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Arc length of petal of polar graph - Intregration, Calculus, Mathematics Video Lecture - JEE
FAQs on Arc length of petal of polar graph - Intregration, Calculus, Mathematics Video Lecture - JEE
| 1. How do you calculate the arc length of a petal in a polar graph? |  |
Ans. To calculate the arc length of a petal in a polar graph, you can use the formula:
\[L = \int_{\theta_1}^{\theta_2} \sqrt{r^2 + \left(\frac{{dr}}{{d\theta}}\right)^2} d\theta\]
where \(r\) is the polar function and \(\frac{{dr}}{{d\theta}}\) is the first derivative of \(r\) with respect to \(\theta\). The integral is taken over the interval from \(\theta_1\) to \(\theta_2\) that corresponds to the petal.
| 2. How can we determine the value of \(\theta\) for a specific point on a polar graph? | |
Ans. To determine the value of \(\theta\) for a specific point on a polar graph, you can use the inverse tangent function (\(\tan^{-1}\)). If the coordinates of the point are \((r, \theta)\), then you can calculate \(\theta\) using the formula:
\[\theta = \tan^{-1}\left(\frac{{y}}{{x}}\right)\]
where \(x\) and \(y\) are the Cartesian coordinates of the point.
| 3. Can we use the arc length formula for any polar graph? | |
Ans. Yes, the arc length formula mentioned earlier can be used for any polar graph. It is a general formula that calculates the arc length of any curve defined by a polar function. However, it is important to determine the appropriate limits of integration (\(\theta_1\) and \(\theta_2\)) based on the desired portion of the curve (e.g., a petal).
| 4. Is it possible to calculate the arc length of a petal using Cartesian coordinates? | |
Ans. While it is technically possible to calculate the arc length of a petal using Cartesian coordinates, it can be more complicated compared to using polar coordinates. The polar coordinate system is specifically designed to represent curves and shapes that exhibit radial symmetry, making it easier to express and calculate quantities such as arc length. Therefore, it is generally recommended to use polar coordinates when dealing with polar graphs.
| 5. Can the arc length of a petal be negative? | |
Ans. No, the arc length of a petal cannot be negative. Arc length is always a positive quantity that represents the distance along a curve. It cannot have a negative value since it measures the length of a path.
Text Transcript from Video
- What I want to do in this video
is find the arc length of one petal,
I guess we could call it,
of the graph of r is equal
to four sine of two theta.
So I want to find the length
of this portion of the curve
that is in red right over here.
We'll do this in two phases.
First of all I want to set
up the definite integral
for finding that arc length,
and then we'll evaluate it.
I'll actually use a
calculator to evaluate it
because it'll be a little
bit more straightforward
at least for this integral.
Let's just start with that.
I encourage you to pause the video.
Knowing what we know about
the formula for arc length,
when we have it in polar form,
see if you can apply it to
figure out this arc length
right over here.
I'm assuming you've had a go at it,
so let's remind ourselves
that the arc length
is going to be the integral
from our starting angle
to our ending angle, we'll
call it from alpha to beta,
of the square root of the
derivative of our function.
Which respect to theta squared,
plus our function squared.
Plus our function squared d theta.
Let's figure out what r prime of theta
and what r of theta actually are.
I'll color code it.
R prime of theta, we
know what r of theta is
let me just rewrite it.
We know r of theta is equal
to four sine of two theta.
Our prime of theta is just
going to be derivative
of two theta with respect
to theta is two times four
is eight, derivative of sine
is cosine, cosine of two theta.
If we wanted to write the
definite integral for arc length
it's going to be equal
to the definite integral.
What's our starting bound?
For the petal that we care about,
we're going to start at
an angle of zero radians.
When we have an angle of
zero radians, r is zero,
that's this point right over here.
We're going to start at zero radians.
And we're going to go all
the way up to pi over two.
When sine of two times pi
over two is sine of pi,
which is zero, so we'll
get back to this point
right over here.
We're going to go from
zero to pi over two,
to pi over two radians.
It's going to be the square root of,
give myself some real estate here,
r prime of theta squared is
going to be this thing squared.
It's going to be 60--
Let me do it in that same blue color,
it's going to be 64 cosine
squared of two theta,
then plus this squared
which is going to be
16 sine squared of two theta,
then of course we have our d theta.
You can attempt to solve this
or evaluate this analytically,
it's not too straightforward
so I'm going to use a calculator.
It just gives us a
practice with knowing that
hey, there are tools out
there that can help us
do these types of things.
I'll go to the definite integral,
I'll go to second calculus on my t 85,
I picked this choice, which
is for the definite integral.
Let me express when I'm taking
the definite integral of,
it's going to be the square root of 64
times cosine of two theta squared.
I'm going to use x instead of theta
just because it's an easier variable
to use on my calculator.
Cosine of two--
Actually let me make sure
my parenthesis are good.
Cosine of two x, alright I'll close that.
Now I want to square that.
I did that first part.
Plus 16 times
plus 16
times
16 times
sine of two x.
Close that, close that.
Squared.
I'm not sure if the calculator
knows to interpret that
as multiplication, so let me
insert a times right over here.
64 times cosine of two
x, that thing squared.
Plus 16 times sine of two
x, that thing squared.
Let me go to the end.
Let me close my radical,
so that's going to close
that right over there,
that closes what I'm
taking the square root of.
Then comma, the variable
that I'm integrating
with respect to is x, in this case.
Everywhere I saw theta here
I'm just replacing with an x there.
I want to do it from,
instead of theta equals zero
to theta equals pi over two,
we'll say x is equal to zero
to x is equal to pi over two.
Hopefully I have not made a
mistake when I've typed this in,
And I get...
It's chugging on a little bit.
And we get...
Will it actually come up?
Oh finally!
If we were to round it
to the nearest thousandth
it'll be nine point six eight eight.
Approximately nine point six eight eight.
Let's just see, does that
make intuitive sense?
It goes as far as four away,
so if you just went out
to four and then back down
that would be about eight.
Of course we have actually gone out some,
so it actually makes intuitive
sense that this arc length
would be nine point six eight eight.
Hopefully you enjoyed that.
is find the arc length of one petal,
I guess we could call it,
of the graph of r is equal
to four sine of two theta.
So I want to find the length
of this portion of the curve
that is in red right over here.
We'll do this in two phases.
First of all I want to set
up the definite integral
for finding that arc length,
and then we'll evaluate it.
I'll actually use a
calculator to evaluate it
because it'll be a little
bit more straightforward
at least for this integral.
Let's just start with that.
I encourage you to pause the video.
Knowing what we know about
the formula for arc length,
when we have it in polar form,
see if you can apply it to
figure out this arc length
right over here.
I'm assuming you've had a go at it,
so let's remind ourselves
that the arc length
is going to be the integral
from our starting angle
to our ending angle, we'll
call it from alpha to beta,
of the square root of the
derivative of our function.
Which respect to theta squared,
plus our function squared.
Plus our function squared d theta.
Let's figure out what r prime of theta
and what r of theta actually are.
I'll color code it.
R prime of theta, we
know what r of theta is
let me just rewrite it.
We know r of theta is equal
to four sine of two theta.
Our prime of theta is just
going to be derivative
of two theta with respect
to theta is two times four
is eight, derivative of sine
is cosine, cosine of two theta.
If we wanted to write the
definite integral for arc length
it's going to be equal
to the definite integral.
What's our starting bound?
For the petal that we care about,
we're going to start at
an angle of zero radians.
When we have an angle of
zero radians, r is zero,
that's this point right over here.
We're going to start at zero radians.
And we're going to go all
the way up to pi over two.
When sine of two times pi
over two is sine of pi,
which is zero, so we'll
get back to this point
right over here.
We're going to go from
zero to pi over two,
to pi over two radians.
It's going to be the square root of,
give myself some real estate here,
r prime of theta squared is
going to be this thing squared.
It's going to be 60--
Let me do it in that same blue color,
it's going to be 64 cosine
squared of two theta,
then plus this squared
which is going to be
16 sine squared of two theta,
then of course we have our d theta.
You can attempt to solve this
or evaluate this analytically,
it's not too straightforward
so I'm going to use a calculator.
It just gives us a
practice with knowing that
hey, there are tools out
there that can help us
do these types of things.
I'll go to the definite integral,
I'll go to second calculus on my t 85,
I picked this choice, which
is for the definite integral.
Let me express when I'm taking
the definite integral of,
it's going to be the square root of 64
times cosine of two theta squared.
I'm going to use x instead of theta
just because it's an easier variable
to use on my calculator.
Cosine of two--
Actually let me make sure
my parenthesis are good.
Cosine of two x, alright I'll close that.
Now I want to square that.
I did that first part.
Plus 16 times
plus 16
times
16 times
sine of two x.
Close that, close that.
Squared.
I'm not sure if the calculator
knows to interpret that
as multiplication, so let me
insert a times right over here.
64 times cosine of two
x, that thing squared.
Plus 16 times sine of two
x, that thing squared.
Let me go to the end.
Let me close my radical,
so that's going to close
that right over there,
that closes what I'm
taking the square root of.
Then comma, the variable
that I'm integrating
with respect to is x, in this case.
Everywhere I saw theta here
I'm just replacing with an x there.
I want to do it from,
instead of theta equals zero
to theta equals pi over two,
we'll say x is equal to zero
to x is equal to pi over two.
Hopefully I have not made a
mistake when I've typed this in,
And I get...
It's chugging on a little bit.
And we get...
Will it actually come up?
Oh finally!
If we were to round it
to the nearest thousandth
it'll be nine point six eight eight.
Approximately nine point six eight eight.
Let's just see, does that
make intuitive sense?
It goes as far as four away,
so if you just went out
to four and then back down
that would be about eight.
Of course we have actually gone out some,
so it actually makes intuitive
sense that this arc length
would be nine point six eight eight.
Hopefully you enjoyed that.
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About this Video
Jan 16, 2025
Last updated
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