# Calculus

## Applications of the Derivative # Planar Motion

## Position and Trajectory

Suppose a particle moves in a plane. Then its position can be expressed as

${\mathbf{r} = x\mathbf{i} + y\mathbf{j}},$

where x, y are the Cartesian coordinates and i, j are the unit vectors along the x- and y- coordinate axes, respectively.

As the coordinates x, y depend on time, we can write the position vector in parametric form:

$\mathbf{r} = x\left( t \right)\mathbf{i} + y\left( t \right)\mathbf{j}.$

This vector equation defines the trajectory of the particle.

## Velocity and Speed

The velocity $$\mathbf{v}$$ for a particle is defined as the time derivative of the particle's position vector, that is

$\mathbf{v} = \frac{{d\mathbf{r}}}{{dt}}.$

In coordinate form, it is given by

$\mathbf{v} = \frac{{dx}}{{dt}}\mathbf{i} + \frac{{dy}}{{dt}}\mathbf{j},$

where $${v_x} = \frac{{dx}}{{dt}},$$ $${v_y} = \frac{{dy}}{{dt}}$$ are components of the velocity vector.

The particle's speed $$v$$ is a scalar quantity. It is defined as the magnitude of the velocity:

${v = \sqrt {v_x^2 + v_y^2} = \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} .}$

## Acceleration

The acceleration $$\mathbf{a}$$ for a particle is defined as the time derivative of the particle's velocity vector:

$\mathbf{a} = \frac{{d\mathbf{v}}}{{dt}} = \frac{{{d^2}\mathbf{r}}}{{d{t^2}}}.$

In case when $$\mathbf{v}$$ is written in coordinate form, the acceleration is given by

$\mathbf{a} = \frac{{d{v_x}}}{{dt}}\mathbf{i} + \frac{{d{v_y}}}{{dt}}\mathbf{j} = \frac{{{d^2}x}}{{d{t^2}}}\mathbf{i} + \frac{{{d^2}y}}{{d{t^2}}}\mathbf{j},$

where $${a_x} = \frac{{d{v_x}}}{{dt}} = \frac{{{d^2}x}}{{d{t^2}}},$$ $${a_y} = \frac{{d{v_y}}}{{dt}} = \frac{{{d^2}y}}{{d{t^2}}}$$ are components of the acceleration vector.

The magnitude of the acceleration is computed as

$a = \sqrt {a_x^2 + a_y^2} = \sqrt {{{\left( {\frac{{d{v_x}}}{{dt}}} \right)}^2} + {{\left( {\frac{{d{v_y}}}{{dt}}} \right)}^2}} = \sqrt {{{\left( {\frac{{{d^2}x}}{{d{t^2}}}} \right)}^2} + {{\left( {\frac{{{d^2}y}}{{d{t^2}}}} \right)}^2}} .$

## Motion with Constant Acceleration

Suppose a particle or object moves with a constant acceleration $$\mathbf{a}:$$

$\mathbf{a} = \frac{{d\mathbf{v}}}{{dt}} = \text{const}.$

Then the velocity at time $$t$$ is written as

$\mathbf{v}\left( t \right) = {\mathbf{v}_0} + \mathbf{a}t,$

where $${\mathbf{v}_0}$$ is the initial velocity vector at $$t = 0.$$

The position at time $$t$$ is described by the equation

$\mathbf{r}\left( t \right) = {\mathbf{r}_0} + {\mathbf{v}_0}t + \frac{1}{2}\mathbf{a}{t^2},$

where $${\mathbf{r}_0}$$ is the initial position at $$t = 0.$$

## Free Fall Motion

Free fall is motion of a body with a constant acceleration caused by gravity. Such an object moves with the downward acceleration

$\mathbf{a} = - g\mathbf{j},$

where $$g = 9.8\,\frac{\text{m}}{\text{s}^2}$$ is the acceleration due to gravity, $$\mathbf{j}$$ is a unit vector pointing vertically upward.

The vertical velocity $$v\left( t \right)$$ and the position (or altitude) $$y\left( t \right)$$ during a free fall are given by the equations

$v\left( t \right) = {v_0} - gt\;\;\left({\frac{\text{m}}{\text{s}}}\right),$
$y\left( t \right) = {y_0} + {v_0}t - \frac{1}{2}g{t^2}\;\;\left({\text{m}}\right),$

where $${v_0}$$ is the initial velocity.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

A particle is moving in $$xy-$$plane according to the equations $x = t, y = {t^3},$ where $$x,y$$ are measured in meters. Find the particle's velocity and speed at $$t = 1\,\text{s}.$$

### Example 2

Assume that an object moves on a trajectory according to the equations $x\left( t \right) = t + \cos t, y\left( t \right) = t - \sin t.$ Find the magnitude of the acceleration vector.

### Example 3

A particle moves along the hyperbola $y = \frac{{12}}{x},$ so that the $$x-$$coordinate is increasing at a constant rate of $$3$$ meters per second. Find the speed of the particle when it is at the point $$\left( {3,4} \right).$$

### Example 4

A particle is moving along the curve given by the parametric equations $x = 1 + t, y = 1 - t.$ Determine the $$xy-$$equation of the trajectory and speed along it.

### Example 5

A particle moves on a trajectory described by the parametric equations $x\left( t \right) = {2^t}, y\left( t \right) = {8^t},$ where $$t \ge 0.$$ What is the shape of the trajectory?

### Example 6

A particle's position is given by the equations $x\left( t \right) = \frac{1}{2} - {t^2}, y\left( t \right) = \sqrt 2 t,$ where $$x,y$$ are measured in meters. Determine the particle's distance $$d$$ from the origin as a function of time $$t.$$

### Example 1.

A particle is moving in $$xy-$$plane according to the equations $x = t, y = {t^3},$ where $$x,y$$ are measured in meters. Find the particle's velocity and speed at $$t = 1\,\text{s}.$$

Solution.

From the definition of velocity,

$\mathbf{v} = \frac{{dx}}{{dt}}\mathbf{i} + \frac{{dy}}{{dt}}\mathbf{j},$

where

$\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( t \right) = 1,\;\;\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {{t^3}} \right) = 3{t^2}.$

Hence, the velocity vector at $$t = 1$$ is given by

$\mathbf{v}\left( {t = 1} \right) = \mathbf{i} + 3\mathbf{j}.$

The speed at this time is equal to

${\left| {\mathbf{v}} \right| = \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} } = \sqrt {{1^2} + {3^2}} = \sqrt {10}\,\frac{\text{m}}{\text{s}}.$

### Example 2.

Assume that an object moves on a trajectory according to the equations $x\left( t \right) = t + \cos t, y\left( t \right) = t - \sin t.$ Find the magnitude of the acceleration vector.

Solution.

We differentiate $$x\left( t \right)$$ and $$y\left( t \right)$$ twice to find the acceleration of the object:

$x^\prime\left( t \right) = \left( {t + \cos t} \right)^\prime = 1 - \sin t,$
$x^{\prime\prime}\left( t \right) = \left( {1 - \sin t} \right)^\prime = - \cos t,$
$y^\prime\left( t \right) = \left( {t - \sin t} \right)^\prime = 1 + \cos t,$
$y^{\prime\prime}\left( t \right) = \left( {1 + \cos t} \right)^\prime = - \sin t.$

Now we can compute the magnitude of the acceleration vector:

$a = \left| \mathbf{a} \right| = \sqrt {{{\left( {x^{\prime\prime}\left( t \right)} \right)}^2} + {{\left( {y^{\prime\prime}\left( t \right)} \right)}^2}} = \sqrt {{{\left( { - \cos t} \right)}^2} + {{\left( { - \sin t} \right)}^2}} = \sqrt {{{\cos }^2}t + {{\sin }^2}t} = 1.$

### Example 3.

A particle moves along the hyperbola $y = \frac{{12}}{x},$ so that the $$x-$$coordinate is increasing at a constant rate of $$3$$ meters per second. Find the speed of the particle when it is at the point $$\left( {3,4} \right).$$

Solution.

First, we determine the velocity vector of the particle using the formula

$\mathbf{v} = \frac{{dx}}{{dt}}\mathbf{i} + \frac{{dy}}{{dt}}\mathbf{j}.$

The derivative $$\frac{{dx}}{{dt}}$$ is known:

$\frac{{dx}}{{dt}} = 3\,\frac{\text{m}}{\text{s}}.$

Compute the derivative $$\frac{{dy}}{{dt}}$$ by the chain rule:

$\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {\frac{{12}}{x}} \right) = - \frac{{12}}{{{x^2}}}\frac{{dx}}{{dt}}.$

At the given point we have

$\frac{{dy}}{{dt}} = - \frac{{12}}{{{3^2}}} \cdot 3 = - 4\,\frac{\text{m}}{\text{s}}.$

Now we can find the particle's speed:

$v = \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} = \sqrt {{3^2} + {{\left( { - 4} \right)}^2}} = 5\,\frac{\text{m}}{\text{s}}.$

### Example 4.

A particle is moving along the curve given by the parametric equations $x = 1 + t, y = 1 - t.$ Determine the $$xy-$$equation of the trajectory and speed along it.

Solution.

We solve the first equation for $$t$$ and plug it into the second equation:

$x = 1 + t,\;\; \Rightarrow t = x - 1,\;\; \Rightarrow y = 1 - \left( {x - 1} \right),\;\; \Rightarrow y = 2 - x.$

Hence, the particle's trajectory is the straight line $$y = 2 - x.$$ Find the velocity vector:

$\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {1 + t} \right) = 1,$
$\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {1 - t} \right) = - 1.$

Then

$\mathbf{v} = \frac{{dx}}{{dt}}\mathbf{i} + \frac{{dy}}{{dt}}\mathbf{j} = 1 \cdot \mathbf{i} + \left( { - 1} \right) \cdot \mathbf{j} = \mathbf{i} - \mathbf{j}.$

The speed is equal to

$v = \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} = \sqrt {{1^2} + {{\left( { - 1} \right)}^2}} = \sqrt 2 .$

$y = 2 - x,\;v = \sqrt 2 .$

### Example 5.

A particle moves on a trajectory described by the parametric equations $x\left( t \right) = {2^t}, y\left( t \right) = {8^t},$ where $$t \ge 0.$$ What is the shape of the trajectory?

Solution.

From the first equation, we find:

$x\left( t \right) = {2^t},\;\; \Rightarrow t = {\log _2}x.$

Substitute this into the second equation to obtain the $$xy-$$equation of the curve:

$y = {8^t},\;\; \Rightarrow y = {8^{{{\log }_2}x}} = {\left( {{2^3}} \right)^{{{\log }_2}x}} = {2^{{{\log }_2}{x^3}}} = {x^3}.$

As $$t \ge 0,$$ then $$x \ge {2^0} = 1.$$

Hence, the given curve is the cubic parabola $$y = {x^3}$$ where $$x \ge 1.$$

### Example 6.

A particle's position is given by the equations $x\left( t \right) = \frac{1}{2} - {t^2}, y\left( t \right) = \sqrt 2 t,$ where $$x,y$$ are measured in meters. Determine the particle's distance $$d$$ from the origin as a function of time $$t.$$

Solution.

The distance from the origin is defined as

$d\left(t\right) = \sqrt {{x^2}\left( t \right) + {y^2}\left( t \right)} .$

Substituting $$x\left( t \right),$$ $$y\left( t \right),$$ we obtain:

$d\left(t\right) = \sqrt {{{\left( {\frac{1}{2} - {t^2}} \right)}^2} + {{\left( {\sqrt 2 t} \right)}^2}} = \sqrt {\frac{1}{4} - {t^2} + {t^4} + 2{t^2}} = \sqrt {{t^4} + {t^2} + \frac{1}{4}} = \sqrt {{{\left( {{t^2} + \frac{1}{2}} \right)}^2}} = {{t^2} + \frac{1}{2}}\,\left(\text{m}\right).$

See more problems on Page 2.