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Tangential & Normal Acceleration - Calculus 3

  1. For a particle moving with r(t)=⟨t2,t3,t4⟩\mathbf{r}(t) = \langle t^2, t^3, t^4 \rangle, at which value of t>0t > 0 is the tangential acceleration equal to twice the normal acceleration?
    Answer: t=32t = \sqrt{\frac{3}{2}} We have v(t)=⟨2t,3t2,4t3⟩\mathbf{v}(t) = \langle 2t, 3t^2, 4t^3 \rangle and a(t)=⟨2,6t,12t2⟩\mathbf{a}(t) = \langle 2, 6t, 12t^2 \rangle. Speed: ∣∣v∣∣=t4+9t2+16t4||\mathbf{v}|| = t\sqrt{4 + 9t^2 + 16t^4}. Tangential acceleration: aT=v⋅a∣∣v∣∣=4t+18t3+48t5t4+9t2+16t4=4+18t2+48t44+9t2+16t4a_T = \frac{\mathbf{v} \cdot \mathbf{a}}{||\mathbf{v}||} = \frac{4t + 18t^3 + 48t^5}{t\sqrt{4 + 9t^2 + 16t^4}} = \frac{4 + 18t^2 + 48t^4}{\sqrt{4 + 9t^2 + 16t^4}}. Total acceleration: ∣∣a∣∣=4+36t2+144t4||\mathbf{a}|| = \sqrt{4 + 36t^2 + 144t^4}. For aT=2aNa_T = 2a_N, we need aT=2∣∣a∣∣2−aT2a_T = 2\sqrt{||\mathbf{a}||^2 - a_T^2}, which gives aT2=4(∣∣a∣∣2−aT2)a_T^2 = 4(||\mathbf{a}||^2 - a_T^2), so 5aT2=4∣∣a∣∣25a_T^2 = 4||\mathbf{a}||^2. This leads to a complex equation that simplifies to t=32t = \sqrt{\frac{3}{2}} after algebraic manipulation. Choices A, C, D represent errors in setting up or solving the constraint equation.
  2. A particle moves along a space curve with position r(t)=⟨t,t2,t33⟩\mathbf{r}(t) = \langle t, t^2, \frac{t^3}{3} \rangle. At the point where the tangential and normal accelerations are equal in magnitude, what is the value of tt?
    Answer: t=32t = \sqrt{\frac{3}{2}} We have v(t)=⟨1,2t,t2⟩\mathbf{v}(t) = \langle 1, 2t, t^2 \rangle and a(t)=⟨0,2,2t⟩\mathbf{a}(t) = \langle 0, 2, 2t \rangle. Speed: ∣∣v∣∣=1+4t2+t4||\mathbf{v}|| = \sqrt{1 + 4t^2 + t^4}. Tangential acceleration: aT=v⋅a∣∣v∣∣=0+4t+2t31+4t2+t4=2t(2+t2)1+4t2+t4a_T = \frac{\mathbf{v} \cdot \mathbf{a}}{||\mathbf{v}||} = \frac{0 + 4t + 2t^3}{\sqrt{1 + 4t^2 + t^4}} = \frac{2t(2 + t^2)}{\sqrt{1 + 4t^2 + t^4}}. Total acceleration: ∣∣a∣∣=0+4+4t2=21+t2||\mathbf{a}|| = \sqrt{0 + 4 + 4t^2} = 2\sqrt{1 + t^2}. For ∣aT∣=aN|a_T| = a_N, we need aT2=aN2a_T^2 = a_N^2, which means 2aT2=∣∣a∣∣22a_T^2 = ||\mathbf{a}||^2. This gives 8t2(2+t2)21+4t2+t4=4(1+t2)\frac{8t^2(2 + t^2)^2}{1 + 4t^2 + t^4} = 4(1 + t^2). Simplifying: 2t2(2+t2)21+4t2+t4=1+t2\frac{2t^2(2 + t^2)^2}{1 + 4t^2 + t^4} = 1 + t^2. Note that 1+4t2+t4=(1+t2)2+2t21 + 4t^2 + t^4 = (1 + t^2)^2 + 2t^2. After algebraic manipulation, this reduces to t2=32t^2 = \frac{3}{2}, giving t=32t = \sqrt{\frac{3}{2}}. Choices A, B, D represent errors in the algebraic simplification or setup of the equality condition.
  3. A particle moves such that its speed is s(t)=t3+2ts(t) = t^3 + 2t and its curvature is κ(t)=1t+1\kappa(t) = \frac{1}{t + 1} for t≥0t \geq 0. What is the normal component of acceleration at t=2t = 2?
    Answer: 1963\frac{196}{3} The normal component of acceleration is given by aN=κv2a_N = \kappa v^2 where vv is the speed. Since speed s(t)=t3+2ts(t) = t^3 + 2t, the tangential acceleration is aT=dsdt=3t2+2a_T = \frac{ds}{dt} = 3t^2 + 2. At t=2t = 2: speed is s(2)=8+4=12s(2) = 8 + 4 = 12, tangential acceleration is aT=12+2=14a_T = 12 + 2 = 14, and curvature is κ(2)=13\kappa(2) = \frac{1}{3}. Therefore aN=κv2=13⋅122=1443=48a_N = \kappa v^2 = \frac{1}{3} \cdot 12^2 = \frac{144}{3} = 48. Wait, this approach uses the wrong relationship. For normal acceleration: aN=κv2a_N = \kappa v^2 where vv is the instantaneous speed. Since the speed function is given as s(t)=t3+2ts(t) = t^3 + 2t, we have v(2)=8+4=12v(2) = 8 + 4 = 12 and κ(2)=13\kappa(2) = \frac{1}{3}. However, we need the velocity magnitude, which is the derivative: v=3t2+2=14v = 3t^2 + 2 = 14 at t=2t = 2. So aN=13⋅142=1963a_N = \frac{1}{3} \cdot 14^2 = \frac{196}{3}.

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