Vector Differential Calculus Questions and Answers

This set of Vector Calculus Multiple Choice Questions & Answers (MCQs) focuses on “Vector Differentiation”.

1. The Gradient of a constant is a null vector.
a) True
b) False
View Answer

Answer: a
Explanation:
If ∅(x,y,z)is a constant, then
\(\frac {∂∅}{∂x}, \frac {∂∅}{∂y}\) and \(\frac {∂∅}{∂z}\) are zeroes.
∴ ∇∅ = \(\overrightarrow {i} \frac {∂∅}{∂x} + \overrightarrow {j} \frac {∂∅}{∂y} + \overrightarrow {i} \frac {∂∅}{∂z}\)
= \(\overrightarrow {i}\)(0) + \(\overrightarrow {j}\)(0) + \(\overrightarrow {k}\)(0) = \(\overrightarrow {0}\)

2. If ∅₁, ∅₂ are scalar point functions, then ∇(c₁∅₁ ± c₂∅₂) = c₁∇∅₁ ± c₂∇∅₂ where c₁ and c₂ are constants.
a) True
b) False
View Answer

Answer: a
Explanation:
∇(c₁∅₁ ± c₂∅₂) = (\(\overrightarrow {i} \frac {∂}{∂x} + \overrightarrow {j} \frac {∂}{∂y} + \overrightarrow {i} \frac {∂}{∂z}\))(c₁∅₁ ± c₂∅₂)
= \(\overrightarrow {i} \frac {∂}{∂x}\)[c₁∅₁ ± c₂∅₂] + \(\overrightarrow {j} \frac {∂}{∂y}\) [c₁∅₁ ± c₂∅₂] + \(\overrightarrow {k} \frac {∂}{∂z}\)[c₁∅₁ ± c₂∅₂]
= \(\overrightarrow {i} \bigg [ \)c₁\(\frac {∂∅_1}{∂x}\) ± c₂\(\frac {∂∅_2}{∂x} \bigg ] \) + \(\overrightarrow {j} \bigg [ \)c₁\(\frac {∂∅_1}{∂y}\) ± c₂\(\frac {∂∅_2}{∂y} \bigg ] \) + \(\overrightarrow {k} \bigg [ \)c₁\(\frac {∂∅_1}{∂z}\) ± c₂\(\frac {∂∅_2}{∂z} \bigg ] \)
= c₁\(\bigg [ \overrightarrow {i} \frac {∂∅_1}{∂x} + \overrightarrow {j} \frac {∂∅_2}{∂y} + \overrightarrow {k} \frac {∂∅_3}{∂z} \bigg ] \) ± c₂\(\bigg [ \overrightarrow {i} \frac {∂∅_1}{∂x} + \overrightarrow {j} \frac {∂∅_2}{∂y} + \overrightarrow {k} \frac {∂∅_3}{∂z} \bigg ]\)
∇(c₁∅₁ ± c₂∅₂) = c₁∇∅₁ ± c₂∇∅₂

3. Which of the following is ∇∅ if ∅ = xyz?
a) yz\(\overrightarrow {i}\) + xz\(\overrightarrow {j}\) + xy\(\overrightarrow {k}\)
b) y\(\overrightarrow {i}\) + x\(\overrightarrow {j}\) + x\(\overrightarrow {k}\)
c) z\(\overrightarrow {i}\) + z\(\overrightarrow {j}\) + y\(\overrightarrow {k}\)
d) \(\overrightarrow {i} + \overrightarrow {j} + \overrightarrow {k}\)
View Answer

Answer: a
Explanation:
Given: ∅ = xyz
∇∅ = \(\overrightarrow {i} \frac {∂∅}{∂x} + \overrightarrow {j} \frac {∂∅}{∂y} + \overrightarrow {k} \frac {∂∅}{∂z}\)
= \(\overrightarrow {i}\)yz + \(\overrightarrow {j}\)xz + \(\overrightarrow {k}\)xy = yz\(\overrightarrow {i}\) + xz\(\overrightarrow {j}\) + xy\(\overrightarrow {k}\)

4. Which of the following is ∇∅ if ∅ = log⁡(x² + y² + z²)?
a) \(\frac {2}{r^2} \overrightarrow {r}\)
b) \(\frac {2}{r} \overrightarrow {r}\)
C) \(\frac {2}{r^3} \overrightarrow {r}\)
d) \(\frac {1}{r^2} \overrightarrow {r}\)
View Answer

Answer: a
Explanation:
Given: ∅ = xyz
∇∅ = \(\overrightarrow {i}\frac {∂∅}{∂x} + \overrightarrow {j}\frac {∂∅}{∂y} + \overrightarrow {k}\frac {∂∅}{∂z}\)
= \(\overrightarrow {i} \bigg [ \frac {2x}{x^2 + y^2 + z^2} \bigg ] + \overrightarrow {j} \bigg [ \frac {2y}{x^2 + y^2 + z^2} \bigg ] + \overrightarrow {k} \bigg [ \frac {2z}{x^2 + y^2 + z^2} \bigg ] \)
= \(\frac {2}{x^2 + y^2 + z^2}\)[x\(\overrightarrow {i}\) + y\(\overrightarrow {j}\) + z\(\overrightarrow {k}\)]
= \(\frac {2}{x^2 + y^2 + z^2} [\overrightarrow {r} ] = \frac {2}{r^2}\overrightarrow {r}\)

5. Which of the following are the values of ∇(r) and ∇\((\frac {1}{r})\)?
a) \(\frac {\overrightarrow {r}}{r}, \frac { – \overrightarrow {r}}{r^3}\)
b) \(\frac {\overrightarrow {r}}{r^2} ,\frac { – 1}{r^2}\)
c) \(\frac {\overrightarrow {r}}{r^4} ,\frac { – 1}{r^4}\)
d) \(\frac {\overrightarrow {r}}{r^3} ,\frac { – 1}{r^3}\)
View Answer

Answer: a
Explanation:
We know that \(\overrightarrow {r}\) = x\(\overrightarrow {i}\) + y\(\overrightarrow {j}\) + z\(\overrightarrow {k}\)
r = |\(\overrightarrow {r}\)| = \(\sqrt {x^2 + y^2 + z^2}\), r² = x² + y² + z²
\(\frac {∂r}{∂x} = \frac {x}{r}; \frac {∂r}{∂y} = \frac {y}{r}; \frac {∂r}{∂z} = \frac {z}{r}\)
(i) ∇r = \(\overrightarrow {i} \frac {∂r}{∂x} + \overrightarrow {j} \frac {∂r}{∂y} + \overrightarrow {k} \frac {∂z}{∂r}\)
= \(\overrightarrow {i} (\frac {x}{r}) + \overrightarrow {j}(\frac {y}{r}) + \overrightarrow {k}(\frac {z}{r})\)
= \(\frac {x\overrightarrow {i} + y\overrightarrow {j} + z\overrightarrow {k}}{r} = \frac {\overrightarrow {r}}{r}\)
(ii) ∇\((\frac {1}{r}) = \sum \overrightarrow {i} \frac {∂}{∂x} (\frac {1}{r})\)
= ∑\(\overrightarrow {i}(\frac { – 1}{r^2}) \frac {∂r}{∂x}\)
= ∑\(\overrightarrow {i}(\frac { – 1}{r^2}) (\frac {x}{r}) = – \frac {1}{r^3}\) ∑x\(\overrightarrow {i}\)
= – \(\frac {1}{r^3}\)[x\(\overrightarrow {i}\) + y\(\overrightarrow {j}\) + z\(\overrightarrow {k}\)]
∴ ∇\((\frac {1}{r}) = – \frac {\overrightarrow {r}}{r^3}\)

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6. The temperature at a point (x,y,z) in space is given by T(x,y,z) = x² + y² – z. A mosquito located at (1,1,2) desires to fly in such a direction that it gets cooled faster. Which of the following direction it should fly?
a) 2\(\overrightarrow {i}\) + 2\(\overrightarrow {j}\) – \(\overrightarrow {k}\)
b) 2\(\overrightarrow {i}\) – 2\(\overrightarrow {j}\) – \(\overrightarrow {k}\)
c) 2\(\overrightarrow {i}\) + 2\(\overrightarrow {j}\) + \(\overrightarrow {k}\)
d) 2\(\overrightarrow {i}\) – 2\(\overrightarrow {j}\) + \(\overrightarrow {k}\)
View Answer

Answer: a
Explanation:
Given: T = x² + y² – z
∇T = \(\overrightarrow {i}\frac {∂T}{∂x} + \overrightarrow {j} \frac {∂T}{∂y} + \overrightarrow {k} \frac {∂T}{∂z}\)
= 2x \(\overrightarrow {i}\) + 2y \(\overrightarrow {j}\) – \(\overrightarrow {k}\)
∇T_(1,1,2) = 2\(\overrightarrow {i}\) + 2\(\overrightarrow {j}-\overrightarrow {k}\)
∴ Mosquito will fly in the direction of maximum rate of decrease, which is (2\(\overrightarrow {i}\) + 2\(\overrightarrow {j}\) – \(\overrightarrow {k}\)).

7. Which of the following is the value of ∇(rⁿ)?
a) nr^n-2\(\overrightarrow {r}\)
b) r^n-2\(\overrightarrow {r}\)
c) nr^n-2
d) nr^n-1\(\overrightarrow {r}\)
View Answer

Answer: a
Explanation:
∇(rⁿ) = ∑\(\overrightarrow {i} \frac {∂}{∂x}\)(rⁿ) = ∑\(\overrightarrow {i}\) nr^n-1\(\frac {∂r}{∂x}\)
= ∑\(\overrightarrow {i}\)nr^n-1\(\frac {x}{r}\)
= ∑\(\overrightarrow {i}\)nr^n-2x
= nr^n-2[x\(\overrightarrow {i}\) + y\(\overrightarrow {j}\) + z\(\overrightarrow {k}\)]
= nr^n-2\(\overrightarrow {r}\)

8. Which of the following values is obtained by evaluating ∇(log⁡r)?
a) \(\frac {\overrightarrow {r}}{r^2}\)
b) \(\frac {\overrightarrow {r}}{r^3}\)
c) \(\frac {\overrightarrow {r}}{r^4}\)
d) \(\frac {\overrightarrow {r}}{r^5}\)
View Answer

Answer: a
Explanation:
∇(log⁡r) = ∑\(\overrightarrow {i}\frac {∂}{∂x}\)(log⁡r) = ∑\(\overrightarrow {i}\frac {1}{r} \frac {∂r}{∂x}\)
= ∑\(\overrightarrow {i}\frac {1}{r} \frac {x}{r}\) = ∑\(\overrightarrow {i}\frac {x}{r^2} = \frac {1}{r^2}\) [x\(\overrightarrow {i}\) + y\(\overrightarrow {j}\) + z\(\overrightarrow {k}\)] = \(\frac {\overrightarrow {r}}{r^2}\)

9. Which of the following is the value of ∇(e^{x² + y² + z²})?
a) 2e^r²\(\overrightarrow {r}\)
b) e^r²\(\overrightarrow {r}\)
c) 2e^r⁴\(\overrightarrow {r}\)
d) 2\(\overrightarrow {r}\)
View Answer

Answer: a
Explanation:
We know that \(\overrightarrow {r}\) = x\(\overrightarrow {i}\) + y\(\overrightarrow {j}\) + z\(\overrightarrow {k}\)
r² = x² + y² + z²
\(\frac {∂r}{∂x} = \frac {x}{r}, \frac {∂r}{∂y} = \frac {y}{r}, \frac {∂r}{∂z} = \frac {z}{r}\)
∇(e^{x² + y² + z²}) = ∇(e^r²) = ∑\(\overrightarrow {i}\) \(\frac {∂}{∂x}\)(e^r²)
= ∑\(\overrightarrow {i}\)e^r²2r \(\frac {∂r}{∂x}\) = ∑\(\overrightarrow {i}\)e^r²2r \(\frac {x}{r}\)
= ∑\(\overrightarrow {i}\)e^r²2x = 2e^r²[x\(\overrightarrow {i}\) + y \(\overrightarrow {j}\) + z\(\overrightarrow {k}\)]
= 2e^r²\(\overrightarrow {r}\)

10. Which of the following is the directional derivative of ∅ = x²yz + 4xz² at (1, – 2, – 1) in the direction of 2\(\overrightarrow {i}\) – \(\overrightarrow {j}\) + 3\(\overrightarrow {k}\)?
a) \(\frac {37}{3}\)
b) \(\frac {1}{3}\)
c) \(\frac {11}{3}\)
d) \(\frac {7}{3}\)
View Answer

Answer: a
Explanation:
Given: ∅ = x²yz + 4xz²
∴ ∇∅ = \(\overrightarrow {i} \frac {∂∅}{∂x} + \overrightarrow {j}\frac {∂∅}{∂y} + \overrightarrow {k}\frac {∂∅}{∂z}\)
= (2xy + 4xz²) \(\overrightarrow {i}\) + x²z \(\overrightarrow {j}\) + (x²y + 8xz)\(\overrightarrow {k}\)
(∇∅)_{(1, -2, -1)} = 8\(\overrightarrow {i}\) – \(\overrightarrow {j}\) – 10\(\overrightarrow {k}\)
\(\overrightarrow {a}\) = 2\(\overrightarrow {i}\) – \(\overrightarrow {j}\) – 2\(\overrightarrow {k}\)
\(|\overrightarrow {a}| = \sqrt {4 + 1 + 4}\) = 3
D.D. = ∇∅.\(\frac {\overrightarrow {a}}{|\overrightarrow {a}|}\)
= (8\(\overrightarrow {i}\) – \(\overrightarrow {j}\) – 10\(\overrightarrow {k}\)).\(\frac {(2\overrightarrow {i}-\overrightarrow {j} – 2\overrightarrow {k})}{3}\)
= \(\frac {1}{3}\)[16 + 1 + 20] = \(\frac {37}{3}\)

11. Which of the following is the directional derivative of ∅ = x²yz + 4xz² + xyz at (1,2,3) in the direction of 2\(\overrightarrow {i} + \overrightarrow {j} – \overrightarrow {k}\)?
a) \(\frac {1}{\sqrt {6}}\)[86]
b) \(\frac {1}{\sqrt {6}}\)[83]
c) \(\frac {1}{\sqrt {6}}\)[84]
d) \(\frac {1}{\sqrt {6}}\)[96]
View Answer

Answer: a
Explanation:
Given: ∅ = x²yz + 4xz² + xyz
∴ ∇∅ = \(\overrightarrow {i} \frac {∂∅}{∂x} +\overrightarrow {j}\frac {∂∅}{∂y} + \overrightarrow {k}\frac {∂∅}{∂z}\)
= (2xyz + 4z² + yz) \(\overrightarrow {i}\) + (x²z + xz) \(\overrightarrow {j}\) + (x²y + 8xz + xy)\(\overrightarrow {k}\)
∇∅_(1,2,3) = 54\(\overrightarrow {i}\) + 6\(\overrightarrow {j}\) + 28\(\overrightarrow {k}\)
Given: \(\overrightarrow {a}\) = 2\(\overrightarrow {i}\) + \(\overrightarrow {j}\) – \(\overrightarrow {k}\)
\(|\overrightarrow {a}| = \sqrt {4 + 1 + 1}\) = √6
D.D. = ∇∅.\(\frac {\overrightarrow {a}}{|\overrightarrow {a}|}\)
= (54\(\overrightarrow {i}\) + 6\(\overrightarrow {j}\) + 28\(\overrightarrow {k}\)).\(\frac {(2\overrightarrow {i}-\overrightarrow {j} – 2\overrightarrow {k})}{\sqrt 6}\)
= \(\frac {1}{\sqrt 6}\)[(54\(\overrightarrow {i}\) + 6\(\overrightarrow {j}\) + 28\(\overrightarrow {k}\)). (2\(\overrightarrow {i}+\overrightarrow {j} – \overrightarrow {k}\))]
= \(\frac {1}{\sqrt 6}\)[108 + 6 – 28] = \(\frac {1}{\sqrt 6}\)[86]

12. Which of the following is the directional derivative of 4x²z + xy²z at (1, -1, 2) in the
direction of 2\(\overrightarrow {i}-\overrightarrow {j}\) + 3\(\overrightarrow {k}\)?
a) \(\frac {55}{\sqrt {14}}\)
b) \(\frac {75}{\sqrt {14}}\)
c) \(\frac {65}{\sqrt {14}}\)
d) \(\frac {35}{\sqrt {14}}\)
View Answer

Answer: a
Explanation:
∅ = 4x²z + xy²z
∇∅ = \(\overrightarrow {i} \frac {∂∅}{∂x} + \overrightarrow {j}\frac {∂∅}{∂y} + \overrightarrow {k}\frac {∂∅}{∂z}\)
= (8xz + y²z) \(\overrightarrow {i}\) + (2xyz) \(\overrightarrow {j}\) + (4x² + xy²)\(\overrightarrow {k}\)
(∇∅)_{(1, -1, 2)} = (16 + 2) \(\overrightarrow {i}\) + (-4) \(\overrightarrow {j}\) + (4 + 1)\(\overrightarrow {k}\)
= 18\(\overrightarrow {i}\) – 4\(\overrightarrow {j}\) + 5\(\overrightarrow {k}\)
\(\overrightarrow {a}\) = 2\(\overrightarrow {i}\) – \(\overrightarrow {j}\) + 3\(\overrightarrow {k}\); \(|\overrightarrow {a}| = \sqrt {4 + 1 + 9}\) = 14
D.D. = ∇∅.\(\frac {\overrightarrow {a}}{|\overrightarrow {a}|}\) = (18\(\overrightarrow {i}\) – 4\(\overrightarrow {j}\) + 5\(\overrightarrow {k}\)).\(\frac {(2\overrightarrow {i}-\overrightarrow {j} + 3\overrightarrow {k})}{\sqrt {14}}\)
= \(\frac {36 + 4 + 15}{\sqrt {14}} = \frac {55}{\sqrt {14}}\)

13. In what direction from the point (3, 1, -2) is the Directional derivative of ∅ = x²y²z⁴ a maximum?
a) \(\sqrt {175104}\)
b) \(\sqrt {167543}\)
c) \(\sqrt {145678}\)
d) \(\sqrt {176543}\)
View Answer

Answer: a
Explanation:
Given: ∅ = x²y²z⁴
∇∅ = \(\overrightarrow {i} \frac {∂∅}{∂x} + \overrightarrow {j}\frac {∂∅}{∂y} + \overrightarrow {k}\frac {∂∅}{∂z}\)
= 2xy²z⁴\(\overrightarrow {i}\) + 2x²yz⁴ \(\overrightarrow {j}\) + 4x²y²z³\(\overrightarrow {k}\)
(∇∅)_{(3, 1, -2)} = 96 \(\overrightarrow {i}\) + 288 \(\overrightarrow {j}\) – 288\(\overrightarrow {k}\)
|∇∅| = \(\sqrt {9216 + 82944 + 82944} = \sqrt {175104}\)
The directional derivative is maximum in the direction ∇∅ and the magnitude of this maximum
is |∇∅| = \(\sqrt {175104}\).

14. Which of the following is the unit normal to the surface xy = z² at the point(1, 1, -1)?
a) \(\frac {\overrightarrow {i} + \overrightarrow {j} + 2\overrightarrow {k}}{\sqrt 6} \)
b) \(\frac {\overrightarrow {i} – \overrightarrow {j} + 2\overrightarrow {k}}{\sqrt 6} \)
c) \(\frac {\overrightarrow {i} – \overrightarrow {j} – 2\overrightarrow {k}}{\sqrt 6} \)
d) \(\frac {\overrightarrow {i} + \overrightarrow {j} – 2\overrightarrow {k}}{\sqrt 6} \)
View Answer

Answer: a
Explanation:
Given: ∅ = xy – z²
∇∅ = y\(\overrightarrow {i}\) + x\(\overrightarrow {j}\) – 2z \(\overrightarrow {k}\)
(∇∅)_{(1, 1, -1)} = \(\overrightarrow {i}\) + \(\overrightarrow {j}\) + 2\(\overrightarrow {k}\)
|∇∅| = \(\sqrt {1 + 1 + 4}\) = √6
A unit normal to the given surface at the point is \(\frac {∇∅}{|∇∅|}=\frac {\overrightarrow {i} + \overrightarrow {j} + 2\overrightarrow {k}}{\sqrt 6} \).

15. Which of the following is the angle between the surfaces z = x² + y² – 3 and x² + y² + z² = 9 at (2, -1, 2)?
a) cos^-1[ \(\frac {8}{3\sqrt {21}}\) ]
b) cos^-1⁡[ \(\frac {8}{\sqrt {21}}\) ]
c) cos^-1⁡[ \(\frac {1}{3\sqrt {21}}\) ]
d) cos^-1⁡[ \(\frac {2}{3\sqrt {21}}\) ]
View Answer

Answer: a
Explanation:
∅₁ = x² + y² – z – 3 ∅₂ = x² + y² + z² – 9
∇∅₁ = 2x\(\overrightarrow {i}\) + 2y\(\overrightarrow {j}\) – \(\overrightarrow {k}\)∇∅₂ = 2x\(\overrightarrow {i}\) + 2y\(\overrightarrow {j}\) + 2z\(\overrightarrow {k}\)
(∇∅₁)_{(2, -1, 2)} = 4\(\overrightarrow {i}\) – 2\(\overrightarrow {j}\) – \(\overrightarrow {k}\)(∇∅₂)_{(2, -1, 2)} = 4\(\overrightarrow {i}\) – 2\(\overrightarrow {j}\) + 4\(\overrightarrow {k}\)
|∇∅₁| = \(\sqrt {16 + 4 + 1}\) = \(\sqrt {21}\) |∇∅₂| = \(\sqrt {16 + 4 + 16}\) = 6
cos⁡θ = \(\frac {∇∅_1.∇∅_2}{|∇∅_1 | |∇∅_2 |} = \frac {(4\overrightarrow {i} – 2\overrightarrow {j} – \overrightarrow {k})( 4\overrightarrow {i} – 2\overrightarrow {j} + 4\overrightarrow {k})}{\sqrt {21} \sqrt {36}} = \frac {8}{3\sqrt {21}}\)
θ = cos^-1\(\bigg [ \frac {8}{3\sqrt {21}} \bigg ] \)

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