Calc 2 problem

S

sheilaw

New member
Joined
Jun 29, 2009
Messages
16
I don't understand how to solve my homework problem, please help me! Thank you

Find the total arc length of the three-leaved rose
r=3 sin (3theta)
 
sheilaw said:
I don't understand how to solve my homework problem, please help me! Thank you

Find the total arc length of the three-leaved rose
r=3 sin (3theta)
Click to expand...
Plot the function in your graphing calculator.

You'll see there are three symmetric petals of equal size and shape.

So you can find the arc-length of one-half (r = 0 to r = max) petal - then multiply by six(6)

r = 0 at t= 0

At what value of t, you get r = maximum (t[sub:16ru795t]m[/sub:16ru795t])

Then use the arc-length formula and integrate and multiply by 6.

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
 
I tried to graph it, but I found that "three-leaved" are not equal.
Is f(x)=r=3 sin (3theta) ?? Can I just put r=3 sin (3theta) into the Arc Length Formula?
 
sheilaw said:
I tried to graph it, but I found that "three-leaved" are not equal. - they must be - you are seeing distortion due to scaling.

Is f(x)=r=3 sin (3theta) ?? Can I just put r=3 sin (3theta) into the Arc Length Formula?

Yes - but you need to first find limits of integration - read my response above.
Click to expand...
 
To find the limits of integration. set 3sin(3θ)=0\displaystyle 3sin(3{\theta})=0 and solve for theta.

We find that θ=π3\displaystyle {\theta}=\frac{\pi}{3}

the polar arc length formula is L=∫0π3r2+(drdθ)2dθ\displaystyle L=\int_{0}^{\frac{\pi}{3}}\sqrt{r^{2}+\left(\frac{dr}{d\theta}\right)^{2}}d{\theta}

This gives the arc length around one petal. So, multiply by 3.

where drdθ=9cos(3θ)\displaystyle \frac{dr}{d\theta}=9cos(3\theta)

And yes, it may not look like it because of the unconstrained scaling, but the petals are all the same in area and arc length.
 

Attachments

  • rose.jpg
    rose.jpg
    20.3 KB · Views: 395
By the way, you may find the resulting integral quite difficult. I would use tech to solve it, unless you just have to set it up without evaluating.
 
Thanks, galactus, that's what I thought.

Note: When graphing roses in polar. r = acos(n?) or r = asin(n?), n petals if n is odd and 2n petals if n is even (n?2).

Also note for area and arc length: When n is odd area and/or arc length goes from 0 to ? and when n is even area and/or arc length goes from 0 to 2?. If n is odd and you graph from 0 to 2?, you'll get the same graph if you graphed from 0 to ? as the graph repeats itself, but your measurements will be twice what they should be,
 
Yes, these roses are cool. But I always wondered what real-life application they have, if any. Maybe just something cool to learn polar calculus with.
 
Right on, galactus.

After reducing the above integral, I got 12∫06π(5+4cos(u))du.\displaystyle \frac{1}{2}\int_{0}^{6\pi}\sqrt(5+4cos(u))du.

I then whipped out my trusty TI-89 and got 20.0473398313,,, (the total arc length) which makes sense as the "radius" of each petal is three, so three twice times three = 18.

Another note: When dealing with arc length (unless the integral is contrived), one in the past had to usually resort to numerical integration (pure drudgery); fortunately now we have TI's or computer programs to do our grunt work.
 
galactus said:
Yes, these roses are cool. But I always wondered what real-life application they have, if any. Maybe just something cool to learn polar calculus with.
Click to expand...
A gas turbine can be approximated through those petals for first approximation of power calculation.

There are cross-sections of synthetic fibers - that are tri-lobal, in particular nylon fibers for carpets - whose shapes can be approximated (and behaviour modelled) through these.

Generally, these calculations are reduced to elliptic integrals.
 
Yes, I ran this one through Maple and got a solution in terms of an Elliptic integral. Thanks for the info, SK. That's good to know.
 
Can some one help me one more question? thank you

An ellipsoid of revolution is obtained by revolving the ellipse x^2/a^2+y^2/b^2=1, around the x-asis. suppose that a>b. Show that the ellipsoid has surface area
A=2piab[b/a+a/(a^2-b^2)^(-1/2)arcsin((a^2-b^2)^(-1/2)/a)]

I found y(x)=b(1-x^2/a^2), y'=-bx/a^2(1-x^2/a^2)^(-1/2)
put it into the surface area formulas
A=2integral(0toa) 2pi b(1-x^2/a^2){1+[ bx/a^2(1-x^2/a^2)^(-1/2)]^2}^1/2 dx
let x=asin(theta) dx=acos(theta)d(theta)
A=4pibintegral(0topi/2) [1-(sin(theta)^2)/a]^(1/2)*{1+b^2sin(theta)^2/a^2(1-sin(theta)^(-1)]^(1/2)*acos(theta)d(theta)
.
.
A=4piabintegral(0topi/2) cos(theta)/a*[a^2+(b^2-a^2)sin(theta)^2]^(1/2)*d(theta)

I tried to solve it, but i dont know to go on
 
The distance between the foci is 2c. And c=a2−b2\displaystyle c=\sqrt{a^{2}-b^{2}}

Solve the ellipse equation for y:

y=baa2−x2,   dydx=−baxa2−x2,   1+(dydx)2=a4−c2x2a2(a2−x2)\displaystyle y=\frac{b}{a}\sqrt{a^{2}-x^{2}}, \;\ \frac{dy}{dx}=\frac{-b}{a}\frac{x}{\sqrt{a^{2}-x^{2}}}, \;\ 1+\left(\frac{dy}{dx}\right)^{2}=\frac{a^{4}-c^{2}x^{2}}{a^{2}(a^{2}-x^{2})}

S=2(2π)∫0ay1+(dy/dx)2dx=4π∫0abaa2−x2a4−c2x2aa2−x2dx\displaystyle S=2(2\pi)\int_{0}^{a}y\sqrt{1+(dy/dx)^{2}}dx=4{\pi}\int_{0}^{a}\frac{b}{a}\sqrt{a^{2}-x^{2}}\frac{\sqrt{a^{4}-c^{2}x^{2}}}{a\sqrt{a^{2}-x^{2}}}dx

=4πba2∫0aa4−c2x2dx\displaystyle =\frac{4{\pi}b}{a^{2}}\int_{0}^{a}\sqrt{a^{4}-c^{2}x^{2}}dx

Let u=cx:

4πba2c∫0aca4−u2du\displaystyle \frac{4{\pi}b}{a^{2}c}\int_{0}^{ac}\sqrt{a^{4}-u^{2}}du

=4πba2c[u2a4−u2+a42sin−1(ua2)]0ac\displaystyle =\frac{4{\pi}b}{a^{2}c}\left[\frac{u}{2}\sqrt{a^{4}-u^{2}}+\frac{a^{4}}{2}sin^{-1}(\frac{u}{a^{2}})\right]_{0}^{ac}

=4πba2c[12aca4−a2c2+12a4sin−1(ca)]\displaystyle =\frac{4{\pi}b}{a^{2}c}\left[\frac{1}{2}ac\sqrt{a^{4}-a^{2}c^{2}}+\frac{1}{2}a^{4}sin^{-1}(\frac{c}{a})\right]

=2πab[ba+acsin−1(ca)]\displaystyle =\boxed{2{\pi}ab\left[\frac{b}{a}+\frac{a}{c}sin^{-1}(\frac{c}{a})\right]}

It was difficult to read your post, but is this close to what you need?.
 
I don't get that why y=b/a(a^2-x^2)^(1/2)
Sorry for the typing, because i don't know how to type that signs.
 
I get S = 4bπa2∫0a(a4−a2x2+b2x2)dx\displaystyle I \ get \ S \ = \ \frac{4b\pi}{a^{2}}\int_{0}^{a}\sqrt(a^{4}-a^{2}x^{2}+b^{2}x^{2})dx


Hence, S = 2abπ[a(arcsin(a2−b2)a)(a2−b2)+b]\displaystyle Hence, \ S \ = \ 2ab\pi[\frac{a(arcsin\frac{\sqrt(a^{2}-b^{2})}{a})}{\sqrt(a^{2}-b^{2})}+b]
 
You must log in or register to reply here.
Top