Third derivative of R(t).

Question

How do I do third derivative of the following expression:

$R(t) = 7sin(at)\hat x +4e^{-8t}\hat y + 8t^3\hat z$

$(a)$ represents acceleration

my goal is to find what $a$ is equal to when $t=0.27778 sec$

How do I find the $a$ at time t=0.27778 if initial a inside the cos was a=4 ?

Answer 1

I suggest keeping the components of of the vector $R(t)$ and finding the magnitude in the final answer.

We have that:

$$\vec {R(t)} = 7\sin(at)\hat x + 4e^{-8t}\hat y +8t^3\hat z$$

Taking the derivative:

$$\vec {R'(t)} = 7a\cos(at)\hat x -32e^{-8t}\hat y +24t^\hat z$$ $$\vec {R''(t)} = -7a^2\sin(at)\hat x +256e^{-8t}\hat y +48t\hat z$$ $$\vec {R^{(3)}(t)} = -7a^3\cos(at)\hat x -2048e^{-8t}\hat y +48\hat z$$

Use this to find your value of $a$.