Chapter Contents

• Applications of Differentiation
• 1. Tangents and Normals
• 2. Newton's Method
• 3. Curvilinear Motion
• 4. Related Rates
• 5. Curve Sketching Using Differentiation
• 6. More Curve Sketching Using Differentiation
• 7. Applied Maximum and Minimum Problems
• 8. Radius of Curvature
• Applications of Differentiation Problem Solver

• Comments, Questions?

Math activity mode

Recommendation

Easy to understand calculus lessons on DVD. Try before you commit. More info:
MathTutorDVD.com

Online Algebra Solver

Solve your algebra problem step by step!

Online Algebra Solver »

Share

Share this page.

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

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

for t = 0 to 2π 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	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5	6.0
x(t)	0	.48	.84	1.0	.91	.60	.14	-.35	-.76	-.98	-.96	-.71	-.28
y(t)	1	.88	.54	.07	-.42	-.80	-.99	-.94	-.65	-.21	.28	.71	.96

Answer

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.

Answer

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.

Answer

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.

Answer

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 v_x is constant at 3 cm s^-1, find the magnitude and direction of the velocity of the particle at the point (-1, -1).

Answer

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.

Answer

Now let's return to the example we saw at the beginning of the calculus section. We have not seen how to differentiate sin x and cos x yet, but let's get LiveMath to do it for us:

LIVEMath