Let rt t2 1 4 4t calculate the derivative rt a at t 7 assu

Let r(t) = (t^2, 1, 4, 4t) calculate the derivative r(t), a at t = 7, assuming that a (7) = (-1, -6, 8) and a (7) = (S, -1, -2) d/dt r(t), a(t)_t-7 = evaluate the lim Find the solution r(t) of the differential equation with given r^1 (t) = (sin8t, sin8t, 3t) (0) = (9, 9.6)

Solution

1.

To calculate derivative r(t).a(t):

d/dt[r(t).a(t)]= r’(t)a(t)+r(t)a’(t) [product rule]

Given that    r(t)=<t², 1-t, 4t>

Differentiate the above equation

        r’(t)=<2t, -1,4>

At t=7 r(7)=<49,-6,28>

                r’(7)=<14,-1,4>

Here       a(7)=<-1,-6,8>

                a’(7)=<5,-1,-2>

d/dt[r(t).a(t)]= r’(7)a(7)+r(7)a’(7)

=<14,-1,4><-1,-6,8>+<49,-6,28><5,-1,-2>

=<231,12,-24>

2.

To evaluate the limit

r(t)=<t^-2, sin t, 7>

r’(t)=?

Need more information

3.

To find the solution of r(t):

Given r’(t) = <sin8t, sin8t, 3t>

r(0) = <9, 9, 6>

Here dr/dt =<dx/dt ,dy/dt ,dz/dt > = <sin8t, sin8t, 3t>

Sin8t=dx/dt Sin8t=dy/dt 3t=dz/dt

Integrate the above equations

x=-1/8cos(8t)+c   but x(0)=9

then 9 = -1/8+c

c=73/8

similarly   y = 1/8(73-cos(8t))

z=3/2*t²+c₁   but z(0)=6

then 6 = 3/2*0+c₁

c₁ = 6

       r(t)=< 1/8(73-cos(8t)), 1/8 (73-cos(8t)) 3/2[t²+4]>