Applications of the Derivative

Applications of Derivative Logo

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:

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


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 \(t = 0.\)

Velocity vector when moving with a constant acceleration
Figure 1.

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 \(t = 0.\)

Position vector when moving with a constant acceleration
Figure 2.

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

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 \(t = 1\,\text{s}.\)


From the definition of velocity,

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


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

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.


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:

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

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 \(\left( {3,4} \right).\)


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.


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:

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


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

The speed is equal to

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

The answer is

\[y = 2 - x,\;v = \sqrt 2 .\]

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?


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 \(x \ge 1.\)

A part of cubic parabola y=x^3.
Figure 3.

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 \(t.\)


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.

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