2D Velocity Functions with Calculus: Videos & Practice Problems

In 2D Velocity Functions with Calculus, motion is described with a position function \(\vec r(t)\) and a velocity function \(\vec v(t)\). The central relationship is \(\vec v(t)=\frac{d\vec r}{dt}\) , so finding instantaneous velocity means taking the derivative of each vector component. In two dimensions, the \(i\) and \(j\) components are differentiated the same way as ordinary functions of time.

To go backward from velocity to position, use the indefinite integral: \(\vec r(t)=\int \vec v(t)\\,dt+\vec C\) . The constant of integration is a vector and is found from initial conditions. To find displacement over a time interval, use the definite integral: \(\Delta \vec r=\int_{t_i}^{t_f}\vec v(t)\\,dt\) , which gives final minus initial without adding a constant.

When the direction of motion is needed, use the velocity components \(v_x\) and \(v_y\) to determine the angle relative to the x-axis with \(\theta=\tan^{-1}\\!\left(\frac{v_y}{v_x}\right)\) . This connects calculus with vector motion by showing how derivatives and integrals control both magnitude and direction in 2D.

Solving Problems with Position and Velocity Functions in 2D

Understanding the relationship between position and velocity functions in vector form is essential for solving motion problems in two dimensions. The position vector function, denoted as R(t), describes an object's location at any time t, while the velocity vector function, V(t), represents the rate of change of position with respect to time. The fundamental connection between these functions is that velocity is the derivative of position, expressed mathematically as \( \mathbf{V}(t) = \frac{d\mathbf{R}}{dt} \). Conversely, the position function can be obtained by integrating the velocity function over time.

When differentiating a position vector function to find velocity, each component of the vector is differentiated separately. For example, if the position function is given by \( \mathbf{R}(t) = (t^3 - 4)\mathbf{i} + (5t - 3)\mathbf{j} \), the velocity function is found by taking the derivative of each component:

\[\mathbf{V}(t) = \frac{d}{dt} \left( t^3 - 4 \right) \mathbf{i} + \frac{d}{dt} \left( 5t - 3 \right) \mathbf{j} = 3t^2 \mathbf{i} + 5 \mathbf{j}\]

Here, the power rule for differentiation is applied: the exponent is brought down as a coefficient, and the exponent is reduced by one. Constants differentiate to zero, and unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) remain unchanged as they indicate direction.

To find the position function from a given velocity function, integration is used. The indefinite integral of the velocity function yields the position function plus an integration constant vector \(\mathbf{C}\), which accounts for the initial position:

\[\mathbf{R}(t) = \int \mathbf{V}(t) \, dt = \int \left( 3t^2 \mathbf{i} + 5 \mathbf{j} \right) dt = t^3 \mathbf{i} + 5t \mathbf{j} + \mathbf{C}\]

Determining the constant vector \(\mathbf{C}\) requires initial conditions, such as the position at a specific time. For instance, if at \(t=2\) seconds the position is \(\mathbf{R}(2) = 4\mathbf{i} + 7\mathbf{j}\), substituting into the integrated function gives:

\[4\mathbf{i} + 7\mathbf{j} = (2^3)\mathbf{i} + 5 \times 2 \mathbf{j} + \mathbf{C} = 8\mathbf{i} + 10\mathbf{j} + \mathbf{C}\]

Solving for \(\mathbf{C}\) yields:

\[\mathbf{C} = (4 - 8)\mathbf{i} + (7 - 10)\mathbf{j} = -4\mathbf{i} - 3\mathbf{j}\]

Thus, the complete position function is:

\[\mathbf{R}(t) = (t^3 - 4)\mathbf{i} + (5t - 3)\mathbf{j}\]

Displacement, defined as the change in position over a time interval, can be calculated using the definite integral of the velocity function between two time points \(t_i\) and \(t_f\(:

\[\Delta \mathbf{R} = \int_{t_i}^{t_f} \mathbf{V}(t) \, dt = \mathbf{R}(t_f) - \mathbf{R}(t_i)\]

For example, to find the displacement from \)t=2\) to \(t=4\) seconds for the velocity function \( \mathbf{V}(t) = 3t^2 \mathbf{i} + 5 \mathbf{j} \), evaluate the integral:

\[\Delta \mathbf{R} = \left[ t^3 \mathbf{i} + 5t \mathbf{j} \right]_{2}^{4} = (4^3 \mathbf{i} + 5 \times 4 \mathbf{j}) - (2^3 \mathbf{i} + 5 \times 2 \mathbf{j}) = (64 - 8) \mathbf{i} + (20 - 10) \mathbf{j} = 56 \mathbf{i} + 10 \mathbf{j}\]

This vector represents the net change in position over the specified time interval.

Mastering the interplay between derivatives and integrals in vector functions allows for seamless transitions between position, velocity, and displacement calculations in two-dimensional motion. Recognizing when to differentiate or integrate, and correctly applying initial conditions, ensures accurate modeling of an object's trajectory.

The velocity of a plane as a function of time is given by . If the plane started from a point at $t=0$, what is it's position at $t=12s$?

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• B.
• C.

Answer: (A)

Solving Problems with Position and Velocity Functions in 2D Example 1

To determine the direction of a particle's velocity at a specific time when given position functions in both the x and y directions, we start by understanding that velocity components are the derivatives of the position functions with respect to time. If the position functions are given as \( x(t) \) and \( y(t) \), then the velocity components \( v_x(t) \) and \( v_y(t) \) are found by differentiating these functions:

\[ v_x(t) = \frac{dx}{dt} \]

\[ v_y(t) = \frac{dy}{dt} \]

For example, if the position functions are \( x(t) = 5t + 6t^2 \) and \( y(t) = 7 - 3t^2 \), their derivatives yield the velocity components:

\[ v_x(t) = 5 + 12t \]

\[ v_y(t) = -6t \]

To find the velocity direction at a specific time, such as \( t = 4 \), substitute this value into the velocity component functions:

\[ v_x(4) = 5 + 12 \times 4 = 53 \, \text{m/s} \]

\[ v_y(4) = -6 \times 4 = -24 \, \text{m/s} \]

The direction of the velocity vector is the angle \( \theta \) it makes with the positive x-axis. This angle can be calculated using the inverse tangent function, which relates the y-component and x-component of the velocity:

\[ \theta = \tan^{-1} \left( \frac{v_y}{v_x} \right) \]

Substituting the values at \( t = 4 \):

\[ \theta = \tan^{-1} \left( \frac{-24}{53} \right) \approx -24.4^\circ \]

The negative angle indicates that the velocity vector points below the positive x-axis by 24.4 degrees. This method of finding the velocity direction by differentiating position functions and then calculating the angle using the inverse tangent is essential in kinematics for understanding particle motion in two dimensions.

A dog running around a park has a velocity described by $v=(0.03t^2-2.5)\hat{ı}+(-0.1t+2)\hat{ȷ}$. What is the dog's displacement vector in unit-vector notation from $t=15s$ to $t=30s$?

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• B.
• C.
• D.

Answer: (C)

Here's what students ask on this topic:

How do you find the velocity function from a given 2D position function using calculus?

To find the velocity function $\mathrm{\backslash (\backslash vec\{v\}\backslash )(t)}$ from a given 2D position function $\mathrm{\backslash (\backslash vec\{r\}\backslash )(t)}$, you take the derivative of each component of the position vector with respect to time. If $\mathrm{\backslash (\backslash vec\{r\}\backslash )(t)\; =\; x(t)\backslash (\backslash hat\{i\}\backslash )\; +\; y(t)\backslash (\backslash hat\{j\}\backslash )}$, then the velocity is $\mathrm{\backslash (\backslash vec\{v\}\backslash )(t)\; =\; \backslash (\backslash frac\{d\backslash vec\{r\}\backslash )\}\{dt\}\; =\; \backslash (\backslash frac\{dx\}\{dt\}\backslash [\backslash hat\{i\}\backslash )\; +\; \backslash (\backslash frac\{dy\}\{dt\}\backslash ]\backslash hat\{j\}\backslash )}$. This means differentiating the x-component and y-component functions separately. For example, if $\mathrm{x(t)\; =\; t^3\; -\; 4}$ and $\mathrm{y(t)\; =\; 5t\; -\; 3}$, then $\mathrm{v\_x(t)\; =\; 3t^2}$ and $\mathrm{v\_y(t)\; =\; 5}$. The velocity vector is then $\mathrm{3t^2\backslash (\backslash hat\{i\}\backslash )\; +\; 5\backslash (\backslash hat\{j\}\backslash )}$. This process uses the same derivative rules as in single-variable calculus but applies them component-wise to vectors.

How do you determine the position function from a given 2D velocity function?

To find the position function $\mathrm{\backslash (\backslash vec\{r\}\backslash )(t)}$ from a given velocity function $\mathrm{\backslash (\backslash vec\{v\}\backslash )(t)}$, you integrate each component of the velocity vector with respect to time. The position is given by the indefinite integral $\mathrm{\backslash (\backslash vec\{r\}\backslash )(t)\; =\; \backslash (\backslash int\backslash )\; \backslash (\backslash vec\{v\}\backslash )(t)\; dt\; +\; \backslash (\backslash vec\{C\}\backslash )}$, where $\mathrm{\backslash (\backslash vec\{C\}\backslash )}$ is the vector constant of integration. For example, if $\mathrm{\backslash (\backslash vec\{v\}\backslash )(t)\; =\; 3t^2\backslash (\backslash hat\{i\}\backslash )\; +\; 5\backslash (\backslash hat\{j\}\backslash )}$, then integrating gives $\mathrm{\backslash (\backslash vec\{r\}\backslash )(t)\; =\; t^3\backslash (\backslash hat\{i\}\backslash )\; +\; 5t\backslash (\backslash hat\{j\}\backslash )\; +\; \backslash (\backslash vec\{C\}\backslash )}$. To find $\mathrm{\backslash (\backslash vec\{C\}\backslash )}$, use initial conditions such as the position at a specific time. Plugging in the known time and position values allows you to solve for $\mathrm{\backslash (\backslash vec\{C\}\backslash )}$. This method ensures the position function matches the given velocity and initial position.

What is the difference between definite and indefinite integrals when calculating displacement and position in 2D motion?

In 2D motion, the indefinite integral of velocity $\mathrm{\backslash (\backslash vec\{v\}\backslash )(t)}$ gives the position function $\mathrm{\backslash (\backslash vec\{r\}\backslash )(t)\; =\; \backslash (\backslash int\backslash )\; \backslash (\backslash vec\{v\}\backslash )(t)\; dt\; +\; \backslash (\backslash vec\{C\}\backslash )}$, where $\mathrm{\backslash (\backslash vec\{C\}\backslash )}$ is the constant vector determined by initial conditions. This integral includes an arbitrary constant because it represents a family of functions. In contrast, the definite integral calculates displacement over a time interval $\mathrm{[t\_i,\; t\_f]}$ as $\mathrm{\backslash (\backslash Delta\backslash )\; \backslash (\backslash vec\{r\}\backslash )\; =\; \backslash (\backslash int\backslash )\_\{t\_i\}^\{t\_f\}\; \backslash (\backslash vec\{v\}\backslash )(t)\; dt\; =\; \backslash (\backslash vec\{r\}\backslash )(t\_f)\; -\; \backslash (\backslash vec\{r\}\backslash )(t\_i)}$. Here, no constant is added because the integral evaluates the net change in position between two times. Thus, indefinite integrals find the full position function, while definite integrals find the displacement vector between two points in time.

How do you calculate displacement from a velocity function in two dimensions over a specific time interval?

To calculate displacement $\mathrm{\backslash (\backslash Delta\backslash )\; \backslash (\backslash vec\{r\}\backslash )}$ from a velocity function $\mathrm{\backslash (\backslash vec\{v\}\backslash )(t)}$ over a time interval $\mathrm{t\_i}$ to $\mathrm{t\_f}$, you compute the definite integral: $\mathrm{\backslash (\backslash Delta\backslash )\; \backslash (\backslash vec\{r\}\backslash )\; =\; \backslash (\backslash int\backslash )\_\{t\_i\}^\{t\_f\}\; \backslash (\backslash vec\{v\}\backslash )(t)\; dt}$. This integral gives the net change in position vector without an integration constant. For example, if $\mathrm{\backslash (\backslash vec\{v\}\backslash )(t)\; =\; 3t^2\backslash (\backslash hat\{i\}\backslash )\; +\; 5\backslash (\backslash hat\{j\}\backslash )}$, then integrating from $\mathrm{t=2}$ to $\mathrm{t=4}$ yields $\mathrm{\backslash (\backslash Delta\backslash )\; \backslash (\backslash vec\{r\}\backslash )\; =\; [t^3\backslash (\backslash hat\{i\}\backslash )\; +\; 5t\backslash (\backslash hat\{j\}\backslash )]\_2^4\; =\; (64\backslash (\backslash hat\{i\}\backslash )\; +\; 20\backslash (\backslash hat\{j\}\backslash ))\; -\; (8\backslash (\backslash hat\{i\}\backslash )\; +\; 10\backslash (\backslash hat\{j\}\backslash ))\; =\; 56\backslash (\backslash hat\{i\}\backslash )\; +\; 10\backslash (\backslash hat\{j\}\backslash )}$. This vector represents the displacement during that time interval.

How do you find the integration constant vector when determining position from velocity in 2D motion?

When integrating a velocity function $\mathrm{\backslash (\backslash vec\{v\}\backslash )(t)}$ to find position $\mathrm{\backslash (\backslash vec\{r\}\backslash )(t)}$, the integration constant $\mathrm{\backslash (\backslash vec\{C\}\backslash )}$ is a vector that accounts for initial position. To find $\mathrm{\backslash (\backslash vec\{C\}\backslash )}$, use given initial conditions such as $\mathrm{\backslash (\backslash vec\{r\}\backslash )(t\_0)\; =\; \backslash (\backslash vec\{r\}\backslash )\_0}$. Substitute $\mathrm{t\_0}$ into the integrated position function and set it equal to $\mathrm{\backslash (\backslash vec\{r\}\backslash )\_0}$. Solve the resulting vector equation for $\mathrm{\backslash (\backslash vec\{C\}\backslash )}$. For example, if $\mathrm{\backslash (\backslash vec\{r\}\backslash )(t)\; =\; t^3\backslash (\backslash hat\{i\}\backslash )\; +\; 5t\backslash (\backslash hat\{j\}\backslash )\; +\; \backslash (\backslash vec\{C\}\backslash )}$ and $\mathrm{\backslash (\backslash vec\{r\}\backslash )(2)\; =\; 4\backslash (\backslash hat\{i\}\backslash )\; +\; 7\backslash (\backslash hat\{j\}\backslash )}$, then plugging in gives $\mathrm{8\backslash (\backslash hat\{i\}\backslash )\; +\; 10\backslash (\backslash hat\{j\}\backslash )\; +\; \backslash (\backslash vec\{C\}\backslash )\; =\; 4\backslash (\backslash hat\{i\}\backslash )\; +\; 7\backslash (\backslash hat\{j\}\backslash )}$. Solving yields $\mathrm{\backslash (\backslash vec\{C\}\backslash )\; =\; -4\backslash (\backslash hat\{i\}\backslash )\; -\; 3\backslash (\backslash hat\{j\}\backslash )}$. This vector is then added to complete the position function.

How do you determine the direction of motion from velocity components in 2D?

To find the direction of motion from velocity components $\mathrm{v\_x}$ and $\mathrm{v\_y}$ in 2D, calculate the angle $\mathrm{\backslash (\backslash theta\backslash )}$ relative to the x-axis using the inverse tangent function: $\mathrm{\backslash (\backslash theta\backslash )\; =\; \backslash (\backslash tan\backslash )^\{-1\}\backslash (\backslash left\backslash )(\backslash (\backslash frac\{v\_y\}\{v\_x\}\backslash )\backslash (\backslash right\backslash ))}$. This angle indicates the direction of the velocity vector. For example, if $\mathrm{v\_x\; =\; 3t^2}$ and $\mathrm{v\_y\; =\; 5}$, at a specific time $t$, plug in the values to find $\mathrm{\backslash (\backslash theta\backslash )}$. This method connects calculus-derived velocity components to the geometric direction of motion in the plane.