# Vector Calculus Questions and Answers – Vector Differentiation

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

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 ∇(c11 ± c22) = c1∇∅1 ± c2∇∅2 where c1 and c2 are constants.
a) True
b) False

Explanation:
∇(c11 ± c22) = ($$\overrightarrow {i} \frac {∂}{∂x} + \overrightarrow {j} \frac {∂}{∂y} + \overrightarrow {i} \frac {∂}{∂z}$$)(c11 ± c22)
= $$\overrightarrow {i} \frac {∂}{∂x}$$[c11 ± c22] + $$\overrightarrow {j} \frac {∂}{∂y}$$ [c11 ± c22] + $$\overrightarrow {k} \frac {∂}{∂z}$$[c11 ± c22]
= $$\overrightarrow {i} \bigg [$$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$$\bigg [ \overrightarrow {i} \frac {∂∅_1}{∂x} + \overrightarrow {j} \frac {∂∅_2}{∂y} + \overrightarrow {k} \frac {∂∅_3}{∂z} \bigg ]$$
∇(c11 ± c22) = 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) $$\overrightarrow {i} + \overrightarrow {j} + \overrightarrow {k}$$

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) $$\frac {1}{r^2} \overrightarrow {r}$$

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) $$\frac {\overrightarrow {r}}{r^3} ,\frac { – 1}{r^3}$$

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$$\overrightarrow {j}$$ + $$\overrightarrow {k}$$

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) nrn-2$$\overrightarrow {r}$$
b) rn-2$$\overrightarrow {r}$$
c) nrn-2
d) nrn-1$$\overrightarrow {r}$$

Explanation:
∇(rn) = ∑$$\overrightarrow {i} \frac {∂}{∂x}$$(rn) = ∑$$\overrightarrow {i}$$ nrn-1$$\frac {∂r}{∂x}$$
= ∑$$\overrightarrow {i}$$nrn-1$$\frac {x}{r}$$
= ∑$$\overrightarrow {i}$$nrn-2x
= nrn-2[x$$\overrightarrow {i}$$ + y$$\overrightarrow {j}$$ + z$$\overrightarrow {k}$$]
= nrn-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) $$\frac {\overrightarrow {r}}{r^5}$$

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 ∇(ex2 + y2 + z2)?
a) 2er2$$\overrightarrow {r}$$
b) er2$$\overrightarrow {r}$$
c) 2er4$$\overrightarrow {r}$$
d) 2$$\overrightarrow {r}$$

Explanation:
We know that $$\overrightarrow {r}$$ = x$$\overrightarrow {i}$$ + y$$\overrightarrow {j}$$ + z$$\overrightarrow {k}$$
r2 = x2 + y2 + z2
$$\frac {∂r}{∂x} = \frac {x}{r}, \frac {∂r}{∂y} = \frac {y}{r}, \frac {∂r}{∂z} = \frac {z}{r}$$
∇(ex2 + y2 + z2) = ∇(er2) = ∑$$\overrightarrow {i}$$ $$\frac {∂}{∂x}$$(er2)
= ∑$$\overrightarrow {i}$$er22r $$\frac {∂r}{∂x}$$ = ∑$$\overrightarrow {i}$$er22r $$\frac {x}{r}$$
= ∑$$\overrightarrow {i}$$er22x = 2er2[x$$\overrightarrow {i}$$ + y $$\overrightarrow {j}$$ + z$$\overrightarrow {k}$$]
= 2er2$$\overrightarrow {r}$$

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) $$\frac {7}{3}$$

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]

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) $$\frac {35}{\sqrt {14}}$$

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) $$\sqrt {176543}$$

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) $$\frac {\overrightarrow {i} + \overrightarrow {j} – 2\overrightarrow {k}}{\sqrt 6}$$

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⁡[ $$\frac {2}{3\sqrt {21}}$$ ]

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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