# 3. Curvilinear Motion

by M. Bourne

In the last Chapter (in The Derivative as an Instantaneous Rate of Change), we found out how to find the velocity from the displacement function using:

v=(ds)/(dt)

and the acceleration from the velocity function (or displacement function), using:

a=(dv)/dt=(d^2s)/(dt^2)

These formulae are only appropriate for rectilinear motion (i.e. velocity and acceleration in a straight line). This is inadequate for most real situations, so we introduce here the concept of curvilinear motion, where an object is moving in a plane along a specified curved path.

We generally express the x and y components of the motion as functions of time. This form is called parametric form. (See another example using parametric form in Lissajous Figures.)

### Example 1: Parametric Equations

Draw the curve

x(t) = sin t,
y(t) = cos t

for t = 0 to in 0.5 intervals.

First, we need to set up a table of values which we obtain by substituting various values of t:

 t 0 0.5 1 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 x(t) 0 0.48 0.84 1 .91 .60 .14 −.35 −.76 −.98 −.96 −.71 −.28 y(t) 1 0.88 0.54 0.07 −.42 −.80 −.99 −.94 −.65 −.21 .28 .71 .96

We plot our 13 points, starting at (0, 1) as follows (we move clockwise in this example, and I have numbered the points):

If we were to use a smaller interval for t, we would obtain a smooth parametric curve, which in this case is a circle.

We see that we have formed a circle, centre (0,0), radius 1 unit.

Notice that the variable t does not appear in the axes of this graph, just the variables x and y.

## Horizontal and Vertical Components of Velocity

The horizontal component of the velocity is written:

v_x=(dx)/(dt)

and the vertical component is written:

v_y=(dy)/(dt)

We want to find the magnitude of the resultant velocity v once we know the horizontal and vertical components. We use:

v=sqrt((v_x)^2+(v_y)^2

The direction θ that the object is moving in, is found using:

tan\ theta_v=(v_y)/(v_x

### Example 2

If x = 5t3 and y = 4t2 at time t, find the magnitude and direction of the velocity when t = 10.

When t = 10, the particle is at (5000, 400).

Here is the graph of the motion.

Note:

• The axes are x and y (and do not involve t).
• The particle is accelerating as time goes on (the numbered pink dots are at one second intervals)

We are told that

x = 5t3

So

dx/dt=15t^2

At t = 10, the velocity in the x-direction is given by:

dx/dt = v_x=15(10)^2=1500\ "ms"^-1

Also, y = 4t2, so the velocity in the y-direction is:

dy/dt=8t

When t = 10, the velocity in the y-direction is:

dy/dt=v_y=8(10)=80\ "ms"^-1

So the magnitude of the velocity will be:

v=sqrt((v_x)^2+(v_y)^2)

=sqrt(1500^2+80^2)

=1502.1\ "ms"^-1

Now for the direction of the velocity (it is an angle, relative to the positive x-axis):

tan\ theta_v=v_y/v_x=80/1500

So θ_v= 0.053 radians = 3.05^@.

### Example 3

If

x=(20t)/(2t+1)

and

y=0.1(t^2+t)

at time t, find the magnitude and direction of the velocity when t = 2. Plot the curve.

When t = 2, the particle is at (8, 0.6).

x=(20t)/(2t+1)

so

(dx)/(dt)=((2t+1)20-20t(2))/(2t+1)^2

=20/((2t+1)^2)

At t = 2,

dx/dt = v_x= 20/25 = 0.8\ "ms"^-1.

Also, y = 0.1(t2 + t) so

(dy)/(dt)=0.1(2t+1)

When t = 2,

dy/dt = v_y= 0.5\ "ms"^-1.

So

v=sqrt((v_x)^2+(v_y)^2)

=sqrt(0.8^2+0.5^2)

=0.943\ text(ms)^-1

Now for the direction:

tan\ theta_v=(v_y)/(v_x)=0.5/0.8 = 0.625

So θ_v= arctan(0.625) = 0.558 radians

## Acceleration of a Body in Curvilinear Motion

The expressions for acceleration are very similar to those for velocity:

Horizontal component of acceleration:

a_x=(dv_x)/(dt)

Vertical component of acceleration:

a_y=(dv_y)/(dt)

Magnitude of acceleration:

a=sqrt((a_x)^2+(a_y)^2)

Direction of acceleration:

tan\ theta_a=(a_y)/(a_x)

### Example 4

A car on a test track goes into a turn described by x = 20 + 0.2t3, y = 20t − 2t2, where x and y are measured in metres and t in seconds.

(i) Sketch the curve for 0 ≤ t ≤ 8.

(ii) Find the acceleration of the car at t = 3.0 seconds.

(i) Sketch: We substitute values of t from 0 to 8 into the expressions for x and y and obtain the following after joining the dots.

Note the axes are x- and y-based, and do not involve t.

(ii) Acceleration:

Horizontal acceleration:

x = 20 + 0.2t3

v_x=(dx)/(dt)=0.6t^2

a_x=(d^2x)/(dt^2)=1.2t

At t = 3.0, ax = 3.6

Vertical acceleration:

y = 20t − 2t2

v_y=(dy)/(dt)=20-4t

a_y=(d^2y)/(dt^2)=-4

At t = 3.0, ay = -4

Now

a=sqrt((a_x)^2+(a_y)^2) =sqrt(3.6^2+(-4)^2) =5.38

and

theta_a=arctan((a_y)/(a_x)) =arctan\ ((-4)/3.6) =312^"o" [4th quadrant]

So the car's acceleration has magnitude 5.38 ms-2, and direction 312^@ from the positive x-axis.

## What if x and y are NOT given as functions of t?

### Example 5

A particle moves along the path y = x2 + 4x + 2 where units are in centimetres. If the horizontal velocity vx is constant at 3 cm s-1, find the magnitude and direction of the velocity of the particle at the point (-1, -1).

This is a different situation to the other examples. This time we have y in terms of x, and there are no expressions given in terms of "t" at all.

To be able to find magnitude and direction of velocity, we will need to know

v_x=(dx)/(dt)

and

v_y=(dy)/(dt)

But the question already gives us

v_x=(dx)/(dt)=3

so all we need to find is (dy)/(dt).

To find this, we differentiate the given function with respect to t throughout using the techniques we learned back in implicit differentiation:

y = x2 + 4x + 2

(dy)/(dt)=2x(dx)/(dt)+4(dx)/(dt)+0

Since

(dx)/(dt)=3

and we want to know the velocity at x = -1, we substitute these two values and get:

(dy)/(dt)=2(-1)(3)+4(3)=6

So now we have vy = 6 cm s-1.

So the magnitude of the velocity is given by:

v=sqrt((v_x)^2+(v_y)^2)=sqrt(3^2+6^2) =6.7082

The direction of the velocity is given by:

theta_v=arctan((v_y)/(v_x))=arctan(6/3) =63.432^"o"

So the velocity is 6.7 cm s-1, in the direction 63.4^@.

### Example 6

A rocket follows a path given by (distances in km):

y=x-x^3/90

If the horizontal velocity is given by V(x) = x, find the magnitude and direction of the velocity when the rocket hits the ground (assume level terrain) if time is in minutes.

Let's first see a graph of the motion, to better understand what is going on.

We can see that the rocket hits the ground again somewhere around x = 9.5 km. At this point, the horizontal velocity is positive (the rocket is going left to right) and the vertical velocity is negative (the rocket is going down).

"V(x) = x" means that as x increases, the horizontal velocity also increases with the same number (different units, of course). So for example, at x = 2 km, the horizontal speed is 2 km/min, and at x = 7 km, the horizontal speed is 7 km/min, and so on.

To calculate the magnitude of the velocity when the rocket hits the ground, we need to know the vertical and horizontal components of the velocity at that point.

(1) Horizontal velocity. We just need to solve the following equation to find the exact point the rocket hits the ground:

x-x^3/90=0

Factoring gives:

x-x^3/90=x(1-x^2/90)

And solving for 0 gives us x = 0, x = -3sqrt(10), x = 3sqrt(10)

We only need the last value, x = 3sqrt(10)~~ 9.4868\ "km" (This value is consistent with the graph above).

So the horizontal speed when the rocket hits the ground is 9.4868 km/min (since V(x) = x).

(2) Vertical velocity. We now need to use implicit differentiation with respect to t (not x!) to find the vertical velocity.

y=x-x^3/90

(dy)/(dt)=(dx)/(dt)-1/30x^2(dx)/(dt)

But we already know dx/(dt) and x at impact, so we simply substitute:

(dy)/(dt) =(9.48683298)- 1/30(9.48683298)^2(9.48683298)

This gives us a negative velocity, as we expected before:

(dy)/(dt)=-18.97366596

So now we need to find the magnitude of the velocity. This takes into account both the horizontal and vertical components.

"Magnitude"=sqrt(((dx)/(dt))^2+((dy)/(dt))^2

Substituting, we have:

sqrt((9.48683298)^2+(-18.97366596)^2) =21.21320344

Velocity has magnitude and direction. Now for the direction part.

"angle of motion" = arctan\ (dy"/"dt)/(dx"/"dt)

Substituting our vertical and horizontal components, we have:

arctan\ (-18.97366596)/9.48683298 =-1.107148718

In degrees, this is equivalent to

-1.107148718 × 57.25578  = -63.3907^@

We can see that this answer is reasonable by zooming in on the portion of the graph where the rocket hits the ground (with equal-axis scaling):

So in summary, the velocity of the rocket when it hits the ground is 21.2 km/min in the direction 63.4^@ below the horizontal.