QCE Vault / Specialist Maths Vector calculus — Question 194 Original QCE Vault practice · 5 marks
Browse all questions Vectors revision Original practice exam Report an issue Q194 · Practice question Technology-free Complex familiar 5 marks
QUESTION 194 (5 marks) For 0 ≤ t ≤ 3 0\le t\le3 0 ≤ t ≤ 3 , a particle has position r ( t ) = ( t + 1 ) i + ( t + 1 ) − 1 j \mathbf r(t)=(t+1)\mathbf i+(t+1)^{-1}\mathbf j r ( t ) = ( t + 1 ) i + ( t + 1 ) − 1 j metres. a) Determine its Cartesian path and the segment traced.
[2 marks] b) Determine minimum speed and whether acceleration is ever perpendicular to velocity.
[3 marks] WORKED SOLUTION
Practice marking scheme 5 marks ANSWER (a) x y = 1 xy=1 x y = 1 , 1 ≤ x ≤ 4 1\le x\le4 1 ≤ x ≤ 4 . (b) 257 / 16 \sqrt{257}/16 257 /16 m/s at t=3; never perpendicular. Worked solution
(a) Since x=t+1 and y=1/(t+1), eliminating t gives y=1/x with x ranging from 1 to 4.
(b) Velocity is ( 1 , − ( t + 1 ) − 2 ) (1,-(t+1)^{-2}) ( 1 , − ( t + 1 ) − 2 ) and acceleration ( 0 , 2 ( t + 1 ) − 3 ) (0,2(t+1)^{-3}) ( 0 , 2 ( t + 1 ) − 3 ) . Speed squared 1 + ( t + 1 ) − 4 1+(t+1)^{-4} 1 + ( t + 1 ) − 4 decreases, so its minimum is 1 + 1 / 256 = 257 / 16 \sqrt{1+1/256}=\sqrt{257}/16 1 + 1/256 = 257 /16 . Their dot product − 2 ( t + 1 ) − 5 -2(t+1)^{-5} − 2 ( t + 1 ) − 5 is always negative, never zero. Equivalent mathematically justified methods accepted.
Exact answers unless a decimal accuracy is specified.
Part a: Eliminate the parameter.
[1 mark] Part a: Give the traced domain.
[1 mark] Part b: Find velocity and acceleration.
[1 mark] Part b: Use decreasing speed to find the endpoint minimum.
[1 mark] Part b: Use the nonzero dot product to rule out perpendicularity.
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