If F t e2t 3tsin t and Gt t2 t3t find ddtGt ddtFGt ddtF Ti

If F (t) = (e2t ,3t,sin t) and G(t) = (t2 ,t,3t) find d/dt|G(t)|, d/dt(F.G)(t), d/dt(F Times G)(t)

PART 1. The absolute value (length) of vector G is the square root of [ t^4 + t^2 + 9 t^2]

or [t^4 + 10 t^2 ] or [ t^4 + 10 t^2 ]^ (1/2)

so its derivative is (1/2) (4 t^3 + 20 t ) / [t^4 + 20 t^2] which reduces and simplifies to

(2 t^2 + 10) / (t^2 + 20)

PART 2. The scalar dot product of F and G is t^2 e^2t + 3 t^2 + 3t sin t

The derivative of this is t^2 (2) e^2t + e^2t (2t) + 6t + 3t cos t + t sin t

none of these are \"like terms\" so this result does not simplify

PART 3. The results in parts 1 and 2 were both scalars, but we will get a vector here in Part 3.

The vector cross product of F and G is i [9 t^2 - t sin t] - j [3t e^2t - t^2 sin t] + k [t e^2t - 3 t^3 ]

The derivative of this is

i [18t - t cos t - sin t] - j [ 6t e^2t + e e^2t - t^2 cos t - 2t sin t] + k [2t e^2t + e^2t - 9 t^2]