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} )
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^3hat{k} ̂)
([∇ φ]_{(1,-1,1)} = hat{i} – hat{j} – 3hat{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} + 2hat{j} – 2hat{k} )
(a ̂ = frac{(i ̂ +2 j ̂ -2hat{k})}{sqrt{(1+4+4)}} = frac{1}{3} (i ̂ +2 j ̂ -2hat{k}))
∴ Directional derivative = (∇ φ. a ̂ = (i ̂ – j ̂ -3hat{k}) . frac{1}{3} (i ̂ +2 j ̂ -2hat{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 (3sqrt{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} )
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}}) = 3sqrt{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 = -5hat{i} + hat{j} )
Hence directional derivative at A(3,2) towards D(2,4) is
(∇ φ . widehat{AD} = (-5hat{i} + hat{j}) . left(frac{-5hat{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) (-13hat{i} – 24hat{j} )
b) (13hat{i} + 24 hat{j} )
c) (±13hat{i} ∓24hat{j} )
d) (∓13hat{i} ± 24hat{j} )
Answer: c
Explanation: Let the directional derivative is zero along (a1hat{i} + a2hat{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)} = 24hat{i} + 13hat{j} )
([∇ f]_{(1,3)} . (a_1hat{i} +a_2hat{j}) = 0 )
((a_1hat{i} +a_2hat{j}) . (a_1hat{i} + a_2hat{j}) = 0 )
(24a_1 + 13a_2 = 0 ∴ frac{a1}{13}=frac{(-a2)}{24} )
∴ Direction are (overline{u1} = 13hat{i} – 24hat{j}, overline{u2} = -13hat{i} + 24hat{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}}] )
Answer: a
Explanation: A cone is a level surface say f = 0 of f (x, y, z) = 4(x2+y2) – z2
(∴ ∇ f = 8xhat{i} + 8yhat{j} – 2zhat{k} )
or ( ∇ f_{(1,0,2)} = 8hat{i} – 4hat{k} )
(hat{n} = frac{(8hat{i} – 4hat{k})}{sqrt{(8)2+(-4)2}} )
( = frac{(8hat{i} – 4hat{k})}{sqrt{80}} )
( = frac{(8hat{i} – 4hat{k})}{sqrt{(16*5)}} = frac{(2hat{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 )
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} + bhat{j} + bhat{k} )
And ( ∇ f_2 = (8xy) hat{i} + (4x^2)hat{j} + (3z^2)hat{k}, also , ∇ f_2|_{(1,-1,2)} = -8hat{i} + 4hat{j} + 12hat{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} + bhat{j} + bhat{k}] . (-8hat{i} + 4hat{j} + 12hat{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{-2hat{i}}{3}+frac{2hat{j}}{3}+frac{-hat{k}}{3} )
b) (frac{2hat{i}}{3}+frac{-2hat{j}}{3}+frac{-hat{k}}{3} )
c) (frac{2hat{i}}{3}+frac{2hat{j}}{3}+frac{hat{k}}{3} )
d) (frac{2hat{i}}{3}+frac{2hat{j}}{3}+frac{-hat{k}}{3} )
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) )
( = 2xhat{i} + 2yhat{j} – hat{k} )
Now ( ∇ T |_{(1,1,2)} = 2hat{i} + 2hat{j} – hat{k} )
∴ required direction( = frac{∇ T}{|∇ T|} )
(=frac{2hat{i} + 2hat{j} – hat{k}}{sqrt{(4+4+1)}} )
(= frac{2hat{i} +2hat{j} – hat{k}}{3} )
(= frac{2hat{i}}{3}+frac{2hat{j}}{3}+frac{-hat{k}}{3}. )
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