Question

Suppose that a particle following the given path c(t) flies off on a tangent at t = t0. Compute the position of the particle at the given time t1. c(t) = (2t2, t3 − 4t, 0), where t0 = 3, t1 = 4

Answer 1

Answer:

(30,35,0)

Step-by-step explanation:

We are given that

$C(t)=(2t^2,t^3-4t,0)$

$t_0=3,t_1=4$

The velocity of particle,$v(t)=c'(t)=(4t,3t^2-4,0)$

Substitute $t_0=3$

$C(3)=(2(3)^2,(3)^3-4(3),0))=(18,15,0)$

$v(3)=(4(3),3(3)^2-4,0)=(12,23,0)$

The tangent line at t=3 is given by

$l(t)=c(2)+(t-3)v(3)=(18,15,0)+(t-3)(12,23,0)$

Now, substitute t=4

$l(4)=(18,15,0)+(4-3)+(12,23,0)=(18,12,0)+(12,23,0)$

$l(4)=(30,35,0)$

Hence, the position of particle  at $t_1=4$=(30,35,0)