Question

r(t)= <t^3, ln(3-t), sqrt(t) >
find domain of r and answer in interval notation

Answer 1

the domain of r is [0,3)
we do not care whether it is differentiable or not at t=0

Answer 2

The domain of r is all values of t that ensure that the components of r are real numbers:

- t^3: defined for all t (i.e. domain is (-infinity, infinity))
- ln(3-t): defined only for t < 3, because ln(x) is only defined for x > 0
- sqrt(t): defined only for t ≥ 0, because the square root of a negative number is not real.

Therefore, the domain of r is:

(0, 3)

Note that we exclude 0 because the square root of 0 is defined but not differentiable, which violates the assumptions of the problems we will consider. The domain in interval notation is:

(0, 3)

Answer 3

If the domain of r is [0,3), then the components of r must be defined and real for all values of t in that interval. Let's check:

- t^3: defined for all t in [0,3)
- ln(3-t): defined only for t < 3, so it is defined for all t in [0,3)
- sqrt(t): defined only for t ≥ 0, so it is defined for all t in [0,3)

Therefore, the domain of r is indeed [0,3). Note that the function r is not differentiable at t=0 because the square root function is not differentiable at t=0. However, this may or may not be relevant depending on the context of the problem.