Answer:
the unit vector at t=1 is <15/√334 , 0 , -3/√334 >
Step-by-step explanation:
for the parameterized curve
r(t) = < 15 t ,10 , 3/t > for -20 < t < 2 ( i have corrected the -20 for the 20 so the inequality has sense and the t=1 falls in the range provided )
the derivative of r with respect to t is
r'(t) = < 15 , 0 , -3/t² >
then the unit tangent vector is defined as
u(t) = r'(t) / |r(t)|
for t=1
u(t=1) = r'(1) / |r(1)| = < 15 , 0 , -3/1² > / √[(15*1)²+(10)²+(3/1)²] = < 15 , 0 , -3 > / √334 = <15/√334 , 0 , -3/√334 >
then the unit vector at t=1 is <15/√334 , 0 , -3/√334 >
