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For more information, please see full course syllabus of Calculus BC

For more information, please see full course syllabus of Calculus BC

## Discussion

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## Practice Questions

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## Table of Contents

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### Parametric Differentiation

- Differentiate the 2 parametric parts separately
- Divide
- Simplify

### Parametric Differentiation

A parametric curve is defined by:

x(t) = 3t − 1

y(t) = 7 − 5t

Find [dy/dx]

x(t) = 3t − 1

y(t) = 7 − 5t

Find [dy/dx]

- Find [dx/dt]
- x(t) = 3t − 1
- x′(t) = 3
- Find [dy/dt]
- y(t) = 7 − 5t
- y′(t) = − 5

Divide y′(t) by x′(t) to find [dy/dx][dy/dx] = [( − 5)/3]

A parametric curve is defined by:

x(t) = [2/t] − 2

y(t) = [1/((t − 1)

Find [dy/dx]

x(t) = [2/t] − 2

y(t) = [1/((t − 1)

^{2})]Find [dy/dx]

- Find [dx/dt]
- x(t) = [2/t] − 2
- x′(t) = − 2(t)
^{ − 2} - Find [dy/dt]
- y(t) = [1/((t − 1)
^{2})] - y′(t) = − (t − 1)
^{ − 2} - Divide y′(t) by x′(t) to find [dy/dx]
- [dy/dx] = [( − (t − 1)
^{ − 2})/( − 2(t)^{ − 2})]

= [(2t

^{2})/((t − 1)^{2})]A parametric curve is defined by:

x(t) = 5t + 2

y(t) = √{6t}

Find [dy/dx], for t = 3

x(t) = 5t + 2

y(t) = √{6t}

Find [dy/dx], for t = 3

- Find [dx/dt]
- x(t) = 5t + 2
- x′(t) = 5
- Find [dy/dt]
- y(t) = √{6t}
- y′(t) = [1/(6√{6t} )]
- Divide y′(t) by x′(t) to find [dy/dx]
- [dy/dx] = [([1/(6√{6t} )])/5]
- = [1/(30√{6t} )]

Solve for t = 3[1/(30√{6t} )] = [1/(30√{6(3)} )] = [1/(30√{18} )]

A parametric curve is defined by:

y(t) = t

x(t) = cos([t/2] − 1)

Find [dy/dx] for t = 4

y(t) = t

^{2}+ 8t − 3x(t) = cos([t/2] − 1)

Find [dy/dx] for t = 4

- Find [dx/dt]
- x(t) = cos([t/2] − 1)
- x′(t) = − 2sin([t/2] − 1)
- Find [dy/dt]
- y(t) = t
^{2}+ 8t − 3 - y′(t) = 2t + 8
- Divide y′(t) by x′(t) to find [dy/dx]
- [dy/dx] = [(2t + 8)/( − 2sin([t/2] − 1))]
- = [(t + 4)/( − sin([t/2] − 1))]
- Solve for t = 4
- [(t + 4)/( − sin([t/2] − 1))] = [(4 + 4)/( − sin([4/2] − 1))]
- = [8/( − sin(1))]

= −9.51 (in rads)

If a function is defined parametrically as (3t

^{3,},t^{2}− 2t + 8), what is its [dy/dx]?- Find [dx/dt]
- x(t) = 3t
^{3,} - x′(t) = 9t
^{2} - Find [dy/dt]
- y(t) = t
^{2}− 2t + 8 - y′(t) = 2t − 2

Divide y′(t) by x′(t) to find [dy/dx][dy/dx] = [(2t − 2)/(9t

^{2})]If a particle's position is defined parametrically as (3t

^{3,},t^{2}− 2t + 8), what is it's speed at t = 6?- Use the definition of speed
- speed = √{([dx/dt])
^{2}+ ([dy/dt])^{2}} - = √{(9t
^{2})^{2}+ (2t − 2)^{2}}

Solve for t = 6√{(9t

^{2})^{2}+ (2t − 2)^{2}} = √{(9(6)^{2})^{2}+ (2(6) − 2)^{2}} = 20.59If a particle's position is defined parametrically as (3t

^{3,},t^{2}− 2t + 8), what is it's acceleration at t = 3?- Find [(d
^{2}x)/(dt^{2})] - [dx/dt] = 9t
^{2} - [(d
^{2}x)/(dt^{2})] = 18t - Find [(d
^{2}y)/(dt^{2})] - [dy/dt] = 2t − 2
- [(d
^{2}y)/(dt^{2})] = 2 - Use the definition of acceleration
- acceleration = < [(d
^{2}y)/(dt^{2})],[(d^{2}x)/(dt^{2})] > - = < 18t,2 >
- Solve for t = 3
- < 18t,2 > = < 18(3),2 >

= < 44,2 >

If a function is defined parametrically as ([(t

^{2})/2],t^{2}− 7t + 10), what is its [dy/dx]?- Find [dx/dt]
- x(t) = [(t
^{2})/2] - x′(t) = t
- Find [dy/dt]
- y(t) = t
^{2}− 7t + 10 - y′(t) = 2t − 7

Divide y′(t) by x′(t) to find [dy/dx][dy/dx] = [(2t − 7)/t]

If a particle's position is defined parametrically as ([(t

^{2})/2],t^{2}− 7t + 10), what is it's speed at t = 2?- Use the definition of speed
- speed = √{([dx/dt])
^{2}+ ([dy/dt])^{2}} - = √{(t)
^{2}+ (2t − 7)^{2}} - Solve for t = 2
- √{(t)
^{2}+ (2t − 7)^{2}} = √{(2)^{2}+ (2(2) − 7)^{2}}

=3.61

If a particle's position is defined parametrically as ([(t

^{2})/2],t^{2}− 7t + 10), what is it's acceleration at t = 3?- Find [(d
^{2}x)/(dt^{2})] - [dx/dt] = t
- [(d
^{2}x)/(dt^{2})] = 0 - Find [(d
^{2}y)/(dt^{2})] - [dy/dt] = 2t − 7
- [(d
^{2}y)/(dt^{2})] = 2 - Use the definition of acceleration
- acceleration = < [(d
^{2}y)/(dt^{2})],[(d^{2}x)/(dt^{2})] > - = < 0,2 >
- Solve for t = 3
- < 0,2 > = < 0,2 >

= < 0,2 >

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Parametric Differentiation

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Parametric Differentiation 0:15
- Example 1 1:16
- Example 2 1:54
- Example 3 3:15
- Example 4 4:59
- Parametric Differentiation: Position, Speed & Acceleration 7:45
- Example 5 8:32
- Example 6 10:37

1 answer

Last reply by: Acme Wang

Sun Jun 12, 2016 3:11 AM

Post by Luvivia Chang on April 1, 2015

In example 3, the question is asking for a tangent line, not necessary a tangent line that is horizontal. So the line should be an equation with variable t, not an exact point.

1 answer

Last reply by: Craig McRae

Wed Jul 31, 2013 2:51 PM

Post by chetachi ohagi on June 14, 2013

Where did he get the y prime derivative from in example 2?

2 answers

Last reply by: Luvivia Chang

Wed Apr 1, 2015 7:30 AM

Post by Christopher Lagos on May 3, 2012

Why is it that when solving for the acceleration you leave it as a coordinate rather than a single quantity.