AP Calculus BC Lesson 9.3

Elizabeth Fein
14 Apr 202310:45
EducationalLearning
32 Likes 10 Comments

TLDRThis video lesson delves into the concept of arc length for parametric equations, emphasizing the integral formula for calculating it. The script walks through several examples, demonstrating how to set up and solve integrals for the arc length of curves defined by parametric equations over given time intervals. It also touches on using calculators for these calculations and hints at future topics like velocity vectors.

Takeaways
  • πŸ“š The arc length formula for parametric equations is discussed, which is similar to the one from lesson 8.13: L = ∫(√(1 + (dx/dt)^2 + (dy/dt)^2)) dt from time A to time B.
  • πŸŒ€ The script provides an example of finding the arc length of a curve defined by parametric equations x(t) = 5t^3 - t and y(t) = t^2/3 from t = 0 to t = 2, resulting in an arc length of 38.382.
  • πŸ” The process of setting up and solving the integral for the arc length is demonstrated, including finding the derivatives with respect to the parameter t.
  • πŸ“ˆ Another example is given for a curve defined by x(t) = 5t and y(t) = 2t^3 from t = 0 to t = 4, which matches answer choice A when set up correctly.
  • 🌐 The script explains how to find the arc length of a path described by x(t) = 2cos(t) and y(t) = 3sin(t) from Ο€/2 to Ο€, with the correct answer being choice C.
  • πŸ“Š The length of a curve traced by x(t) = 4e^t and y(t) = 5.4cos(e^t) from t = 0 to t = 2 is calculated to be 34.875.
  • πŸš€ A calculator-free response question is discussed, involving a particle moving along a curve with given dx/dt and dy/dt, from time T = 4 to T = 6, resulting in a distance of 12.136.
  • 🌟 The concept of velocity vectors is introduced, which will be formally discussed in a future video.
  • πŸ›€οΈ The total distance traveled by a particle is synonymous with arc length when dealing with parametric equations.
  • πŸ“ The script concludes with an example of finding the total distance traveled by a particle from time T = 0 to T = 1, with the answer being 4.073.
  • πŸ“‹ The script emphasizes the importance of setting up the integral correctly and using the arc length formula to find the length of curves defined by parametric equations.
Q & A
  • What is the arc length formula for parametric equations?

    -The arc length formula for parametric equations is given by L = ∫ from A to B of the square root of (DX/DT)^2 + (DY/DT)^2, where A and B represent different times for the parametric functions.

  • How is the arc length formula derived?

    -The arc length formula for parametric equations is derived from the general arc length formula for functions, which is similar in form. However, the exact derivation is not covered in the video transcript provided.

  • What are the parametric equations for the curve whose length is to be found in the first example?

    -The parametric equations for the curve in the first example are X(T) = 5T^3 - T and Y(T) = T^2/3, from T = 0 to T = 2.

  • What is the integral set up to calculate the arc length of the curve in the first example?

    -The integral set up to calculate the arc length is ∫ from 0 to 2 of the square root of (15T^2 - 1)^2/T^2 + (2/3T)^2 with respect to T.

  • What is the numerical result for the arc length of the curve in the first example?

    -The numerical result for the arc length of the curve in the first example is approximately 38.382 units.

  • What are the parametric equations for the curve in the second example?

    -The parametric equations for the curve in the second example are X(T) = 5T and Y(T) = 2T^3, from T = 0 to T = 4.

  • How do you find the derivatives for the arc length integral in the second example?

    -The derivatives for the arc length integral in the second example are found by differentiating the parametric equations with respect to T, resulting in DX/DT = 5 and DY/DT = 6T^2.

  • What is the integral set up to calculate the arc length of the curve in the second example?

    -The integral set up to calculate the arc length is ∫ from 0 to 4 of the square root of 5^2 + (6T^2)^2 with respect to T.

  • What are the parametric equations for the curve in the third example?

    -The parametric equations for the curve in the third example are X(T) = 4e^T and Y(T) = 5.4cos(e^T), from T = 0 to T = 2.

  • What is the numerical result for the arc length of the curve in the third example?

    -The numerical result for the arc length of the curve in the third example is approximately 34.875 units.

  • How do you calculate the total distance traveled by a particle along a curve?

    -The total distance traveled by a particle along a curve is calculated using the arc length formula for parametric equations, where the functions represent the particle's position with respect to time.

  • What is the velocity vector given for the particle in the last example?

    -The velocity vector given for the particle in the last example is V(T) = sin(e^0.36T), 7.8T, representing the rates of change of the x and y positions with respect to time T.

Outlines
00:00
πŸ“š Calculating Arc Length of Parametric Equations

This paragraph introduces the concept of arc length for parametric equations, referencing the arc length formula from a previous lesson. It explains that the formula for parametric equations is similar to the one for functions, with the integral running from time A to time B. An example is provided where the arc length is calculated for a specific set of parametric equations from time 0 to 2. The process involves setting up the integral and using a calculator to find the numerical answer, which is given as 38.382.

05:02
πŸ“ Arc Length Formula Application for Different Parametric Equations

The paragraph continues with the application of the arc length formula for parametric equations. It provides two additional examples: one involving the parametric equations X of T equals 5T and Y of T equals 2T cubed from T equals 0 to T equals 4, and another involving X equals 2 times the cosine of T and Y equals 3 times the sine of T from T equals Ο€/2 to T equals Ο€. For each, the paragraph outlines the steps to find the derivatives and set up the integral to calculate the arc length, matching the results with given answer choices.

10:03
🌐 Length of a Curve Traced by Parametric Equations

This section discusses the calculation of the length of a curve traced by parametric equations X of T equals 4e to the power of T and Y of T equals 5.4 times the cosine of e to the power of T from T equals 0 to T equals 2. The integral is set up for this interval, and the derivatives are calculated to find the arc length, which is given as 34.875. The paragraph also includes a calculator-free response question related to the topic, where the arc length formula is used to find the distance a particle travels along a curve from time T equals 4 to T equals 6.

πŸš€ Total Distance Traveled by a Moving Particle

The final paragraph addresses the calculation of the total distance traveled by a particle moving along a curve in the XY plane. Given the position X of t y of t and the velocity vector V of T, the paragraph explains how to use the arc length formula to find the distance from time T equals 0 to T equals 1. The integral is set up using the velocity components as derivatives, and the calculation yields a distance of 4.073. The paragraph emphasizes that the units are not specified, so the answer is presented as a numerical value.

Mindmap
Solution
Problem Statement
Introduction
Solution
Problem Statement
Example 4
Example 3
Example 2
Example 1
Formula Application
Definition
Arc Length Formula
Lesson Context
Velocity Vectors
Calculator-Free Response
Examples and Calculations
Parametric Equations and Arc Length
Introduction
Arc Length of Parametric Equations
Alert
Keywords
πŸ’‘arc length
Arc length is the distance along the curve of a function between two points. In the context of the video, it refers to the formula used to calculate the length of a curve defined by parametric equations. The formula is integral from A to B of the square root of 1 plus F'(x)^2 dx, where A and B are the starting and ending points on the curve.
πŸ’‘parametric equations
Parametric equations are a set of equations that determine the coordinates of a point moving in a plane or space. They are used when the relationship between the x and y coordinates cannot be expressed directly. In the video, parametric equations are used to define curves for which the arc length is being calculated.
πŸ’‘integral
An integral is a mathematical concept used to calculate the area under a curve or the accumulation of a quantity over an interval. In the video, integrals are used to find the arc length of curves defined by parametric equations.
πŸ’‘derivative
A derivative is a mathematical operation that finds the rate of change or slope of a function at a given point. In the context of the video, derivatives are used to find the rate of change of the x and y coordinates with respect to time in parametric equations, which is necessary for calculating arc length.
πŸ’‘square root
The square root of a number is a value that, when multiplied by itself, gives the original number. In the video, the square root is used in the arc length formula to find the magnitude of the change in the function's values over the interval.
πŸ’‘time
In the context of parametric equations, time is the independent variable that parameterizes the system. It represents the progression along the curve, and different values of time correspond to different points on the curve.
πŸ’‘calculus
Calculus is a branch of mathematics that deals with rates of change and accumulation. It is the foundation for understanding the concepts of derivatives and integrals, which are used in the video to calculate arc lengths of parametric curves.
πŸ’‘graphing calculator
A graphing calculator is a specialized electronic device used for mathematical calculations, including those involving integrals and derivatives. In the video, a graphing calculator is used to perform the calculations necessary to find the arc lengths of parametric curves.
πŸ’‘chain rule
The chain rule is a technique in calculus used to find the derivative of a composite function. It states that the derivative of a function composed of two or more functions is the product of the derivative of the outer function and the derivative of the inner function.
πŸ’‘natural log
The natural log, or logarithm base e, is a mathematical function that finds the power to which e (Euler's number, approximately 2.71828) must be raised to obtain a given value. It is used in various mathematical and scientific contexts, including in the calculation of arc lengths in the video.
πŸ’‘velocity vector
A velocity vector is a vector that represents the speed and direction of an object's motion. It is used in physics and related fields to describe the motion of particles or objects over time. In the video, the concept is mentioned in relation to a particle moving along a curve in the XY plane.
Highlights

Discussion of arc length for parametric equations.

Recall of the arc length formula from a previous lesson.

Explanation of the integral formula for calculating arc length.

Use of different variables (A and B) to represent starting and ending x values in the arc length formula.

Transition to the arc length formula for parametric equations involving different times (A and B).

Example calculation of the arc length of a curve defined by specific parametric equations.

Integration of the square root of the sum of derivatives squared over a time interval.

Demonstration of how to set up a problem for calculator integration.

Mention of using calculator's parametric mode for solving problems.

Another example involving finding the arc length of a curve defined by different parametric equations.

Explanation of the process to find the arc length of a path described by parametric equations involving trigonometric functions.

Calculation of the arc length of a curve traced by exponential and trigonometric functions.

Illustration of how to set up and solve a calculator-free response question involving arc length.

Description of a particle's movement along a curve and the calculation of its total distance traveled.

Use of the arc length formula to find the distance a particle travels between two time points.

Inference of velocity vector components from given parametric equations.

Calculation of the total distance traveled by a particle using its velocity vector components.

Final answer provided for the distance traveled by a particle over a given time interval.

Transcripts
00:04

welcome to lesson 9.3 in this video

00:07

we'll be discussing arc length for

00:09

parametric equations

00:10

recall the arc length formula from

00:12

lesson 8.13 L is equal to the integral

00:14

from A to B of the square root of 1 plus

00:17

F Prime of x squared DX

00:20

now a would be your starting x value and

00:22

B would be your ending x value for

00:24

parametric equations the arc length

00:26

formula is very similar and this formula

00:28

actually comes from this formula we

00:30

don't have time to go over that one in

00:31

this video but that's where it comes

00:33

from for parametric equations the arc

00:35

length formula is L is equal to the

00:37

integral from A to B and in this case A

00:40

and B represent different times so we

00:42

could label this one t equals time one

00:45

and this one t equals time two now we

00:49

would be integrating from time one to

00:50

time 2 of the square root of DX to t

00:53

squared plus d y DT squared and that all

00:56

has a DT on the end now as we mentioned

00:58

a represents the starting time and B

01:00

represents the ending time let's try an

01:03

example find the length of the curve

01:05

defined by the parametric equations X of

01:07

T equals five T cubed minus t and Y of T

01:10

equals t squared over 3 from T equals 0

01:13

to T equals 2.

01:15

since this is a calculator question all

01:17

we have to do is really set up this

01:18

problem and then the calculator will do

01:19

the integrating for us we would say that

01:22

the length is equal to the integral from

01:24

0 to 2 since that's our starting and

01:26

ending times respectively of the square

01:29

root of and then what's DX DT the

01:32

derivative of x with respect to T we

01:34

could make the calculator do this for us

01:36

but it's really pretty easy we have 5 T

01:38

cubed minus t which means that the

01:40

derivative is going to be 15 t squared

01:42

minus 1. so that's our DX DT squared

01:45

plus our d y DT and in this case if

01:49

we're taking the derivative of t squared

01:51

over 3 that's going to be two-thirds T

01:55

then we square that because we take d y

01:57

d t squared and we stick a DT on the end

01:59

of this entire integral then we can get

02:01

out the graphing calculator and it will

02:03

give us a numerical answer for that now

02:05

if you prefer to work through these in

02:07

parametric mode you can hit the mode

02:09

button and then your calculator you can

02:10

switch it over to parametric instead of

02:12

just regular functions if you're using

02:14

regular functions you would just have to

02:16

put an X in place of a t if you're okay

02:17

with that you can leave it as a function

02:19

it's going to be the same answer no

02:21

matter what

02:22

so to get the integral we'll say math

02:25

and then hit number nine that's going to

02:26

give us our integral from 0 to 2 then we

02:29

put in the square root of and then we

02:31

just enter all of this stuff again it's

02:34

putting the T As an X in there but

02:36

that's completely fine

02:39

and it says the length is 38.382

02:44

that would be our answer

02:46

which of the following gives the length

02:47

of the curve defined by the parametric

02:49

equations X of T equals 5T and Y of T

02:52

equals 2T cubed from T equals 0 to T

02:54

equals 4. in order to get this problem

02:56

correct you have to know the arc length

02:58

formula for parametric equations that

03:00

length is going to be the the integral

03:02

from Time 1 to time two

03:04

of the square root of our X Prime of t

03:08

or you could say DX DT either one of

03:10

those works here I'll go with prime so

03:12

you have X Prime of t squared plus y

03:14

Prime of T also squared with a DT on the

03:17

end so then it's just a matter of

03:19

plugging everything in right here in

03:21

this case our starting time is zero and

03:22

our ending time is 4. so we would say

03:24

integral from 0 to 4. of the square root

03:27

of and then what's our X Prime of T the

03:30

derivative of this x of T that would be

03:32

a 5 since the derivative of 5T is a 5.

03:35

so we say DX DT squared plus and then y

03:38

Prime of t or d y DT what's the

03:40

derivative of 2T cubed that's going to

03:43

be 6 t squared and we're going to square

03:45

that entire thing as well and stick a DT

03:47

on the end then we just have to clean

03:49

this up a little bit to make it match

03:50

one of our answer choices we're still

03:52

integrating from zero to four five

03:54

squared is going to be 25 6t squared

03:57

squared is going to be 36 t to the

03:59

fourth and we have that DT on the end

04:01

this matches answer Choice a

04:05

which of the following gives the length

04:07

of the path described by the parametric

04:08

equations x equals 2 times the cosine of

04:11

T and Y equals 3 times the sine of T

04:13

over 3. for pi over 2 is less than or

04:16

equal to T is less than or equal to Pi

04:18

in this case this is our starting time

04:20

T1 and this is our ending time T2 so

04:23

when we set up our integral we're going

04:24

to say our length is equal to the

04:26

integral from pi over 2 to Pi since that

04:28

is our lower bound and our upper bound

04:30

and then we're going to take the square

04:32

root of DX DT squared plus d y d t

04:35

squared DX DT means that we have to find

04:37

the derivative of this equation with

04:39

respect to T that's going to be negative

04:41

2 times the sine of T and we square that

04:44

and then we take the derivative of this

04:46

equation and we Square it the derivative

04:48

of 3 times the sine of T over 3 we're

04:51

going to have to use the chain rule a

04:52

bit there that would be 3 times the

04:54

cosine of T over 3 multiplied by the

04:56

derivative of that inside function the T

04:58

over 3 which is a one-third then we can

05:01

square that as well and we stick a DT on

05:03

the end of the integral now in the

05:05

situation since we have a 3 times 1 3

05:07

both of those are going to cancel and we

05:08

just have cosine of T over 3 left then

05:11

we can clean this up a bit to make it

05:12

match one of our answer choices

05:14

negative 2 squared makes a 4 and the

05:17

sine of t squared we can just write that

05:19

as sine squared of T then to clean up

05:22

this one if we have cosine of T over 3

05:24

squared we can just write that as cosine

05:26

squared of T over 3. and we still have

05:29

that DT on the end of the integral then

05:31

we just figure out this matches which

05:33

answer choice and it looks like in this

05:34

case it matches answer Choice C so C is

05:36

our correct answer here

05:38

what is the length of the curve traced

05:40

by the parametric equations X of T

05:42

equals 4 e to the power of T and Y of T

05:44

equals 5.4 times the cosine of e to the

05:47

power of T from T equals 0 to T equals

05:49

two let's begin by setting up our

05:51

integral since we're going from T equals

05:53

zero to T equals two we'll do the

05:54

integral from 0 to 2. then since we're

05:57

plugging this one into the calculator we

05:59

could make the calculator get these

06:01

derivatives for us but neither of the

06:03

functions are so bad that we can't find

06:04

the derivatives ourselves so let's make

06:06

it a little bit easier for plugging it

06:07

into the calculator and we'll take this

06:09

DX DT right now what's X Prime of t or

06:12

DX DT well 4 e to the power of T is its

06:15

own derivative so we're going to have 4

06:17

e to the power of t squared plus the

06:19

derivative of y of T which is going to

06:22

be Let's see we would have negative 5.4

06:25

times the sine of e to the power of T

06:27

times the derivative of the inside

06:29

function which is going to be that e to

06:31

the power of T and that entire thing

06:33

would be squared then we would stick a

06:35

DT on the end then we can get out the

06:38

calculator and plug this in

06:40

use math9 to get the integral and we're

06:43

going from zero to two don't forget the

06:44

square root and then just plug all of

06:46

this in

06:51

that gives us an answer of 34.875 so in

06:55

this case answer Choice C would be our

06:56

correct response

06:59

let's write part of a calculator free

07:00

response question involving this topic

07:03

the stem of this question is the same

07:05

the one that we worked through in a

07:06

different video but the question it's

07:07

asking us is a bit different now we're

07:09

doing Part D a particle moving along a

07:11

curve in an X Y plane is at position X

07:13

of t y of T at time T is greater than

07:15

zero the particle moves in such a way

07:18

that DX DT is equal to the square root

07:20

of one plus t squared and d y DT equals

07:22

the natural log of 2 plus t squared at

07:25

time T equals four the particle is at

07:27

the point one comma five

07:28

Part D says find the total distance the

07:31

particle travels along the curve from

07:33

time T equals four to T equals six

07:35

now if the particle is moving along this

07:37

curve the distance it travels is going

07:39

to be the length of the curve from T

07:41

equals 4 to T equals six so what we're

07:43

going to do in this case I'm going to

07:45

use D instead of L but it's the same

07:46

thing we're going to use that arc length

07:48

formula we're looking for the integral

07:50

from T equals 4 to T equals 6 of the

07:53

square root of DX DT squared plus d y DT

07:57

squared so we would take our DX DT which

08:00

is the square root of 1 plus t squared

08:02

and square that plus d y DT which is the

08:05

natural log of 2 plus t squared and we

08:08

square that and then we stick our DT on

08:10

the end then we can plug this into the

08:13

calculator fairly easily and get our and

08:15

get our response again use math number

08:17

nine and integrate from four to six

08:20

don't forget the square root and then

08:22

plug this all in now for the square root

08:24

of 1 plus t squared squared you can just

08:26

plug in 1 plus and in this case it's

08:28

going to put it in as X but it's the

08:29

same thing we don't need to do the

08:31

square root and then Square it that's

08:32

kind of redundant

08:34

that gives us 12.136 so 12.136 is the

08:39

distance that that particle traveled

08:40

along the curve from time T equals 4 to

08:42

T equals six

08:44

since they didn't give us a unit here

08:46

our answer is just going to be that

08:49

at time T is greater than or equal to

08:51

zero a particle moving along a curve in

08:53

the X Y plane has position X of t y of t

08:56

with velocity Vector given by V of T

08:58

equals sine of e to the power of 0.36 T

09:02

comma 7.8 T I know that we haven't

09:04

formally discussed velocity vectors

09:06

that's coming up in a future video but

09:08

we can infer pretty easily what it means

09:10

in this situation now we're trying to

09:12

find the total distance traveled by the

09:14

particle from time T equals 0 to T

09:16

equals one so when it says this is our

09:18

velocity Vector the velocity of our x

09:20

coordinate is going to be given by this

09:22

and the velocity of our y coordinate is

09:24

going to be given by this so really what

09:26

we could do to make it easier we could

09:27

think X Prime of T is going to be this

09:31

this right here the sine of e to the

09:34

power of 0.36 T and then our y Prime of

09:38

t or d y DT is going to be this 7.8 T

09:41

since that's what they've plugged in for

09:43

the velocity for X and Y now if we're

09:46

trying to find the total distance

09:47

traveled remember total distance

09:49

traveled is going to be synonymous with

09:50

Arc Length when we're talking about

09:52

parametric equations and we're looking

09:54

to find that total distance from time T

09:55

equals 0 to T equals one so we'll say

09:58

the distance

09:59

is equal to the integral from 0 to 1 of

10:02

the square root of and then we take our

10:04

X Prime of T which is the sine of e to

10:07

the power of 0.36 T and that gets

10:10

squared plus our y Prime of t or our d y

10:14

DT which is 7.8 t squared and then we

10:18

just stick our DT on the end then we can

10:21

make our calculator spit out a value for

10:22

us right here again use math 9 to

10:25

integrate go from 0 to 1 and don't

10:27

forget the square root then just plug

10:29

all this in

10:31

that's equal to 4.073 so 4.073 is going

10:36

to be our final answer that's the

10:37

distance traveled by the particle from

10:39

time T equals 0 to T equals one and

10:41

since they didn't give us any units

10:42

we're good like that

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