← 2023 Paper 1
UPSC 2023 Maths Optional Paper 1 Q5e — Step-by-Step Solution 10 marks · Section B
Differentiation of a vector function of a scalar variable · Vector Analysis · asked 4× in 13 yrs · Read the full method →
Question
If a ⃗ = sin θ i ^ + cos θ j ^ + θ k ^ \vec a=\sin\theta\,\hat i+\cos\theta\,\hat j+\theta\,\hat k a = sin θ i ^ + cos θ j ^ + θ k ^ b ⃗ = cos θ i ^ − sin θ j ^ − 3 k ^ \vec b=\cos\theta\,\hat i-\sin\theta\,\hat j-3\hat k b = cos θ i ^ − sin θ j ^ − 3 k ^ c ⃗ = 2 i ^ + 3 j ^ − 3 k ^ \vec c=2\hat i+3\hat j-3\hat k c = 2 i ^ + 3 j ^ − 3 k ^ a ⃗ × ( b ⃗ × c ⃗ ) \vec a\times(\vec b\times\vec c) a × ( b × c ) θ \theta θ θ = π 2 \theta=\dfrac{\pi}{2} θ = 2 π θ = π \theta=\pi θ = π
Technique
Compute b ⃗ × c ⃗ \vec b\times\vec c b × c a ⃗ × ( b ⃗ × c ⃗ ) \vec a\times(\vec b\times\vec c) a × ( b × c ) θ \theta θ
Solution
Step 1 — Compute b ⃗ × c ⃗ \vec b\times\vec c b × c
b ⃗ × c ⃗ = det ∣ i ^ j ^ k ^ cos θ − sin θ − 3 2 3 − 3 ∣ . \vec b\times\vec c=\det\begin{vmatrix}\hat i & \hat j & \hat k\\ \cos\theta & -\sin\theta & -3\\ 2 & 3 & -3\end{vmatrix}. b × c = det i ^ cos θ 2 j ^ − sin θ 3 k ^ − 3 − 3 .
i ^ \hat i i ^ ( − sin θ ) ( − 3 ) − ( − 3 ) ( 3 ) = 3 sin θ + 9. (-\sin\theta)(-3)-(-3)(3)=3\sin\theta+9. ( − sin θ ) ( − 3 ) − ( − 3 ) ( 3 ) = 3 sin θ + 9. j ^ \hat j j ^ − [ ( cos θ ) ( − 3 ) − ( − 3 ) ( 2 ) ] = − [ − 3 cos θ + 6 ] = 3 cos θ − 6. -[(\cos\theta)(-3)-(-3)(2)]=-[-3\cos\theta+6]=3\cos\theta-6. − [( cos θ ) ( − 3 ) − ( − 3 ) ( 2 )] = − [ − 3 cos θ + 6 ] = 3 cos θ − 6. k ^ \hat k k ^ ( cos θ ) ( 3 ) − ( − sin θ ) ( 2 ) = 3 cos θ + 2 sin θ . (\cos\theta)(3)-(-\sin\theta)(2)=3\cos\theta+2\sin\theta. ( cos θ ) ( 3 ) − ( − sin θ ) ( 2 ) = 3 cos θ + 2 sin θ .
b ⃗ × c ⃗ = ( 3 sin θ + 9 , 3 cos θ − 6 , 3 cos θ + 2 sin θ ) . \vec b\times\vec c=(3\sin\theta+9,\,3\cos\theta-6,\,3\cos\theta+2\sin\theta). b × c = ( 3 sin θ + 9 , 3 cos θ − 6 , 3 cos θ + 2 sin θ ) .
Step 2 — Compute a ⃗ × ( b ⃗ × c ⃗ ) \vec a\times(\vec b\times\vec c) a × ( b × c )
With a ⃗ = ( sin θ , cos θ , θ ) \vec a=(\sin\theta,\cos\theta,\theta) a = ( sin θ , cos θ , θ ) d ⃗ = ( 3 sin θ + 9 , 3 cos θ − 6 , 3 cos θ + 2 sin θ ) \vec d=(3\sin\theta+9,\,3\cos\theta-6,\,3\cos\theta+2\sin\theta) d = ( 3 sin θ + 9 , 3 cos θ − 6 , 3 cos θ + 2 sin θ )
i ^ \hat i i ^ cos θ ⋅ ( 3 cos θ + 2 sin θ ) − θ ⋅ ( 3 cos θ − 6 ) = 3 cos 2 θ + 2 sin θ cos θ − 3 θ cos θ + 6 θ . \cos\theta\cdot(3\cos\theta+2\sin\theta)-\theta\cdot(3\cos\theta-6)=3\cos^2\theta+2\sin\theta\cos\theta-3\theta\cos\theta+6\theta. cos θ ⋅ ( 3 cos θ + 2 sin θ ) − θ ⋅ ( 3 cos θ − 6 ) = 3 cos 2 θ + 2 sin θ cos θ − 3 θ cos θ + 6 θ . j ^ \hat j j ^ − [ sin θ ( 3 cos θ + 2 sin θ ) − θ ( 3 sin θ + 9 ) ] = − 3 sin θ cos θ − 2 sin 2 θ + 3 θ sin θ + 9 θ . -[\sin\theta(3\cos\theta+2\sin\theta)-\theta(3\sin\theta+9)]=-3\sin\theta\cos\theta-2\sin^2\theta+3\theta\sin\theta+9\theta. − [ sin θ ( 3 cos θ + 2 sin θ ) − θ ( 3 sin θ + 9 )] = − 3 sin θ cos θ − 2 sin 2 θ + 3 θ sin θ + 9 θ . k ^ \hat k k ^ sin θ ( 3 cos θ − 6 ) − cos θ ( 3 sin θ + 9 ) = 3 sin θ cos θ − 6 sin θ − 3 sin θ cos θ − 9 cos θ = − 6 sin θ − 9 cos θ . \sin\theta(3\cos\theta-6)-\cos\theta(3\sin\theta+9)=3\sin\theta\cos\theta-6\sin\theta-3\sin\theta\cos\theta-9\cos\theta=-6\sin\theta-9\cos\theta. sin θ ( 3 cos θ − 6 ) − cos θ ( 3 sin θ + 9 ) = 3 sin θ cos θ − 6 sin θ − 3 sin θ cos θ − 9 cos θ = − 6 sin θ − 9 cos θ .
So F ⃗ ( θ ) = a ⃗ × ( b ⃗ × c ⃗ ) \vec F(\theta)=\vec a\times(\vec b\times\vec c) F ( θ ) = a × ( b × c )
F 1 = 3 cos 2 θ + 2 sin θ cos θ − 3 θ cos θ + 6 θ = 3 2 ( 1 + cos 2 θ ) + sin 2 θ − 3 θ cos θ + 6 θ . F_1=3\cos^2\theta+2\sin\theta\cos\theta-3\theta\cos\theta+6\theta=\tfrac{3}{2}(1+\cos 2\theta)+\sin 2\theta-3\theta\cos\theta+6\theta. F 1 = 3 cos 2 θ + 2 sin θ cos θ − 3 θ cos θ + 6 θ = 2 3 ( 1 + cos 2 θ ) + sin 2 θ − 3 θ cos θ + 6 θ . F 2 = − 3 sin θ cos θ − 2 sin 2 θ + 3 θ sin θ + 9 θ = − 3 2 sin 2 θ − ( 1 − cos 2 θ ) + 3 θ sin θ + 9 θ . F_2=-3\sin\theta\cos\theta-2\sin^2\theta+3\theta\sin\theta+9\theta=-\tfrac{3}{2}\sin 2\theta-(1-\cos 2\theta)+3\theta\sin\theta+9\theta. F 2 = − 3 sin θ cos θ − 2 sin 2 θ + 3 θ sin θ + 9 θ = − 2 3 sin 2 θ − ( 1 − cos 2 θ ) + 3 θ sin θ + 9 θ . F 3 = − 6 sin θ − 9 cos θ . F_3=-6\sin\theta-9\cos\theta. F 3 = − 6 sin θ − 9 cos θ .
Step 3 — Differentiate w.r.t. θ \theta θ
d F 1 d θ = − 3 sin 2 θ + 2 cos 2 θ − 3 cos θ + 3 θ sin θ + 6. \dfrac{dF_1}{d\theta}=-3\sin 2\theta+2\cos 2\theta-3\cos\theta+3\theta\sin\theta+6. d θ d F 1 = − 3 sin 2 θ + 2 cos 2 θ − 3 cos θ + 3 θ sin θ + 6. d F 2 d θ = − 3 cos 2 θ − 2 sin 2 θ + 3 sin θ + 3 θ cos θ + 9. \dfrac{dF_2}{d\theta}=-3\cos 2\theta-2\sin 2\theta+3\sin\theta+3\theta\cos\theta+9. d θ d F 2 = − 3 cos 2 θ − 2 sin 2 θ + 3 sin θ + 3 θ cos θ + 9. d F 3 d θ = − 6 cos θ + 9 sin θ . \dfrac{dF_3}{d\theta}=-6\cos\theta+9\sin\theta. d θ d F 3 = − 6 cos θ + 9 sin θ .
(Where I used d ( 3 θ cos θ ) / d θ = 3 cos θ − 3 θ sin θ d(3\theta\cos\theta)/d\theta=3\cos\theta-3\theta\sin\theta d ( 3 θ cos θ ) / d θ = 3 cos θ − 3 θ sin θ
Wait let me re-examine F 1 F_1 F 1 − 3 θ cos θ + 6 θ -3\theta\cos\theta+6\theta − 3 θ cos θ + 6 θ − 3 cos θ + 3 θ sin θ + 6 -3\cos\theta+3\theta\sin\theta+6 − 3 cos θ + 3 θ sin θ + 6
Hmm, my F 1 F_1 F 1 3 2 ( 1 + cos 2 θ ) \tfrac{3}{2}(1+\cos 2\theta) 2 3 ( 1 + cos 2 θ ) 3 cos 2 θ 3\cos^2\theta 3 cos 2 θ − 3 sin 2 θ -3\sin 2\theta − 3 sin 2 θ sin 2 θ \sin 2\theta sin 2 θ 2 cos 2 θ 2\cos 2\theta 2 cos 2 θ
Step 4 — Evaluate at θ = π / 2 \theta=\pi/2 θ = π /2
sin ( π / 2 ) = 1 , cos ( π / 2 ) = 0 , sin π = 0 , cos π = − 1. \sin(\pi/2)=1,\,\cos(\pi/2)=0,\,\sin\pi=0,\,\cos\pi=-1. sin ( π /2 ) = 1 , cos ( π /2 ) = 0 , sin π = 0 , cos π = − 1.
d F 1 / d θ = − 3 ( 0 ) + 2 ( − 1 ) − 3 ( 0 ) + 3 ( π / 2 ) ( 1 ) + 6 = − 2 + 3 π / 2 + 6 = 4 + 3 π / 2. dF_1/d\theta=-3(0)+2(-1)-3(0)+3(\pi/2)(1)+6=-2+3\pi/2+6=4+3\pi/2. d F 1 / d θ = − 3 ( 0 ) + 2 ( − 1 ) − 3 ( 0 ) + 3 ( π /2 ) ( 1 ) + 6 = − 2 + 3 π /2 + 6 = 4 + 3 π /2. d F 2 / d θ = − 3 ( − 1 ) − 2 ( 0 ) + 3 ( 1 ) + 3 ( π / 2 ) ( 0 ) + 9 = 3 + 3 + 9 = 15. dF_2/d\theta=-3(-1)-2(0)+3(1)+3(\pi/2)(0)+9=3+3+9=15. d F 2 / d θ = − 3 ( − 1 ) − 2 ( 0 ) + 3 ( 1 ) + 3 ( π /2 ) ( 0 ) + 9 = 3 + 3 + 9 = 15. d F 3 / d θ = − 6 ( 0 ) + 9 ( 1 ) = 9. dF_3/d\theta=-6(0)+9(1)=9. d F 3 / d θ = − 6 ( 0 ) + 9 ( 1 ) = 9.
d F ⃗ d θ ∣ θ = π / 2 = ( 4 + 3 π 2 ) i ^ + 15 j ^ + 9 k ^ . \boxed{\;\frac{d\vec F}{d\theta}\bigg|_{\theta=\pi/2}=(4+\tfrac{3\pi}{2})\hat i+15\hat j+9\hat k.\;} d θ d F θ = π /2 = ( 4 + 2 3 π ) i ^ + 15 j ^ + 9 k ^ . Step 5 — Evaluate at θ = π \theta=\pi θ = π
sin π = 0 , cos π = − 1 , sin 2 π = 0 , cos 2 π = 1. \sin\pi=0,\,\cos\pi=-1,\,\sin 2\pi=0,\,\cos 2\pi=1. sin π = 0 , cos π = − 1 , sin 2 π = 0 , cos 2 π = 1.
d F 1 / d θ = − 3 ( 0 ) + 2 ( 1 ) − 3 ( − 1 ) + 3 π ( 0 ) + 6 = 2 + 3 + 6 = 11. dF_1/d\theta=-3(0)+2(1)-3(-1)+3\pi(0)+6=2+3+6=11. d F 1 / d θ = − 3 ( 0 ) + 2 ( 1 ) − 3 ( − 1 ) + 3 π ( 0 ) + 6 = 2 + 3 + 6 = 11. d F 2 / d θ = − 3 ( 1 ) − 2 ( 0 ) + 3 ( 0 ) + 3 π ( − 1 ) + 9 = − 3 − 3 π + 9 = 6 − 3 π . dF_2/d\theta=-3(1)-2(0)+3(0)+3\pi(-1)+9=-3-3\pi+9=6-3\pi. d F 2 / d θ = − 3 ( 1 ) − 2 ( 0 ) + 3 ( 0 ) + 3 π ( − 1 ) + 9 = − 3 − 3 π + 9 = 6 − 3 π . d F 3 / d θ = − 6 ( − 1 ) + 9 ( 0 ) = 6. dF_3/d\theta=-6(-1)+9(0)=6. d F 3 / d θ = − 6 ( − 1 ) + 9 ( 0 ) = 6.
Answer
d F ⃗ d θ ∣ θ = π = 11 i ^ + ( 6 − 3 π ) j ^ + 6 k ^ . \boxed{\;\frac{d\vec F}{d\theta}\bigg|_{\theta=\pi}=11\hat i+(6-3\pi)\hat j+6\hat k.\;} d θ d F θ = π = 11 i ^ + ( 6 − 3 π ) j ^ + 6 k ^ .