For the following parameterized curve, find the unit tangent vector at the given value of t.
r(t) = < 15 t ,10 , 3/t > for 20 < t < 2, t = 1

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 >

a) 32pie square meters

step-by-step explanation: