Question

For the following parameterized curve, find the unit tangent vector at the given value of t.

Answer 1

To find the unit tangent vector at t = 25 s, differentiate the position vector and normalize the derivative.

Step-by-step explanation:

The unit tangent vector of a parameterized curve is the derivative of the position vector with respect to time, divided by its magnitude. To find the unit tangent vector at t = 25 s, we first need to find the derivative of the position vector, and then normalize it.

Let's suppose the position vector is r(t) = <x(t), y(t)>. To find the derivative, we differentiate each component with respect to time:

&frac{dr(t)}{dt} = <x'(t), y'(t)>.

Finally, we normalize the derivative to find the unit tangent vector:

T(t) = <x'(t), y'(t)>/|<x'(t), y'(t)>|.