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Arc length question

mathsheadache

Member
Oct 10, 2014
52
The length of the minor arc of a circle is 10cm, while the area of the sector AOB is 150cm2.

a) Form two equations involving r and θ, where θ is measured in radians.

b) Solve these equations simultaneously to find r and θ.

Help to solve? Cant understand the question very well.

I think the arc length formula was

length=\(\displaystyle \frac{n}{360}\cdot2\pi(r)\)

\(\displaystyle \therefore10=\frac{n}{360}\cdot2\pi(r)\)

The question states, Form two equations involving r and θ, where θ is measured in radians.

So would we have to arrange the formula to find the r and \(\displaystyle \theta\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The formulas you want are:

Circular arc-length:

\(\displaystyle s=r\theta\tag{1}\)

Area of circular sector

\(\displaystyle A=\frac{1}{2}r^2\theta\tag{2}\)

Now, if we solve (1) for $\theta$, we find:

\(\displaystyle \theta=\frac{s}{r}\)

Now, substituting this into (2), we obtain:

\(\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)=\frac{rs}{2}\implies r=\frac{2A}{s}\implies\theta=\frac{s^2}{2A}\)

Now, just plug in the given values for $A$ and $s$. :D
 

mathsheadache

Member
Oct 10, 2014
52
The formulas you want are:

Circular arc-length:

\(\displaystyle s=r\theta\tag{1}\)

Area of circular sector

\(\displaystyle A=\frac{1}{2}r^2\theta\tag{2}\)

Now, if we solve (1) for $\theta$, we find:

\(\displaystyle \theta=\frac{s}{r}\)

Now, substituting this into (2), we obtain:

\(\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)=\frac{rs}{2}\implies r=\frac{2A}{s}\implies\theta=\frac{s^2}{2A}\)

Now, just plug in the given values for $A$ and $s$. :D
Im finding trouble understanding the question.
"Form two equations involving r and \(\displaystyle \theta\)" Basically means transpose the formula or make the equation equal to r and \(\displaystyle \theta\)?

Since the arc length = \(\displaystyle s=r\theta\) (I thought arc length was length=\(\displaystyle \frac{n}{360}\cdot2\pi(r)\)?

r=\(\displaystyle s\theta\)?

For b) we just sub in the values? Where s is the length?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You are given values for the arc-length $s$ and the area $A$, and so you want to use the formulas relating $r$ and $\theta$ to these given values (which I gave as (1) and (2)) to be able to express both $r$ and $\theta$ as functions of $s$ and $A$ (which I did using some algebra), so that you can use these given values to determine $r$ and $\theta$.

Once you have $r$ and $\theta$ as functions of $s$ and $A$, it is simply a matter of using the given values to evaluate $r$ and $\theta$.
 

SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249
Im finding trouble understanding the question.
"Form two equations involving r and \(\displaystyle \theta\)" Basically means transpose the formula or make the equation equal to r and \(\displaystyle \theta\)?
It's asking to make two equations for the length of an arc and the area of a sector. As long as your equations have \(\displaystyle r\) and \(\displaystyle /theta\) in them then you're fine for this step.

MarkFL has given you the equations you need (his TeX is also better than mine! (Smile))


\(\displaystyle \displaystyle s=r\theta\tag{1}\)

\(\displaystyle \displaystyle A=\frac{1}{2}r^2\theta\tag{2}\)

As you see both of these equations have \(\displaystyle r\) and \(\displaystyle \theta\) in them so Part A of the question has been solved already :)


Since the arc length = \(\displaystyle s=r\theta\) (I thought arc length was length=\(\displaystyle \frac{n}{360}\cdot2\pi(r)\)?

r=\(\displaystyle s\theta\)?
\(\displaystyle \frac{n}{360}\cdot 2\pi r\) is the equation for arc length where \(\displaystyle n\) is in degrees. Since the question asks for radians you want \(\displaystyle s = r\theta\). See the spoiler below for why they are equivalent

By definition a full circle has 360 degrees and $$2\pi$$ radians so you can say $$360^{\circ} = 2\pi$$ (we omit the c symbol for radians)

Your formula for degrees is $$s = \frac{\theta}{360} \cdot 2\pi r$$ but since we know that $$360^{\circ} = 2\pi$$ it can be subbed in

$$s = \frac{\theta}{2\pi} \cdot 2\pi r = r \theta$$ - the radian expression



For b) we just sub in the values? Where s is the length?
Sub in the values given (10cm and 150cm²) for s and A respectively so you have two equations and two unknowns (\(\displaystyle r\) and \(\displaystyle \theta\)).

You can now solve simultaneously using either the substitution method MarkFL showed in post 2. He found \(\displaystyle \theta\) in terms of \(\displaystyle r\) which means he was able to substitute \(\displaystyle \frac{s}{r}\) wherever \(\displaystyle \theta\) appeared.

He did in terms of \(\displaystyle s\) in post 2 but if you find it easier you may use \(\displaystyle s=10\) for your example
 

mathsheadache

Member
Oct 10, 2014
52
It's asking to make two equations for the length of an arc and the area of a sector. As long as your equations have \(\displaystyle r\) and \(\displaystyle /theta\) in them then you're fine for this step.

MarkFL has given you the equations you need (his TeX is also better than mine! (Smile))


\(\displaystyle \displaystyle s=r\theta\tag{1}\)

\(\displaystyle \displaystyle A=\frac{1}{2}r^2\theta\tag{2}\)

As you see both of these equations have \(\displaystyle r\) and \(\displaystyle \theta\) in them so Part A of the question has been solved already :)




\(\displaystyle \frac{n}{360}\cdot 2\pi r\) is the equation for arc length where \(\displaystyle n\) is in degrees. Since the question asks for radians you want \(\displaystyle s = r\theta\). See the spoiler below for why they are equivalent

By definition a full circle has 360 degrees and $$2\pi$$ radians so you can say $$360^{\circ} = 2\pi$$ (we omit the c symbol for radians)

Your formula for degrees is $$s = \frac{\theta}{360} \cdot 2\pi r$$ but since we know that $$360^{\circ} = 2\pi$$ it can be subbed in

$$s = \frac{\theta}{2\pi} \cdot 2\pi r = r \theta$$ - the radian expression





Sub in the values given (10cm and 150cm²) for s and A respectively so you have two equations and two unknowns (\(\displaystyle r\) and \(\displaystyle \theta\)).

You can now solve simultaneously using either the substitution method MarkFL showed in post 2. He found \(\displaystyle \theta\) in terms of \(\displaystyle r\) which means he was able to substitute \(\displaystyle \frac{s}{r}\) wherever \(\displaystyle \theta\) appeared.

He did in terms of \(\displaystyle s\) in post 2 but if you find it easier you may use \(\displaystyle s=10\) for your example
Thanks this really did help :D
 

mathsheadache

Member
Oct 10, 2014
52
The formulas you want are:

Circular arc-length:

\(\displaystyle s=r\theta\tag{1}\)

Area of circular sector

\(\displaystyle A=\frac{1}{2}r^2\theta\tag{2}\)

Now, if we solve (1) for $\theta$, we find:

\(\displaystyle \theta=\frac{s}{r}\)

Now, substituting this into (2), we obtain:

\(\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)=\frac{rs}{2}\implies r=\frac{2A}{s}\implies\theta=\frac{s^2}{2A}\)

Now, just plug in the given values for $A$ and $s$. :D
Beginning to understand this a lot easier.
Just a question on the last part.
How did you rearrange to get \(\displaystyle \theta\)?

It went from, \(\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)\) which I understand to \(\displaystyle =\frac{rs}{2}\)?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Beginning to understand this a lot easier.
Just a question on the last part.
How did you rearrange to get \(\displaystyle \theta\)?
\(\displaystyle s=r\theta\)

Divide through by $r$

\(\displaystyle \frac{s}{r}=\frac{\cancel{r}\theta}{\cancel{r}}\)

\(\displaystyle \theta=\frac{s}{r}\)

It went from, \(\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)\) which I understand to \(\displaystyle =\frac{rs}{2}\)?
\(\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)=\frac{\cancel{r}\cdot r\cdot s}{2\cancel{r}}=\frac{rs}{2}\)