax + yx2 ay, F is a not a conservative vector.
a) True
b) False
View Answer

Answer: b
Explanation: Q = xy2 and P = yx2
\(\frac{∂P}{∂y} = 2xy\) and \(\frac{∂Q}{∂x} = 2xy \)
Since, both are equal, F is a conservative vector.

14. State whether the given equation is a conservative vector.
G = (x3y) ax + xy3 ay
a) True
b) False
View Answer

Answer: b
Explanation: P = x3 y and Q = xy3
\(\frac{∂P}{∂y} = x^3\) and \(\frac{∂Q}{∂x}= y^3 \)
Now since they aren’t equal, the vector is not a conservative vector or field.

15. Find a unit vector normal to the surface of the ellipsoid at (2,2,1) if the ellipsoid is defined as f(x,y,z) = x2 + y2 + z2 – 10.
a) \(\frac{2}{3} a_x + \frac{2}{3} a_y + \frac{1}{3} a_z \)
b) \(\frac{1}{3} a_x + \frac{1}{3} a_y + \frac{1}{3} a_z \)
c) \(\frac{2}{3} a_x + \frac{2}{3} a_y + \frac{2}{3} a_z \)
d) \(\frac{2}{3} a_x + \frac{1}{3} a_y + \frac{1}{3} a_z \)
View Answer

Answer: a
Explanation: First we have to find the gradient of the function, which is equal to 2x ax + 2y ay + 2z az.
Gradient of f at (2,2,1) is (4,4,2).
\(a_n= \frac{±(4,4,2)}{6} = \frac{2}{3} a_x + \frac{2}{3} a_y + \frac{1}{3} a_z. \)

Sanfoundry Global Education & Learning Series – Vector Calculus.

To practice basic questions and answers on all areas of Vector Calculus, here is complete set of 1000+ Multiple Choice Questions and Answers.

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