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
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 = \(∇ φ. a ̂ = (i ̂ – j ̂ -3\hat{k}) . \frac{1}{3} (i ̂ +2 j ̂ -2\hat{k})\)
\(= \frac{1}{3} (1-2+6) = \frac{5}{3} .\)
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
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
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
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 – \(\hat{n} \) \(∴ \hat{n} = [±\frac{2}{\sqrt{5}}, 0, ∓\frac{1}{\sqrt{5}}]. \)
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
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
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\( = \frac{∇ T}{|∇ T|} \) \(=\frac{2\hat{i} + 2\hat{j} – \hat{k}}{\sqrt{(4+4+1)}} \)
\(= \frac{2\hat{i} +2\hat{j} – \hat{k}}{3} \)
\(= \frac{2\hat{i}}{3}+\frac{2\hat{j}}{3}+\frac{-\hat{k}}{3}. \)
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