If r(t) = 3e^3t, 2e^-3t, 2te^3t, find t(0), r''(0), and r'(t) · r''(t).

Solution:

Given, r(t) = 3e^3t, 2e^-3t, 2te^3t

We have to find t(0), r’’(0) and r’(t).r’’(t).

Put t = 0

r(0) = 3e³⁽⁰⁾, 2e^-3(0), 2te³⁽⁰⁾

r(0) = (3, 2, 0)

Therefore, at t(0) the value of r(0) = (3, 2, 0)

On differentiating,

r’(t) = 9e^3t, -6e^-3t, 6te^3t

Put t = 0

r’(0) = 9e³⁽⁰⁾, -6e^-3(0), 6te³⁽⁰⁾

r’(0) = (9, -6, 0)

On differentiating,

r’’(t) = 27e^3t, 18e^-3t, 18te^3t

Put t = 0

r’’(0) = 27e³⁽⁰⁾, 18e^-3(0), 18te³⁽⁰⁾

r’’(0) = (27, 18, 0)

Therefore, r’’(0) = (27, 18, 0)

r’(0).r’’(0) = (9, -6, 0).(27, 18, 0)

= 9(27) - 6(18) + 0(0)

= 243

Therefore, r’(0).r’’(0) = 135.

Therefore, t(0) = (3, 2, 0), r''(0) = (27, 18, 0) and r'(0) · r''(0) = 135

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If r(t) = 3e^3t, 2e^-3t, 2te^3t, find t(0), r''(0), and r'(t) · r''(t).

Summary:

If r(t) = 3e^3t, 2e^-3t, 2te^3t, t(0) = (3, 2, 0), r''(0) = (27, 18, 0), r'(0) · r''(0) = 135