AP Calculus BC Lesson 9.3
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
- π The arc length formula for parametric equations is discussed, which is similar to the one from lesson 8.13: L = β«(β(1 + (dx/dt)^2 + (dy/dt)^2)) dt from time A to time B.
- π 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.
- π 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.
- π 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.
- π 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.
- π 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.
- π 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.
- π The concept of velocity vectors is introduced, which will be formally discussed in a future video.
- π€οΈ The total distance traveled by a particle is synonymous with arc length when dealing with parametric equations.
- π 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.
- π 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 = β« 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 β« 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 β« 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
π 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.
π 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.
π 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.
π 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
Keywords
π‘arc length
π‘parametric equations
π‘integral
π‘derivative
π‘square root
π‘time
π‘calculus
π‘graphing calculator
π‘chain rule
π‘natural log
π‘velocity vector
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
welcome to lesson 9.3 in this video
we'll be discussing arc length for
parametric equations
recall the arc length formula from
lesson 8.13 L is equal to the integral
from A to B of the square root of 1 plus
F Prime of x squared DX
now a would be your starting x value and
B would be your ending x value for
parametric equations the arc length
formula is very similar and this formula
actually comes from this formula we
don't have time to go over that one in
this video but that's where it comes
from for parametric equations the arc
length formula is L is equal to the
integral from A to B and in this case A
and B represent different times so we
could label this one t equals time one
and this one t equals time two now we
would be integrating from time one to
time 2 of the square root of DX to t
squared plus d y DT squared and that all
has a DT on the end now as we mentioned
a represents the starting time and B
represents the ending time let's try an
example find the length of the curve
defined by the parametric equations X of
T equals five T cubed minus t and Y of T
equals t squared over 3 from T equals 0
to T equals 2.
since this is a calculator question all
we have to do is really set up this
problem and then the calculator will do
the integrating for us we would say that
the length is equal to the integral from
0 to 2 since that's our starting and
ending times respectively of the square
root of and then what's DX DT the
derivative of x with respect to T we
could make the calculator do this for us
but it's really pretty easy we have 5 T
cubed minus t which means that the
derivative is going to be 15 t squared
minus 1. so that's our DX DT squared
plus our d y DT and in this case if
we're taking the derivative of t squared
over 3 that's going to be two-thirds T
then we square that because we take d y
d t squared and we stick a DT on the end
of this entire integral then we can get
out the graphing calculator and it will
give us a numerical answer for that now
if you prefer to work through these in
parametric mode you can hit the mode
button and then your calculator you can
switch it over to parametric instead of
just regular functions if you're using
regular functions you would just have to
put an X in place of a t if you're okay
with that you can leave it as a function
it's going to be the same answer no
matter what
so to get the integral we'll say math
and then hit number nine that's going to
give us our integral from 0 to 2 then we
put in the square root of and then we
just enter all of this stuff again it's
putting the T As an X in there but
that's completely fine
and it says the length is 38.382
that would be our answer
which of the following gives the length
of the curve defined by the parametric
equations X of T equals 5T and Y of T
equals 2T cubed from T equals 0 to T
equals 4. in order to get this problem
correct you have to know the arc length
formula for parametric equations that
length is going to be the the integral
from Time 1 to time two
of the square root of our X Prime of t
or you could say DX DT either one of
those works here I'll go with prime so
you have X Prime of t squared plus y
Prime of T also squared with a DT on the
end so then it's just a matter of
plugging everything in right here in
this case our starting time is zero and
our ending time is 4. so we would say
integral from 0 to 4. of the square root
of and then what's our X Prime of T the
derivative of this x of T that would be
a 5 since the derivative of 5T is a 5.
so we say DX DT squared plus and then y
Prime of t or d y DT what's the
derivative of 2T cubed that's going to
be 6 t squared and we're going to square
that entire thing as well and stick a DT
on the end then we just have to clean
this up a little bit to make it match
one of our answer choices we're still
integrating from zero to four five
squared is going to be 25 6t squared
squared is going to be 36 t to the
fourth and we have that DT on the end
this matches answer Choice a
which of the following gives the length
of the path described by the parametric
equations x equals 2 times the cosine of
T and Y equals 3 times the sine of T
over 3. for pi over 2 is less than or
equal to T is less than or equal to Pi
in this case this is our starting time
T1 and this is our ending time T2 so
when we set up our integral we're going
to say our length is equal to the
integral from pi over 2 to Pi since that
is our lower bound and our upper bound
and then we're going to take the square
root of DX DT squared plus d y d t
squared DX DT means that we have to find
the derivative of this equation with
respect to T that's going to be negative
2 times the sine of T and we square that
and then we take the derivative of this
equation and we Square it the derivative
of 3 times the sine of T over 3 we're
going to have to use the chain rule a
bit there that would be 3 times the
cosine of T over 3 multiplied by the
derivative of that inside function the T
over 3 which is a one-third then we can
square that as well and we stick a DT on
the end of the integral now in the
situation since we have a 3 times 1 3
both of those are going to cancel and we
just have cosine of T over 3 left then
we can clean this up a bit to make it
match one of our answer choices
negative 2 squared makes a 4 and the
sine of t squared we can just write that
as sine squared of T then to clean up
this one if we have cosine of T over 3
squared we can just write that as cosine
squared of T over 3. and we still have
that DT on the end of the integral then
we just figure out this matches which
answer choice and it looks like in this
case it matches answer Choice C so C is
our correct answer here
what is the length of the curve traced
by the parametric equations X of T
equals 4 e 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
two let's begin by setting up our
integral since we're going from T equals
zero to T equals two we'll do the
integral from 0 to 2. then since we're
plugging this one into the calculator we
could make the calculator get these
derivatives for us but neither of the
functions are so bad that we can't find
the derivatives ourselves so let's make
it a little bit easier for plugging it
into the calculator and we'll take this
DX DT right now what's X Prime of t or
DX DT well 4 e to the power of T is its
own derivative so we're going to have 4
e to the power of t squared plus the
derivative of y of T which is going to
be Let's see we would have negative 5.4
times the sine of e to the power of T
times the derivative of the inside
function which is going to be that e to
the power of T and that entire thing
would be squared then we would stick a
DT on the end then we can get out the
calculator and plug this in
use math9 to get the integral and we're
going from zero to two don't forget the
square root and then just plug all of
this in
that gives us an answer of 34.875 so in
this case answer Choice C would be our
correct response
let's write part of a calculator free
response question involving this topic
the stem of this question is the same
the one that we worked through in a
different video but the question it's
asking us is a bit different now we're
doing Part D a particle moving along a
curve in an X Y plane is at position X
of t y of T at time T is greater than
zero the particle moves in such a way
that DX DT is equal to the square root
of one plus t squared and d y DT equals
the natural log of 2 plus t squared at
time T equals four the particle is at
the point one comma five
Part D says find the total distance the
particle travels along the curve from
time T equals four to T equals six
now if the particle is moving along this
curve the distance it travels is going
to be the length of the curve from T
equals 4 to T equals six so what we're
going to do in this case I'm going to
use D instead of L but it's the same
thing we're going to use that arc length
formula we're looking for the integral
from T equals 4 to T equals 6 of the
square root of DX DT squared plus d y DT
squared so we would take our DX DT which
is the square root of 1 plus t squared
and square that plus d y DT which is the
natural log of 2 plus t squared and we
square that and then we stick our DT on
the end then we can plug this into the
calculator fairly easily and get our and
get our response again use math number
nine and integrate from four to six
don't forget the square root and then
plug this all in now for the square root
of 1 plus t squared squared you can just
plug in 1 plus and in this case it's
going to put it in as X but it's the
same thing we don't need to do the
square root and then Square it that's
kind of redundant
that gives us 12.136 so 12.136 is the
distance that that particle traveled
along the curve from time T equals 4 to
T equals six
since they didn't give us a unit here
our answer is just going to be that
at time T is greater than or equal to
zero a particle moving along a curve in
the X Y plane has position X of t y of t
with velocity Vector given by V of T
equals sine of e to the power of 0.36 T
comma 7.8 T I know that we haven't
formally discussed velocity vectors
that's coming up in a future video but
we can infer pretty easily what it means
in this situation now we're trying to
find the total distance traveled by the
particle from time T equals 0 to T
equals one so when it says this is our
velocity Vector the velocity of our x
coordinate is going to be given by this
and the velocity of our y coordinate is
going to be given by this so really what
we could do to make it easier we could
think X Prime of T is going to be this
this right here the sine of e to the
power of 0.36 T and then our y Prime of
t or d y DT is going to be this 7.8 T
since that's what they've plugged in for
the velocity for X and Y now if we're
trying to find the total distance
traveled remember total distance
traveled is going to be synonymous with
Arc Length when we're talking about
parametric equations and we're looking
to find that total distance from time T
equals 0 to T equals one so we'll say
the distance
is equal to the integral from 0 to 1 of
the square root of and then we take our
X Prime of T which is the sine of e to
the power of 0.36 T and that gets
squared plus our y Prime of t or our d y
DT which is 7.8 t squared and then we
just stick our DT on the end then we can
make our calculator spit out a value for
us right here again use math 9 to
integrate go from 0 to 1 and don't
forget the square root then just plug
all this in
that's equal to 4.073 so 4.073 is going
to be our final answer that's the
distance traveled by the particle from
time T equals 0 to T equals one and
since they didn't give us any units
we're good like that
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