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 1 answerLast reply by: Acme WangSun Jun 12, 2016 3:11 AMPost by Luvivia Chang on April 1, 2015In 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 answerLast reply by: Craig McRaeWed Jul 31, 2013 2:51 PMPost by chetachi ohagi on June 14, 2013Where did he get the y prime derivative from in example 2? 2 answersLast reply by: Luvivia ChangWed Apr 1, 2015 7:30 AMPost by Christopher Lagos on May 3, 2012Why is it that when solving for the acceleration you leave it as a coordinate rather than a single quantity.

### 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]
• 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)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)]
= [(2t2)/((t − 1)2)]
A parametric curve is defined by:
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) = t2 + 8t − 3
x(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) = t2 + 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))]
If a function is defined parametrically as (3t3,,t2 − 2t + 8), what is its [dy/dx]?
• Find [dx/dt]
• x(t) = 3t3,
• x′(t) = 9t2
• Find [dy/dt]
• y(t) = t2 − 2t + 8
• y′(t) = 2t − 2
Divide y′(t) by x′(t) to find [dy/dx][dy/dx] = [(2t − 2)/(9t2)]
If a particle's position is defined parametrically as (3t3,,t2 − 2t + 8), what is it's speed at t = 6?
• Use the definition of speed
• speed = √{([dx/dt])2 + ([dy/dt])2}
• = √{(9t2)2 + (2t − 2)2}
Solve for t = 6√{(9t2)2 + (2t − 2)2} = √{(9(6)2)2 + (2(6) − 2)2} = 20.59
If a particle's position is defined parametrically as (3t3,,t2 − 2t + 8), what is it's acceleration at t = 3?
• Find [(d2x)/(dt2)]
• [dx/dt] = 9t2
• [(d2x)/(dt2)] = 18t
• Find [(d2y)/(dt2)]
• [dy/dt] = 2t − 2
• [(d2y)/(dt2)] = 2
• Use the definition of acceleration
• acceleration = < [(d2y)/(dt2)],[(d2x)/(dt2)] >
• = < 18t,2 >
• Solve for t = 3
• < 18t,2 > = < 18(3),2 >
= < 44,2 >
If a function is defined parametrically as ([(t2)/2],t2 − 7t + 10), what is its [dy/dx]?
• Find [dx/dt]
• x(t) = [(t2)/2]
• x′(t) = t
• Find [dy/dt]
• y(t) = t2 − 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 ([(t2)/2],t2 − 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 ([(t2)/2],t2 − 7t + 10), what is it's acceleration at t = 3?
• Find [(d2x)/(dt2)]
• [dx/dt] = t
• [(d2x)/(dt2)] = 0
• Find [(d2y)/(dt2)]
• [dy/dt] = 2t − 7
• [(d2y)/(dt2)] = 2
• Use the definition of acceleration
• acceleration = < [(d2y)/(dt2)],[(d2x)/(dt2)] >
• = < 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.

### 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