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UPSC 2017 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 5× in 14 yrs · Read the full method →
Question
The position vector of a moving point at time t t t r ⃗ = sin t i ^ + cos 2 t j ^ + ( t 2 + 2 t ) k ^ \vec r=\sin t\,\hat i+\cos 2t\,\hat j+(t^2+2t)\hat k r = sin t i ^ + cos 2 t j ^ + ( t 2 + 2 t ) k ^ a ⃗ \vec a a v ⃗ \vec v v r ⃗ \vec r r v ⃗ \vec v v t = 0 t=0 t = 0
Technique
Scalar projection along v ⃗ \vec v v a ⃗ ⋅ v ^ \vec a\cdot\hat v a ⋅ v ^ r ⃗ × v ⃗ \vec r\times\vec v r × v a ⃗ ⋅ ( r ⃗ × v ⃗ ) ^ \vec a\cdot\widehat{(\vec r\times\vec v)} a ⋅ ( r × v )
Solution
Step 1 — Velocity and acceleration; evaluate at t = 0 t=0 t = 0
r ⃗ = ( sin t , cos 2 t , t 2 + 2 t ) , \vec r=(\sin t,\ \cos 2t,\ t^2+2t), r = ( sin t , cos 2 t , t 2 + 2 t ) , v ⃗ = r ⃗ ˙ = ( cos t , − 2 sin 2 t , 2 t + 2 ) , a ⃗ = r ⃗ ¨ = ( − sin t , − 4 cos 2 t , 2 ) . \vec v=\dot{\vec r}=(\cos t,\ -2\sin 2t,\ 2t+2),\qquad \vec a=\ddot{\vec r}=(-\sin t,\ -4\cos 2t,\ 2). v = r ˙ = ( cos t , − 2 sin 2 t , 2 t + 2 ) , a = r ¨ = ( − sin t , − 4 cos 2 t , 2 ) . At t = 0 t=0 t = 0
r ⃗ 0 = ( 0 , 1 , 0 ) , v ⃗ 0 = ( 1 , 0 , 2 ) , a ⃗ 0 = ( 0 , − 4 , 2 ) . \vec r_0=(0,1,0),\qquad \vec v_0=(1,0,2),\qquad \vec a_0=(0,-4,2). r 0 = ( 0 , 1 , 0 ) , v 0 = ( 1 , 0 , 2 ) , a 0 = ( 0 , − 4 , 2 ) . Step 2 — Component of a ⃗ \vec a a v ⃗ \vec v v
This is the scalar projection a ⃗ ⋅ v ⃗ ∣ v ⃗ ∣ \dfrac{\vec a\cdot\vec v}{|\vec v|} ∣ v ∣ a ⋅ v
a ⃗ 0 ⋅ v ⃗ 0 = ( 0 ) ( 1 ) + ( − 4 ) ( 0 ) + ( 2 ) ( 2 ) = 4 , ∣ v ⃗ 0 ∣ = 1 2 + 0 2 + 2 2 = 5 . \vec a_0\cdot\vec v_0=(0)(1)+(-4)(0)+(2)(2)=4,\qquad |\vec v_0|=\sqrt{1^2+0^2+2^2}=\sqrt5. a 0 ⋅ v 0 = ( 0 ) ( 1 ) + ( − 4 ) ( 0 ) + ( 2 ) ( 2 ) = 4 , ∣ v 0 ∣ = 1 2 + 0 2 + 2 2 = 5 . component ∥ v ⃗ = 4 5 = 4 5 5 ≈ 1.789. \text{component}_{\parallel\vec v}=\frac{4}{\sqrt5}=\frac{4\sqrt5}{5}\approx1.789. component ∥ v = 5 4 = 5 4 5 ≈ 1.789. Step 3 — Component of a ⃗ \vec a a r ⃗ \vec r r v ⃗ \vec v v
The normal to that plane is r ⃗ × v ⃗ \vec r\times\vec v r × v a ⃗ ⋅ ( r ⃗ × v ⃗ ) ∣ r ⃗ × v ⃗ ∣ \dfrac{\vec a\cdot(\vec r\times\vec v)}{|\vec r\times\vec v|} ∣ r × v ∣ a ⋅ ( r × v )
r ⃗ 0 × v ⃗ 0 = ∣ i ^ j ^ k ^ 0 1 0 1 0 2 ∣ = ( 1 ⋅ 2 − 0 ⋅ 0 , 0 ⋅ 1 − 0 ⋅ 2 , 0 ⋅ 0 − 1 ⋅ 1 ) = ( 2 , 0 , − 1 ) . \vec r_0\times\vec v_0=\begin{vmatrix}\hat i&\hat j&\hat k\\ 0&1&0\\ 1&0&2\end{vmatrix}=(1\cdot2-0\cdot0,\ 0\cdot1-0\cdot2,\ 0\cdot0-1\cdot1)=(2,0,-1). r 0 × v 0 = i ^ 0 1 j ^ 1 0 k ^ 0 2 = ( 1 ⋅ 2 − 0 ⋅ 0 , 0 ⋅ 1 − 0 ⋅ 2 , 0 ⋅ 0 − 1 ⋅ 1 ) = ( 2 , 0 , − 1 ) . ∣ r ⃗ 0 × v ⃗ 0 ∣ = 2 2 + 0 2 + ( − 1 ) 2 = 5 , |\vec r_0\times\vec v_0|=\sqrt{2^2+0^2+(-1)^2}=\sqrt5, ∣ r 0 × v 0 ∣ = 2 2 + 0 2 + ( − 1 ) 2 = 5 , a ⃗ 0 ⋅ ( r ⃗ 0 × v ⃗ 0 ) = ( 0 ) ( 2 ) + ( − 4 ) ( 0 ) + ( 2 ) ( − 1 ) = − 2. \vec a_0\cdot(\vec r_0\times\vec v_0)=(0)(2)+(-4)(0)+(2)(-1)=-2. a 0 ⋅ ( r 0 × v 0 ) = ( 0 ) ( 2 ) + ( − 4 ) ( 0 ) + ( 2 ) ( − 1 ) = − 2. component ⊥ plane = − 2 5 = − 2 5 5 ≈ − 0.894. \text{component}_{\perp\text{plane}}=\frac{-2}{\sqrt5}=-\frac{2\sqrt5}{5}\approx-0.894. component ⊥ plane = 5 − 2 = − 5 2 5 ≈ − 0.894. The magnitude of this component is 2 5 \dfrac{2}{\sqrt5} 5 2 r ⃗ × v ⃗ \vec r\times\vec v r × v
Result
Answer
along v ⃗ : 4 5 = 4 5 5 ; ⊥ to plane ( r ⃗ , v ⃗ ) : − 2 5 = − 2 5 5 . \boxed{\;\text{along }\vec v:\ \frac{4}{\sqrt5}=\frac{4\sqrt5}{5};\qquad \perp\text{ to plane}(\vec r,\vec v):\ -\frac{2}{\sqrt5}=-\frac{2\sqrt5}{5}.\;} along v : 5 4 = 5 4 5 ; ⊥ to plane ( r , v ) : − 5 2 = − 5 2 5 .