Calculus 2 Lecture Notes, Section 9.2

Calc 2 Lecture NotesSection 9.2Page 1 of 11

Section 9.2: Calculus and Parametric Equations

Big idea: Finding the slope of tangent lines and the area enclosed by a curve described by parametric equations require slightly different formulas than when we had y = f(x).

Big skill: You should be able to calculate the first and second derivatives of a graph defined by parametric equations, the area of a closed graph, and the velocity and acceleration if the parametric equations describe the position of an object.

Let . Then the chain rule tells us that: , which means

First derivative of the graph produced by parametric equations:

If , then .

Note: If , then the graph has a horizontal tangent line.

Note: If , then the graph has a vertical tangent line.

Practice:

1. Find an expression for the slope of the tangent line to a unit circle at the origin described by parametric equations, and identify points with horizontal and vertical tangent lines.

1. Find an expression for the slope of the tangent line to the graph described by ,

and analyze tangent lines at points where or.

Second derivative of the graph produced by parametric equations:

The second derivative tells us what it always told us: the concavity of the graph and the location of critical points where the graph may change concavity.

Practice:

1. Find an expression for the second derivative of a unit circle at the origin described by parametric equations, and then analyze the expression in terms of concavity.

1. Find an expression for the second derivative of the graph described by , and then analyze the expression in terms of concavity.

Theorem 2.1:Vertical and Horizontal Tangent Lines of Parametric Equations

If andare continuous, then for the curve defined by :

(i).If and, then there is a horizontal tangent line at the point.

(ii).If and, then there is a vertical tangent line at the point.

Theorem 2.2: Area Enclosed by a Parametric Curve

Suppose that with describes a curve that is traced out clockwise exactly once as t increases from c to d, and where the curve does not intersect itself, except at the initial and terminal points (x(c) = x(d) and y(c) = y(d)).

Then the enclosed area is:.

If the area is traced out counterclockwise, then the enclosed area is:.

Practice:

1. Derive the formula for the area of a circle using Theorem 2.2.

1. Derive the formula for the area of an ellipse using Theorem 2.2.

1. Find the area enclosed by .

1. For , find the first and second derivatives as well as the enclosed area.

Analysis of the Scrambler Ride:

The Scrambler is an amusement park ride with a large central arm (of radius 2) that rotates about its center point and that has four shorter arms (of radius 1) at the end that rotate at twice the frequency. An acceptable model for the motion of one of the endpoints of the shorter arms is:

Scrambler Parametric Equations:

Scrambler Ride winplot files that can be downloaded:

Table of Positions for the Scrambler Ride:

Meaning of first derivatives when parametric equations represent an object’s position:

The horizontal component of velocity is: (assuming the x axis is “horizontal”).

The vertical component of velocity is: (assuming the y axis is “vertical”).

The direction of the velocity (and tangent to the curve) is: .

The magnitude of the velocity is: .

First Derivative of the Parametric Equations for the Scrambler Ride:

Note: …

Magnitude of the Velocity of the Scrambler Ride:

Note:  zero velocity at the cusp points

Slope of the Tangent Line at the t=/6 cusp for the Scrambler Ride:

 equation of the tangent line is ; this line is inclined at 30 above the x-axis…

Meaning of second derivatives when parametric equations represent an object’s position:

The horizontal component of acceleration is: (assuming the x axis is “horizontal”). This can be used to find the horizontal component of force, since .

The vertical component of acceleration is: (assuming the y axis is “vertical”). This can be used to find the vertical component of force, since .

The direction of the acceleration (and force, since ) is: , where a is the standard angle that the acceleration vector makes with respect to the x-axis.

The magnitude of the acceleration is: . This can be used to find the magnitude of the force, since .

Second Derivative of the Parametric Equations for the Scrambler Ride:

Magnitude of the Acceleration of the Scrambler Ride:

Note: a(t) has its maxima at  greatest force is felt when the ride is “at rest” at the cusps.

Magnitude of the “Jerk” of the Scrambler Ride:

Note: J(t) has its maxima at  greatest jerk is also felt at the cusps.

Direction of the Acceleration of the Scrambler Ride at the t=/6 cusp:

Note: Theacceleration is directed radially inward  a purely centripetal force at this time

Direction of the Acceleration of the Scrambler Ride at the t=/2 midpoint of the top curved leg:

Note: Again, theacceleration is directed radially inward  a purely centripetal force at this time

Comparison of the Direction of the Acceleration of the Scrambler Ride at t=/4 with the direction of the velocity:

Note: Since the directions are not parallel, some of the force is producing a centripetal acceleration, while the remaining force is changing the speed.

Second Derivative of the Graph of the Scrambler Ride:

Note: the graph changes concavity at the cusps where , and the second derivative has critical points when .