Problem 2 Find the domain of \(\mathbf{r}(... [FREE SOLUTION]

Chapter 12: Problem 2

Find the domain of $\mathbf{r}(t)$ and the value of $\mathbf{r}\left(t_{0}\right)$. $$ \mathbf{r}(t)=\left\langle\sqrt{3 t+1}, t^{2}\right\rangle ; t_{0}=1 $$

Short Answer

Expert verified

The domain of $ \mathbf{r}(t) $ is $ t \geq -\frac{1}{3} $ and $ \mathbf{r}(1) = \langle 2, 1 \rangle $.

Step by step solution

Identify components of $ \mathbf{r}(t) $

The vector function $ \mathbf{r}(t) = \left\langle \sqrt{3t+1}, t^2 \right\rangle $ consists of two components: $ \sqrt{3t+1} $ and $ t^2 $.

Determine the domain for $ \sqrt{3t+1} $

The expression $ \sqrt{3t+1} $ is defined when the quantity under the square root is non-negative. Thus, we need $ 3t + 1 \geq 0 $. Solve for $ t $ to find $ t \geq -\frac{1}{3} $.

Determine the domain for $ t^2 $

The function $ t^2 $ is a polynomial and is defined for all real numbers. Thus, its domain is $ \mathbb{R} $.

Combine the domain conditions

To find the domain of $ \mathbf{r}(t) $, combine the domains of the individual components. The condition when both components are defined is $ t \geq -\frac{1}{3} $.

Evaluate $ \mathbf{r}(t_0) $

To find $ \mathbf{r}(1) $, substitute $ t = 1 $ into $ \mathbf{r}(t) = \left\langle \sqrt{3t+1}, t^2 \right\rangle $. This gives $ \mathbf{r}(1) = \left\langle \sqrt{3(1)+1}, 1^2 \right\rangle = \left\langle 2, 1 \right\rangle $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domain of a Function

The domain of a function refers to all the possible input values (usually represented as $ t $ in this context) that will produce valid outputs for a given function. For vector functions like $ \mathbf{r}(t) = \langle \sqrt{3t+1}, t^2 \rangle $, it's crucial to consider the domain of each individual component function separately before determining the domain of the entire vector function.

For the component $ \sqrt{3t+1} $, the domain consists of all values of $ t $ for which the expression under the square root is non-negative. This ensures the square root function is defined, leading us to the condition $ 3t + 1 \geq 0 $. Solving this inequality, we find that $ t \geq -\frac{1}{3} $.
The second component, $ t^2 $, is a polynomial function whose domain is all real numbers, $ \mathbb{R} $, because polynomials are defined for any real number.

Thus, when combining these two domains, the domain of $ \mathbf{r}(t) $ is restricted by the stricter condition, being $ t \geq -\frac{1}{3} $. This ensures both components of the vector function are defined and output valid results.

Square Root Function

The square root function, denoted as $ \sqrt{x} $, is fundamental in mathematics, especially in calculus and algebra. It involves finding a number which, when multiplied by itself, gives the original number. In the context of vector functions, ensuring the expression inside the square root is non-negative is crucial.

Given $ \sqrt{3t+1} $, for the square root to be defined, the expression $ 3t+1 $ must be $ \geq 0 $. This constraint limits the possible values of $ t $.
We solve $ 3t + 1 \geq 0 $, leading us to $ t \geq -\frac{1}{3} $. This result highlights that the square root function's domain depends heavily on the non-negativity requirement.

The square root function's restriction ensures we're only working within the realm where calculations are real and meaningful. This is especially important when incorporating the square root within more complex structures like vector functions.

Polynomial Function

Polynomial functions form one of the simplest yet most flexible classes of functions in mathematics. A polynomial function is expressed as a sum of terms, each consisting of a variable raised to a non-negative integer power and a coefficient. The domain of polynomial functions encompasses all real numbers.

In our vector function $ \mathbf{r}(t) = \langle \sqrt{3t+1}, t^2 \rangle $, the second component $ t^2 $ is a polynomial function. This means it can take any real number as input without restrictions, unlike the square root component.
Mathematically, you can express the domain of $ t^2 $ as $ \mathbb{R} $, indicating that for this component, there are no constraints on the value of $ t $.

Understanding polynomial functions and their unconstrained domains helps in appreciating their utility in modeling phenomena across continuous input ranges. Their use alongside functions with constrained domains, like square roots, often requires careful analysis to establish the final domain of a composite function like a vector function.

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91影视

Find the domain of \(\mathbf{r}(t)\) and the value of \(\mathbf{r}\left(t_{0}\right)\). $$ \mathbf{r}(t)=\left\langle\sqrt{3 t+1}, t^{2}\right\rangle ; t_{0}=1 $$

Short Answer

Step by step solution

Identify components of \( \mathbf{r}(t) \)

Determine the domain for \( \sqrt{3t+1} \)

Determine the domain for \( t^2 \)

Combine the domain conditions

Evaluate \( \mathbf{r}(t_0) \)

Key Concepts

Domain of a Function

Square Root Function

Polynomial Function

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