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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

User SageJustus
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1 Answer

4 votes

Answer:

the unit vector at t=1 is <15/√334 , 0 , -3/√334 >

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 >

User Dale Gerdemann
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