Question

If r(t) = 4e3t, 2e−3t, 4te3t , find t(0), r''(0), and r'(t) · r''(t).

Answer 1

we are given

$r(t)=(4e^{3t} , 2e^{-3t} , 4te^{3t})$

Calculation of r(0):

we can plug t=0

$r(0)=(4e^{3*0} , 2e^{-3*0} , 4*0e^{3*0})$

$r(0)=(4 , 2 , 0)$

Calculation of r'(0):

Firstly, we will find derivative

$r'(t)=(4*3e^{3t} , 2*-3e^{-3t} , 4e^{3t}+4t*3e^{3t})$

we can simplify it

$r'(t)=(12e^{3t} , -6e^{-3t} , 4e^{3t}+12te^{3t})$

now, we can plug t=0

$r'(0)=(12e^{3*0} , -6e^{-3*0} , 4e^{3*0}+12*0e^{3*0})$

$r'(0)=(12 , -6e^{-3*0} , 4e^{3*0}+12*0e^{3*0})$

$r'(0)=(12 , -6 , 4)$

Calculation of r''(0):

we can find second derivative

$r''(t)=(12*3e^{3t} , -6*-3e^{-3t} , 4*3e^{3t}+12e^{3t}+12t*3*e^{3t})$

$r''(t)=(36e^{3t} , 18e^{-3t} , 12e^{3t}+12e^{3t}+36te^{3t})$

now, we can plug t=0

we get

$r''(0)=(36e^{3*0} , 18e^{-3*0} , 12e^{3*0}+12e^{3*0}+36*0*e^{3*0})$

$r''(0)=(36 , 18 , 24)$

Calculation of r'(0) · r''(0):

$r'(0)*r''(0)=(12 , -6 , 4)*(36 , 18 , 24)$

$r'(0)*r''(0)=12*36-6*18+4*24$

$r'(0)*r''(0)=420$............Answer