Find the domain of the vector function r(t)= sqrt(4 - t ^ 2) i + sqrt(t) j - 2 / (sqrt(1 + t)) k.
answer in interval notation
The domain of r(t) is dependent on the range of each of the three component functions:
1. The domain of f(t) = sqrt(4 - t^2) is [-2, 2] because the radicand must be non-negative.
2. The domain of g(t) = sqrt(t) is [0, ∞) because the radicand must be non-negative.
3. The domain of h(t) = -2 / sqrt(1 + t) is (-1, ∞) because the denominator must be positive and non-zero.
Thus, the domain of r(t) is the intersection of the domains of f(t), g(t), and h(t), which is [0, 2].
In interval notation, the domain of r(t) is [0, 2].
To find the domain of the vector function, we need to determine the values of 't' for which the function is defined.
The given vector function is:
r(t) = √(4 - t^2)i + √tj - (2/√(1 + t))k
Let's break down each component of the vector function:
- The term √(4 - t^2) is defined as long as the expression inside the square root is non-negative. Therefore, we need to solve the inequality 4 - t^2 ≥ 0.
Simplifying this inequality, we have -t^2 + 4 ≥ 0, which can be written as: t^2 - 4 ≤ 0.
Factoring the left side, we get: (t - 2)(t + 2) ≤ 0.
So, the possible values for t are: -2 ≤ t ≤ 2.
- The term √t is defined as long as t is non-negative. Hence, t ≥ 0.
- The term 2/√(1 + t) is defined as long as the expression inside the square root is non-zero. Therefore, we need to solve the equation 1 + t ≠ 0.
Solving this equation, we find t ≠ -1.
Now, let's consider all the conditions together:
-2 ≤ t ≤ 2 (from the first term)
t ≥ 0 (from the second term)
t ≠ -1 (from the third term)
Combining these conditions, we find the domain of the vector function 'r(t)' to be:
(-2, -1) U (-1, 0] U [0, 2]
Therefore, the domain of the vector function is (-2, -1) U (-1, 0] U [0, 2] in interval notation.
To find the domain of the vector function r(t), we need to find the values of t for which the function is defined.
First, let's consider the square root terms in the function: sqrt(4 - t^2) and sqrt(t). The square root function is defined only for non-negative real numbers, so we need to ensure that the arguments inside the square root are greater than or equal to zero.
1) For sqrt(4 - t^2) to be non-negative, we have:
4 - t^2 ≥ 0
Solving this inequality, we move -t^2 to the other side:
4 ≥ t^2
Taking the square root on both sides, we get:
2 ≥ t and -2 ≤ t
2) For sqrt(t) to be non-negative, we have:
t ≥ 0
3) For sqrt(1 + t) to be non-zero, it means that 1 + t > 0:
t > -1
Now we have gathered the conditions for the function to be defined:
-2 ≤ t (from condition 1)
t ≥ 0 (from condition 2)
t > -1 (from condition 3)
Combining these conditions, we find that the domain of the vector function r(t) is the intersection of the intervals (-2, ∞) and [0, ∞), which can be written as [0, ∞).
So, the domain of the vector function r(t) is [0, ∞) in interval notation.