Question

if r(t) = 3ti + 3t²j - 5tk, compute the tangential and normal components of the acceleration vector. tangential component a_t(t) = normal component a_n(t) =

Explanation:

Step1: Find the velocity vector v(t)

Differentiate r(t) = 3t\(\mathbf{i}\)+3t\(^{2}\)\(\mathbf{j}\)-5t\(\mathbf{k}\) with respect to t.
v(t)=\(\frac{d\mathbf{r}(t)}{dt}=3\mathbf{i} + 6t\mathbf{j}-5\mathbf{k}\)

Step2: Find the acceleration vector a(t)

Differentiate v(t) with respect to t.
a(t)=\(\frac{d\mathbf{v}(t)}{dt}=0\mathbf{i}+6\mathbf{j}+0\mathbf{k}\)

Step3: Calculate the magnitude of v(t)

\(\vert\mathbf{v}(t)\vert=\sqrt{3^{2}+(6t)^{2}+(- 5)^{2}}=\sqrt{9 + 36t^{2}+25}=\sqrt{36t^{2}+34}\)

Step4: Calculate the tangential component \(a_T(t)\)

\(a_T(t)=\frac{\mathbf{v}(t)\cdot\mathbf{a}(t)}{\vert\mathbf{v}(t)\vert}\), \(\mathbf{v}(t)\cdot\mathbf{a}(t)=(3\mathbf{i} + 6t\mathbf{j}-5\mathbf{k})\cdot(0\mathbf{i}+6\mathbf{j}+0\mathbf{k})=36t\)
So \(a_T(t)=\frac{36t}{\sqrt{36t^{2}+34}}\)

Step5: Calculate the normal component \(a_N(t)\)

We know that \(\vert\mathbf{a}(t)\vert = 6\), and \(a_N(t)=\sqrt{\vert\mathbf{a}(t)\vert^{2}-a_T^{2}(t)}\)
\(a_N(t)=\sqrt{36-\frac{1296t^{2}}{36t^{2}+34}}=\sqrt{\frac{36(36t^{2}+34)-1296t^{2}}{36t^{2}+34}}=\sqrt{\frac{1296t^{2}+1224 - 1296t^{2}}{36t^{2}+34}}=\sqrt{\frac{1224}{36t^{2}+34}}=\frac{6\sqrt{34}}{\sqrt{36t^{2}+34}}\)

Answer:

Tangential component \(a_T(t)=\frac{36t}{\sqrt{36t^{2}+34}}\)
Normal component \(a_N(t)=\frac{6\sqrt{34}}{\sqrt{36t^{2}+34}}\)