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Question

FInd the unit tangent vector T9t) to the curve r(t) at the point with the given value of the parameter t. R9t) = <-2t, e^-t, sqrt(t+5)-t> , t=2 Find T(2)

Answer 1

Your vector-valued function, r(t), has been presented to you as the ordered triple, (x(t),y(t),z(t)).

 

r(t)=(x(t),y(t),z(t))

 

In your case,

x(t) = -2t

y(t) = e^(-t)

z(t) = √(t+5) - t = (t+5)^(1/2) - t  (always express square roots as exponentiation by 1/2 when taking derivatives)

 

To get the tangent vector, r'(t), just take the derivatives of each of the component functions:

 

r'(t) = (x'(t),y'(t),z'(t))

 

You can then substitute in 2 for t to find r'(2).

 

To get the unit tangent vector, T(t), you'll want to divide r'(t) by its magnitude, ||r'(t)||.

 

||r'(t)|| = √(x'(t)^2 + y'(t)^2 + z'(t)^2)

 

T(t) = r'(t)/||r'(t)||

 

T(t) is a vector pointing in the direction of r'(t) and having a magnitude of 1.

 

You can substitute in t=2 to find the T(2), the tangent vector at t=2.