Velocity and Acceleration โ€” Question 5

Section 14-11 ยท Calculus pages 2849โ€“2850

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Question 5

For the helical motion ๐’“(t)=โŸจ3cos2t,3sin2t,4tโŸฉ,\mathbf r(t)=\left\langle 3\cos 2t,3\sin 2t,4t\right\rangle, determine the velocity, acceleration, speed, and angle between velocity and the positive zz-axis.

Tasks

  1. Compute ๐’—\mathbf v and ๐’‚\mathbf a.

  2. Show speed and the requested angle are constant.

  3. Prove acceleration points horizontally toward the helix axis.

Question 5 โ€“ Solution

Strategy. Differentiate, then use the dot-product angle formula with ๐’Œ=โŸจ0,0,1โŸฉ\mathbf k=\left\langle 0,0,1\right\rangle.

Step 1: Motion quantities ๐’—(t)=โŸจโˆ’6sin2t,6cos2t,4โŸฉ,\boxed{\mathbf v(t)=\left\langle -6\sin 2t,6\cos 2t,4\right\rangle}, and ๐’‚(t)=โŸจโˆ’12cos2t,โˆ’12sin2t,0โŸฉ.\boxed{\mathbf a(t)=\left\langle -12\cos 2t,-12\sin 2t,0\right\rangle}. The speed is โˆฅ๐’—โˆฅ=36(sin⁡22t+cos⁡22t)+16=213.\|\mathbf v\|=\sqrt{36(\sin^22t+\cos^22t)+16} =\boxed{2\sqrt{13}}.

Step 2: Direction angle If ฮธ\theta is the angle with the positive zz-axis, cos⁡ฮธ=๐’—โ‹…๐’Œโˆฅ๐’—โˆฅโˆฅ๐’Œโˆฅ=4213=213.\cos\theta=\frac{\mathbf v\cdot\mathbf k}{\|\mathbf v\|\|\mathbf k\|} =\frac 4{2\sqrt{13}}=\frac 2{\sqrt{13}}. Therefore ฮธ=cos⁡โˆ’1(213).\boxed{\theta=\cos^{-1}\left(\frac 2{\sqrt{13}}\right)}. No tt occurs, so both speed and angle are constant.

Step 3: Acceleration geometry Its horizontal component is โˆ’4โŸจ3cos2t,3sin2t,0โŸฉ,-4\left\langle 3\cos 2t,3\sin 2t,0\right\rangle, which is opposite the radial position from the zz-axis. Its zz-component is zero, so acceleration points horizontally inward. Also ๐’—โ‹…๐’‚=0\mathbf v\cdot\mathbf a=0, consistent with constant speed.