Calculus with Vector Functions — Question 3

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Question 3

A curve has velocity r→′(t)=⟨2t,3t2,e−t⟩\vec r'(t)=\left\langle 2t,3t^2,e^{-t}\right\rangle and passes through P=(1,−2,4)P=(1,-2,4) at t=0t=0. Find r→(t)\vec r(t) and the net displacement from t=0t=0 to t=2t=2.

Original worksheet page 1: question and worked solution for 6-7-003
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Question 3 – Solution

Strategy Integrate each velocity component and use the initial condition to determine the constant vector.

See the diagram in the original worksheet below.

Antiderivative r→(t)=⟨t2,t3,−e−t⟩+C→\vec r(t)=\left\langle t^2,t^3,-e^{-t}\right\rangle+\vec C. Since r→(0)=⟨0,0,−1⟩+C→=P\vec r(0)=\left\langle 0,0,-1\right\rangle+\vec C=P, C→=⟨1,−2,5⟩\vec C=\left\langle 1,-2,5\right\rangle. Therefore r→(t)=⟨t2+1,t3−2,5−e−t⟩.\boxed{\vec r(t)=\left\langle t^2+1,t^3-2,5-e^{-t}\right\rangle}.

Displacement r→(2)−r→(0)=∫02r→′(t)dt=⟨4,8,1−e−2⟩\vec r(2)-\vec r(0)=\int_0^2\vec r'(t)dt=\boxed{\left\langle 4,8,1-e^{-2}\right\rangle}.

Verification Differentiation recovers the given velocity and substitution gives PP.

Original worksheet page 2: question and worked solution for 6-7-003

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