Calculus

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 \(\mathbf{i},\mathbf{j}\) 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 \({v_x} = \large{\frac{{dx}}{{dt}}}\normalsize,\) \({v_y} = \large{\frac{{dy}}{{dt}}}\normalsize\) 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 \({a_x} = \large{\frac{{d{v_x}}}{{dt}}}\normalsize = \large{\frac{{{d^2}x}}{{d{t^2}}}}\normalsize,\) \({a_y} = \large{\frac{{d{v_y}}}{{dt}}}\normalsize = \large{\frac{{{d^2}y}}{{d{t^2}}}}\normalsize\) 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 \(g = 9.8\,\large{\frac{\text{m}}{\text{s}^2}}\normalsize\) 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 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 \(y = \large{\frac{{12}}{x}}\normalsize,\) 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 \(x\left( t \right) = {2^t},\) \(y\left( t \right) = {8^t},\) \(t \ge 0.\) What is the shape of the trajectory?

    Example 6

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

    Example 7

    A particle is moving along the curve given by the parametric equations \(x = \tan t,\) \(y = \sec t.\) Find the particle’s speed at \(t = \large{\frac{\pi }{6}}\normalsize.\)

    Example 8

    A particle moves in the \(xy-\)plane, so that its coordinates at time \(t \ge 0\) are \(x\left( t \right) = 3{t^3} – 3{t^2},\) \(y\left( t \right) = 20{t^2} + 2t\) where \(x,y\) are in meters, \(t\) in seconds. Find the magnitude of the particle’s acceleration at \(t = 2\,\text{s}.\)

    Example 9

    A ball is thrown up with a speed of \({v_0} = 19.6\,\large{\frac{\text{m}}{\text{s}}}\normalsize.\)
    1. What is the time it takes to hit the ground? The acceleration of gravity is \(9.8\,\large{\frac{\text{m}}{\text{s}^2}}\normalsize;\)
    2. What is the ball’s maximum height above the ground?

    Example 10

    Two balls are thrown horizontally from the same point in the opposite direction with the initial speeds \({v_{10}} = 4\,\large{\frac{\text{m}}{\text{s}}}\normalsize,\) \({v_{20}} = 9\,\large{\frac{\text{m}}{\text{s}}}\normalsize\) (Figure \(4\)). Find the distance \(d\) between the balls when their velocity vectors are perpendicular to each other.

    Example 11

    A particle moves in the \(xy-\) plane according to the law \(x = at,\) \(y = b{t^2},\) where \(a \gt 0,\) \(b \gt 0.\)
    1. Determine the particle’s trajectory \(y\left( x \right)\) and sketch its graph.
    2. Determine the speed of the particle as a function of time.
    3. Find the angle \(\phi\) between the velocity vector and the \(x-\)axis.

    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

    \[{\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( t \right) = 1,\;\;}\kern0pt{\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {{t^3}} \right) = 3{t^2}.}\]

    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 \(y = \large{\frac{{12}}{x}}\normalsize,\) 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 \(\large{\frac{{dx}}{{dt}}}\normalsize\) is known:

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

    Compute the derivative \(\large{\frac{{dy}}{{dt}}}\normalsize\) 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

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

    The answer is

    \[y = 2 – x,\;v = \sqrt 2 .\]

    Example 5.

    A particle moves on a trajectory described by the parametric equations \(x\left( t \right) = {2^t},\) \(y\left( t \right) = {8^t},\) \(t \ge 0.\) 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 \(x \ge 1.\)

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

    Example 6.

    A particle’s position is given by the equations \(x\left( t \right) = \large{\frac{1}{2}}\normalsize – {t^2},\) \(y\left( t \right) = \sqrt 2 t,\) 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:

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

    Example 7.

    A particle is moving along the curve given by the parametric equations \(x = \tan t,\) \(y = \sec t.\) Find the particle’s speed at \(t = \large{\frac{\pi }{6}}\normalsize.\)

    Solution.

    We take the derivatives of the coordinates \(x\) and \(y\) with respect to time \(t:\)

    \[\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {\tan t} \right) = \frac{1}{{{{\cos }^2}t}},\]

    \[{\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {\sec t} \right) }={ \frac{d}{{dt}}\left( {\frac{1}{{\cos t}}} \right) }={ – \frac{1}{{{{\cos }^2}t}} \cdot \left( { – \sin t} \right) }={ \frac{{\sin t}}{{{{\cos }^2}t}}.}\]

    Calculate the particle’s speed by the formula

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

    This yields:

    \[{v = \sqrt {{{\left( {\frac{1}{{{{\cos }^2}t}}} \right)}^2} + {{\left( {\frac{{\sin t}}{{{{\cos }^2}t}}} \right)}^2}} }={ \frac{{\sqrt {1 + {{\sin }^2}t} }}{{{{\cos }^2}t}}.}\]

    Substituting the time value \(t = \large{\frac{\pi }{6}}\normalsize,\) we get

    \[{v = \frac{{\sqrt {1 + {{\sin }^2}\frac{\pi }{6}} }}{{{{\cos }^2}\frac{\pi }{6}}} }={ \frac{{\sqrt {1 + {{\left( {\frac{1}{2}} \right)}^2}} }}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} }={ \frac{{\sqrt {1 + \frac{1}{4}} }}{{\frac{3}{4}}} }={ \frac{{\frac{{\sqrt 5 }}{2}}}{{\frac{3}{4}}} }={ \frac{{2\sqrt 5 }}{3}.}\]

    Example 8.

    A particle moves in the \(xy-\)plane, so that its coordinates at time \(t \ge 0\) are \(x\left( t \right) = 3{t^3} – 3{t^2},\) \(y\left( t \right) = 20{t^2} + 2t\) where \(x,y\) are in meters, \(t\) in seconds. Find the magnitude of the particle’s acceleration at \(t = 2\,\text{s}.\)

    Solution.

    First, we compute the particle’s velocity:

    \[{\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {3{t^3} – 3{t^2}} \right) }={ 9{t^2} – 6t,}\]

    \[{\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {20{t^2} + 2t} \right) }={ 40t + 2.}\]

    The the velocity vector is given by

    \[{\mathbf{v} = \frac{{dx}}{{dt}}\mathbf{i} + \frac{{dy}}{{dt}}\mathbf{j} }={ \left( {9{t^2} – 6t} \right)\mathbf{i} }+{ \left( {40t + 2} \right)\mathbf{j}.}\]

    Differentiating once more, we get the acceleration:

    \[{\frac{{{d^2}x}}{{d{t^2}}} = \frac{d}{{dt}}\left( {9{t^2} – 6t} \right) }={ 18t – 6,}\]

    \[{\frac{{{d^2}y}}{{d{t^2}}} = \frac{d}{{dt}}\left( {40t + 2} \right) }={ 40.}\]

    Hence

    \[\mathbf{a} = \frac{{{d^2}x}}{{d{t^2}}}\mathbf{i} + \frac{{{d^2}y}}{{d{t^2}}}\mathbf{j} = \left( {18t – 6} \right)\mathbf{i} + 40\mathbf{j}.\]

    Substitute \(t = 2\,\text{s}:\)

    \[\mathbf{a} = 30\mathbf{i} + 40\mathbf{j}.\]

    Then the magnitude of the acceleration vector is equal to

    \[{a = \left| \mathbf{a} \right| = \sqrt {{{30}^2} + {{40}^2}} }={ 50\,\frac{\text{m}}{{{\text{s}^2}}}.}\]

    Example 9.

    A ball is thrown up with a speed of \({v_0} = 19.6\,\large{\frac{\text{m}}{\text{s}}}\normalsize.\)
    1. What is the time it takes to hit the ground? The acceleration of gravity is \(9.8\,\large{\frac{\text{m}}{\text{s}^2}}\normalsize;\)
    2. What is the ball’s maximum height above the ground?

    Solution.

    \(1.\) The \(y-\)coordinate of the ball is defined by the equation

    \[y\left( t \right) = {v_0}t – \frac{{g{t^2}}}{2}.\]

    When the ball hits the ground, the \(y-\)coordinate is equal to zero. Therefore, we can write

    \[{y\left( t \right) = {v_0}t – \frac{{g{t^2}}}{2} }={ 0.}\]

    Solve this equation for \(t:\)

    \[{t\left( {{v_0} – \frac{{gt}}{2}} \right) = 0,\;\;} \Rightarrow {{t_1} = 0,\;}\kern0pt{{t_2} = \frac{{2{v_0}}}{g}.}\]

    The second root gives the time when the ball hits the ground:

    \[{{t_2} = \frac{{2{v_0}}}{g} }={ \frac{{2 \cdot 19.6}}{{9.8}} }={ 4\,\text{s}.}\]

    \(2.\) The time taken to reach the maximum point can be found from the equation \(v\left( t \right) = 0.\) Hence,

    \[{v\left( t \right) = {v_0} – gt = 0,\;\;} \Rightarrow {t = \frac{{{v_0}}}{g}.}\]

    Calculate the maximum height:

    \[{{y_{\max }} = {v_0}t – \frac{{g{t^2}}}{2} }={ {v_0}\left( {\frac{{{v_0}}}{g}} \right) – \frac{g}{2}{\left( {\frac{{{v_0}}}{g}} \right)^2} }={ \frac{{v_0^2}}{g} – \frac{{v_0^2}}{{2g}} }={ \frac{{v_0^2}}{{2g}} }={ \frac{{{{\left( {19.6} \right)}^2}}}{{2 \cdot 9.8}} }={ 19.6\,\text{m}}\]

    Example 10.

    Two balls are thrown horizontally from the same point in the opposite direction with the initial speeds \({v_{10}} = 4\,\large{\frac{\text{m}}{\text{s}}}\normalsize,\) \({v_{20}} = 9\,\large{\frac{\text{m}}{\text{s}}}\normalsize\) (Figure \(4\)). Find the distance \(d\) between the balls when their velocity vectors are perpendicular to each other.

    Solution.

    Two balls thrown in the opposite direction.
    Figure 4.

    First, we write the velocity components for each ball:

    \[{{v_{1x}} = – {v_{10}},\;}\kern0pt{{v_{1y}} = – gt,}\]

    \[{{v_{2x}} = {v_{20}},\;}\kern0pt{{v_{2y}} = – gt.}\]

    Hence, the velocity vectors are given by

    \[{\mathbf{v}_1} = – {v_{10}}\mathbf{i} – gt\mathbf{j},\]

    \[{\mathbf{v}_2} = {v_{20}}\mathbf{i} – gt\mathbf{j}.\]

    When the velocity vectors are perpendicular, their dot product is equal to zero, so that

    \[{{\mathbf{v}_1} \bot {\mathbf{v}_2},\;\;} \Rightarrow {{\mathbf{v}_1} \cdot {\mathbf{v}_2} = 0,\;\;} \Rightarrow {\left( { – {v_{10}}} \right){v_{20}} + \left( { – gt} \right)\left( { – gt} \right) = 0,\;\;} \Rightarrow {{g^2}{t^2} = {v_{10}}{v_{20}},\;\;} \Rightarrow {t = \frac{{\sqrt {{v_{10}}{v_{20}}} }}{g}.}\]

    Note that both the balls are always on the same height as their vertical velocity components are equal. Therefore, the line \(d\) is horizontal.

    The distance between the balls at the time instant \(t = \large{\frac{{\sqrt {{v_{10}}{v_{20}}} }}{g}}\normalsize\) is equal to

    \[{d = {v_{10}}t + {v_{20}}t }={ \left( {{v_{10}} + {v_{20}}} \right)t }={ \left( {{v_{10}} + {v_{20}}} \right)\frac{{\sqrt {{v_{10}}{v_{20}}} }}{g}.}\]

    Substituting the known values, we obtain:

    \[{d = \left( {4 + 9} \right)\frac{{\sqrt {4 \cdot 9} }}{{9.8}} }={ 7.96\,\text{m}}\]

    Example 11.

    A particle moves in the \(xy-\) plane according to the law \(x = at,\) \(y = b{t^2},\) where \(a \gt 0,\) \(b \gt 0.\)
    1. Determine the particle’s trajectory \(y\left( x \right)\) and sketch its graph.
    2. Determine the speed of the particle as a function of time.
    3. Find the angle \(\phi\) between the velocity vector and the \(x-\)axis.

    Solution.

    \(1.\) Let’s solve the first equation for \(t\) and substitute it into the second equation:

    \[{x = at,\;\;} \Rightarrow {t = \frac{x}{a},\;\;} \Rightarrow {y = b{\left( {\frac{x}{a}} \right)^2} }={ \frac{b}{{{a^2}}}{x^2}.}\]

    So the particle’s trajectory is the right branch of the parabola

    \[y = \frac{b}{{{a^2}}}{x^2}.\]

    particle's trajectory
    Figure 5.

    \(2.\) To determine the speed of the particle, we differentiate the coordinates \(x,y\) with respect to time \(t:\)

    \[\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {at} \right) = a,\]

    \[\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {b{t^2}} \right) = 2bt.\]

    Therefore, the particle’s velocity vector is given by

    \[{\mathbf{v} = \frac{{dx}}{{dt}}\mathbf{i} + \frac{{dy}}{{dt}}\mathbf{j} }={ a\mathbf{i} + 2bt\mathbf{j}.}\]

    Since the speed is the absolute value of the velocity, we have

    \[{v = \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} }={ \sqrt {{a^2} + 4{b^2}{t^2}} .}\]

    \(3.\) The velocity vector is written in the form

    \[\mathbf{v} = a\mathbf{i} + 2bt\mathbf{j}.\]

    Therefore, the tangent of the angle \(\varphi\) is given by

    \[\tan \varphi = \frac{{2bt}}{a}.\]

    The angle phi between the velocity vector and the x-axis.
    Figure 6.

    Then

    \[\varphi = \arctan \left( {\frac{{2bt}}{a}} \right).\]