Question

Find the derivative of the given function. $$ \vec{r}(t)=\left(t^{2}\right)\langle\sin t, 2 t+5\rangle $$

Short answer

The derivative is \( \langle t^2 \cos t + 2t \sin t, 6t^2 + 10t \rangle \).

Step-by-step solution

State the Product Rule for Vector Functions

To find the derivative of a vector function that is a product of two functions, such as \( \vec{r}(t) = f(t) \vec{v}(t) \), we use the product rule for derivatives. The rule states that \( \frac{d}{dt} [f(t) \vec{v}(t)] = f(t) \frac{d}{dt} \vec{v}(t) + \vec{v}(t) \frac{d}{dt} f(t) \).

Compute the Derivative of the Scalar Function

Compute the derivative of the scalar function \( f(t) = t^2 \). The derivative, \( \frac{d}{dt} [t^2] \), is \( 2t \).

Compute the Derivative of the Vector Function

Determine the derivative of the vector function \( \vec{v}(t) = \langle \sin t, 2t + 5 \rangle \). The derivative is found by differentiating each component separately: \( \frac{d}{dt} [\langle \sin t, 2t + 5 \rangle] = \langle \cos t, 2 \rangle \).

Substitute into the Product Rule Formula

Substitute \( f(t) = t^2 \), \( \frac{d}{dt} f(t) = 2t \), \( \vec{v}(t) = \langle \sin t, 2t + 5 \rangle \), and \( \frac{d}{dt} \vec{v}(t) = \langle \cos t, 2 \rangle \) into the product rule: \[ f(t) \frac{d}{dt} \vec{v}(t) + \vec{v}(t) \frac{d}{dt} f(t) = t^2 \langle \cos t, 2 \rangle + \langle \sin t, 2t + 5 \rangle (2t) \]

Perform the Scalar and Vector Multiplications

Calculate the products: For \( t^2 \langle \cos t, 2 \rangle = \langle t^2 \cos t, 2t^2 \rangle \).For \( \langle \sin t, 2t + 5 \rangle (2t) = \langle 2t \sin t, 4t^2 + 10t \rangle \)

Add the Resulting Vectors

Add the vectors from Step 5: \( \langle t^2 \cos t, 2t^2 \rangle + \langle 2t \sin t, 4t^2 + 10t \rangle = \langle t^2 \cos t + 2t \sin t, 2t^2 + 4t^2 + 10t \rangle \).The result is \( \langle t^2 \cos t + 2t \sin t, 6t^2 + 10t \rangle \).

Key concepts

Derivative

A derivative is a fundamental concept in calculus that measures how a function changes as its input changes. It represents the rate of change or the slope of the function at a particular point.
When applied to vector functions like \( \vec{r}(t) \), the derivative provides a new vector that shows how each component of the vector function changes with respect to the variable \( t \).
To compute the derivative of a vector function, we differentiate each component separately and then combine these results back into a vector. This method is vital in understanding motion, as it applies to velocity and acceleration in physics.

Product Rule

The product rule is a technique in calculus used to differentiate functions that are products of two other functions.
For a product of a scalar function \( f(t) \) and a vector function \( \vec{v}(t) \), the product rule is defined as:

• \( \frac{d}{dt}[f(t) \vec{v}(t)] = f(t) \frac{d}{dt} \vec{v}(t) + \vec{v}(t) \frac{d}{dt} f(t) \)

This rule helps in finding the derivative of composite functions by simplifying the process into manageable parts.
Each part is differentiated separately and then combined according to the formula, emphasizing the importance of understanding both the scalar and vector components involved.

Vector Function

A vector function, essentially, assigns a vector to each value of the parameter \( t \).
For example, \( \vec{v}(t) = \langle \sin t, 2t + 5 \rangle \) assigns a two-dimensional vector to each \( t \).
Vector functions are used to describe various phenomena such as the trajectory of a moving object, where each vector represents the position of the object at a specific time.
The derivatives of vector functions provide insights into velocity and acceleration, indicating how the motion changes over time. Each dimension or component of the vector function is treated independently when differentiating, resulting in another vector that describes the change in each direction.

Scalar Function

A scalar function is a standard function that returns a single value or scalar for each value of its variable.
In the context of the original problem, the scalar function is \( f(t) = t^2 \).
Scalar functions are straightforward compared to vector functions because they map a single input to a single output.
In calculus, determining the derivative of a scalar function involves standard differentiation techniques familiar from single-variable calculus.
This process of finding the rate of change of the scalar function is crucial when it serves as a component in more complex expressions, like those involving vector functions.