Linear Algebra Questions and Answers – Directional Derivative

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This set of Linear Algebra Objective Questions & Answers focuses on “Directional Derivative”.

1. Find the directional derivative of φ = xy2 + yz3 at (1, -1, 1), towards the point (2, 1, -1).
a) \( \frac{5}{3} \)
b) \( \frac{-5}{3} \)
c) \( \frac{7}{3} \)
d) \( \frac{1}{3} \)
View Answer

Answer: a
Explanation: \(\frac{dφ}{dx} = y^2, \frac{dφ}{dy} =2xy+z^3, \frac{dφ}{dz} = 3yz^3 \)
\(∇ φ = y^2 \hat{i} + (2xy+z^3) \hat{j} + 3yz^3\hat{k} ̂\)
\([∇ φ]_{(1,-1,1)} = \hat{i} – \hat{j} – 3\hat{k} ̂\)
Now \(a ̅ \) is along the line joining(1,-1,1) and (2,1,-1)
\(a ̅ =(2-1) \hat{i} + (1+1) \hat{j} + (-1-1) \hat{k} = \hat{i} + 2\hat{j} – 2\hat{k} \)
\(a ̂ = \frac{(i ̂ +2 j ̂ -2\hat{k})}{\sqrt{(1+4+4)}} = \frac{1}{3} (i ̂ +2 j ̂ -2\hat{k})\)
∴ Directional derivative = \(∇ φ. a ̂ = (i ̂ – j ̂ -3\hat{k}) . \frac{1}{3} (i ̂ +2 j ̂ -2\hat{k})\)
\(= \frac{1}{3} (1-2+6) = \frac{5}{3} .\)
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2. The directional derivative of φ(x,y) at the point A(3,2) towards the point B(2,3). What is \(3\sqrt{2}\) and toward the point (1,0) is \(\sqrt{8}\). What is the directional derivative at the point A towards the point D.
a) \(\frac{6}{5}\)
b) \(\frac{7}{\sqrt{5}} \)
c) \(\frac{6}{\sqrt{5}} \)
d) \(\frac{7}{5} \)
View Answer

Answer: b
Explanation: Here \(\overrightarrow{AB} = (2-3) i ̂ + (3-2) j ̂ = – i ̂ + j ̂ \)
Directional derivative of φ(x,y) toward \(\overrightarrow{AB} \) is
\(∇ φ . \widehat{(AB)} = (\hat{i} \frac{∂φ}{∂x}+ \hat{j}\frac{∂φ}{∂y}) .(\frac{- i ̂ + j ̂ }{\sqrt{2}}) = 3\sqrt{2}\)
\(– \frac{∂φ}{∂x} + \frac{∂φ}{∂y} = 6 …..(i) \)
Directional derivative at A(3,2) towards C(1,0) is
\(∇ φ . \widehat{(AC)} = (i ̂ \frac{∂φ}{∂x}+ j ̂ \frac{∂φ}{∂y}) . \frac{-2 i ̂-2j ̂ }{\sqrt{8}} = \sqrt{8} \)
\(– 2 \frac{∂φ}{∂x} -2 \frac{∂φ}{∂y} = 8 \,or\, \frac{∂φ}{∂x} + \frac{∂φ}{∂y} = 4 …..(ii) \)
From (i) & (ii), \(∇ φ = -5\hat{i} + \hat{j} \)
Hence directional derivative at A(3,2) towards D(2,4) is
\(∇ φ . \widehat{AD} = (-5\hat{i} + \hat{j}) . \left(\frac{-5\hat{i} + \hat{j}}{\sqrt{5}}\right) = \frac{7}{\sqrt{5}}.\)

3. For the function f = x2y + 2y2x, at the point P(1,3), what is the direction in which the directional derivative is zero?
a) \(-13\hat{i} – 24\hat{j} \)
b) \(13\hat{i} + 24 \hat{j} \)
c) \(±13\hat{i} ∓24\hat{j} \)
d) \(∓13\hat{i} ± 24\hat{j} \)
View Answer

Answer: c
Explanation: Let the directional derivative is zero along \(a1\hat{i} + a2\hat{j}\)
\(Then ∇ f = \hat{i} \frac{∂f}{∂x} +\hat{j} \frac{∂f}{∂y} = \hat{i} (2xy+2y^2) + \hat{j}(x^2+4xy) \)
\(i.e.[∇ f]_{(1,3)} = 24\hat{i} + 13\hat{j} \)
\([∇ f]_{(1,3)} . (a_1\hat{i} +a_2\hat{j}) = 0 \)
\((a_1\hat{i} +a_2\hat{j}) . (a_1\hat{i} + a_2\hat{j}) = 0 \)
\(24a_1 + 13a_2 = 0 ∴ \frac{a1}{13}=\frac{(-a2)}{24} \)
∴ Direction are \(\overline{u1} = 13\hat{i} – 24\hat{j}, \overline{u2} = -13\hat{i} + 24\hat{j}. \)

4. The unit normal vector n ̂ of the cone of revolution z2 = 4(x2 + y2) at the Point P (1, 0, 2) is?
a) \([±\frac{2}{\sqrt{5}}, 0, ∓\frac{1}{\sqrt{5}}] \)
b) \([\frac{2}{\sqrt{5}}, 0, \frac{1}{\sqrt{5}}] \)
c) \([-\frac{2}{\sqrt{5}}, 0, -\frac{1}{\sqrt{5}}] \)
d) \([∓\frac{2}{\sqrt{5}}, 0, ±\frac{1}{\sqrt{5}}] \)
View Answer

Answer: a
Explanation: A cone is a level surface say f = 0 of f (x, y, z) = 4(x2+y2) – z2
\(∴ ∇ f = 8x\hat{i} + 8y\hat{j} – 2z\hat{k} \)
or \( ∇ f_{(1,0,2)} = 8\hat{i} – 4\hat{k} \)
\(\hat{n} = \frac{(8\hat{i} – 4\hat{k})}{\sqrt{(8)2+(-4)2}} \)
\( = \frac{(8\hat{i} – 4\hat{k})}{\sqrt{80}} \)
\( = \frac{(8\hat{i} – 4\hat{k})}{\sqrt{(16*5)}} = \frac{(2\hat{i} – \hat{k})}{\sqrt{5}} \, or \, \hat{n} = [\frac{2}{\sqrt{5}}, 0, \frac{-1}{\sqrt{5}}] \)
There are two possible unit vectors, if one is along direction then \(\hat{n} \) other will be along – \(\hat{n} \) \(∴ \hat{n} = [±\frac{2}{\sqrt{5}}, 0, ∓\frac{1}{\sqrt{5}}]. \)
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5. For what value of a & b for which the two surfaces ax2 – byz = (a+2)x & 4x2y + z3 = 4 will be orthogonal to each other at the point (1,-1,2).
a) \(a = 1 \, \& \, b = \frac{3}{2} \)
b) \(a = \frac{-3}{2} \, \& \, b = 1 \)
c) \(a = \frac{3}{2} \, \& \, b = 1 \)
d) \(a = \frac{-3}{2} \, \& \, b = -1 \)
View Answer

Answer: c
Explanation: Let \(f_1 = ax^2 – byz – (a+2)x \, \& \, f_2= 4x^2y+z^3 – 4 \)
\(∴ ∇ f_1 = [2ax – (a+2)]\hat{i} – (-bz)\hat{j} + (-bz)\hat{k} \)
also \( ∇ f_1|_{(1,-1,2)} = (a-2) \hat{i} + b\hat{j} + b\hat{k} \)
And \( ∇ f_2 = (8xy) \hat{i} + (4x^2)\hat{j} + (3z^2)\hat{k}, also \, ∇ f_2|_{(1,-1,2)} = -8\hat{i} + 4\hat{j} + 12\hat{k} \)
Now, if the two surfaces cut each other orthogonally at point P(1,-1,2)
then,
\(∇ f_1 . ∇ f_2 = 0 \)
Or \( [(a-2)\hat{i} + b\hat{j} + b\hat{k}] . (-8\hat{i} + 4\hat{j} + 12\hat{k}) = 0
= -2a + b = -4 …..(i) \)
Also if point P lies on both the curve hence it would also satisfy
Both the curves, i.e substituting P(1,-1,2) in equation
\(= ax^2-byz = (a+2) \)
\(= a+2b = a+2 ⇨ b=1 \)
From eqn (i) \(-2a+1 = -4 => 2a = 3 => a = \frac{3}{2}. \)

6. The temperature of a point in space is given by T = x2 + y2 – z. An insect located at a point (1, 1, 2) desire to fly in such a direction such that it will get warm as soon as possible. In what direction it should move?
a) \(\frac{-2\hat{i}}{3}+\frac{2\hat{j}}{3}+\frac{-\hat{k}}{3} \)
b) \(\frac{2\hat{i}}{3}+\frac{-2\hat{j}}{3}+\frac{-\hat{k}}{3} \)
c) \(\frac{2\hat{i}}{3}+\frac{2\hat{j}}{3}+\frac{\hat{k}}{3} \)
d) \(\frac{2\hat{i}}{3}+\frac{2\hat{j}}{3}+\frac{-\hat{k}}{3} \)
View Answer

Answer: d
Explanation: Here the insect will get warm quickly if it moves normal to the given
Surface in space.
As \(T = x^2+y^2-z \)
\(∴ ∇ T = (\frac{∂}{∂x}\hat{i} + \frac{∂}{∂x}\hat{j} + \frac{∂}{∂x}\hat{k})(x^2+y^2-z) \)
\( = 2x\hat{i} + 2y\hat{j} – \hat{k} \)
Now \( ∇ T |_{(1,1,2)} = 2\hat{i} + 2\hat{j} – \hat{k} \)
∴ required direction\( = \frac{∇ T}{|∇ T|} \) \(=\frac{2\hat{i} + 2\hat{j} – \hat{k}}{\sqrt{(4+4+1)}} \)
\(= \frac{2\hat{i} +2\hat{j} – \hat{k}}{3} \)
\(= \frac{2\hat{i}}{3}+\frac{2\hat{j}}{3}+\frac{-\hat{k}}{3}. \)

Sanfoundry Global Education & Learning Series – Linear Algebra.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn